Using batch processing to make interesting math problems - China DOS Union Forum

#Serial Number ！Solved ？Unsolved

#01 ！(Code provided by lxmxn)
Find narcissistic numbers?


@echo off&&setlocal ENABLEDELAYEDEXPANSION
for %%a in (1 2 3 4 5 6 7 8 9) do (
        for %%b in (0 1 2 3 4 5 6 7 8 9) do (
                for %%c in (0 1 2 3 4 5 6 7 8 9) do (
                        set/a result=%%a*%%a*%%a+%%b*%%b*%%b+%%c*%%c*%%c
                        if "!result!"=="%%a%%b%%c" (
                                echo %%a%%b%%c is a narcissistic number!
                        )

                )
        )
)
pause

Similar type of problem: Find Pythagorean triples? (Code provided by namejm)


@echo off
echo.
echo Pythagorean triples within 100 are as follows:
echo.
setlocal enabledelayedexpansion
for /l %%i in (1,1,100) do (
    for /l %%j in (1,1,100) do (
        for /l %%k in (1,1,100) do (
            set /a a=%%i*%%i
            set /a b=%%j*%%j
            set /a c=%%k*%%k
            set /a sum=!a!+!b!
            if !sum! equ !c! echo %%i %%j %%k
        )
    )
)
pause

#02 ！(Code provided by zouzhxi)
There are four numbers. When any three numbers are added together, the resulting sums are 84, 88, 99, and 110 respectively. Find these four numbers?


@ECHO OFF
echo.
setlocal enabledelayedexpansion
SET A=84
SET B=88
SET C=99
SET D=110
SET /A ABCD=A+B+C+D
SET /A SUM=ABCD/3
SET /A NO1=SUM-A
SET /A NO2=SUM-B
SET /A NO3=SUM-C
SET /A NO4=SUM-D
ECHO.
ECHO.NO1^=%NO1%
ECHO.NO2^=%NO2%
ECHO.NO3^=%NO3%
ECHO.NO4^=%NO4%
PAUSE

#03 ！
Miss Zhao's age has the following characteristics:
1. Its cube is a four-digit number, and its fourth power is a six-digit number;
2. These four-digit number and six-digit number are exactly composed of the ten digits from 0 to 9.
Ask, what number should this be?

【Problem-solving idea】
Take an unknown number, first obtain the cube and fourth power of this number respectively; then divide the ten digits respectively and define them, and finally check whether all from 0 to 9 have been defined. If all have been defined, then this number meets the requirements; otherwise, there are repeated digits and it does not meet the requirements.


@echo off
setlocal enabledelayedexpansion
for /l %%i in (10 1 30) do (
::Clear variables
     set flag=
     for /l %%a in (0 1 9) do set %%a=
::Obtain cube and fourth power
     set /a cube=%%i*%%i*%%i
     set /a s=!cube!*%%i
     if !cube! geq 1000 if !cube! lss 10000 (
         if !s! geq 100000 if !s! lss 1000000 (
            set num=!s!!cube!
            call :test !num!
            if not defined flag echo %%i !num!
       )
    )
)

pause>nul
:test 
for /l %%a in (0 1 9) do (
      set var=%1
::Assign the first digit of the parameter to var_; check whether the variable value !var_! has been defined as a variable.
      set var_=!var:~%%a,1!
      if defined !var_! set flag=1 & goto :eof
      set !var_!=A 
)

#04 ！
A total of 4889 digits were used to number the pages of a dictionary. How many pages does this dictionary have? Answer: 1499

【Problem-solving idea】
The following method is relatively direct. Directly check the size of the number. If it is less than 10, it is a 1-digit number, 10~99 is 2 digits,.... Then add up all the digits.


@echo off
set /a t_num=0
setlocal enabledelayedexpansion

echo Checking data.....
for /l %%i in (1 1 4889) do (
    if !t_num! lss 4889 (
         if %%i lss 10 set /a t_num+=1
         if %%i geq 10 if %%i lss 100 set /a t_num+=2
         if %%i geq 100 if %%i lss 1000 set /a t_num+=3
         if %%i geq 1000 set /a t_num+=4
      )  else set /a num=%%i-1 & goto :exit
)
:exit
echo The required number is: %num%
pause>nul

#05 ！
Ah Cong said that he saw a group of camels in the northwest this time. There are 23 humps and 60 feet. How many single-humped and double-humped camels are there respectively?

【Problem-solving idea】
First determine the total number of camels from the number of feet of camels (divided by 4), and then successively check between 1 and 15 for data that meets the requirements;


@echo off
set /a F_num=23
set /a J_num=60/4

setlocal enabledelayedexpansion
for /l %%i in (1 1 %J_num%) do (
   set /a S_num=%%i
   set /a B_num=%J_num%-%%i
   set /a num=!S_num!+2*!B_num!
   call :test !num! !S_num! !B_num!
)
pause>nul

:test
   if %1 EQU %F_num% echo Possible combination: Number of single-humped camels=%2 Number of double-humped camels=%3
   goto :eof

#06 ！
There is a five-digit odd number. Replace all 2s with 5s and all 5s with 2s in this five-digit odd number, and other numbers remain unchanged, to get a new five-digit number. If half of the new five-digit number is still 1 larger than the original five-digit number, what is the original five-digit number?

【Problem-solving idea】
The clear idea is: the first digit must be 2, and the last digit must be 5 (because it was originally odd, but later can have "half", indicating it is even, so it can be determined that the last digit is 5, which was later replaced with 2.


@echo off & echo Checking data...
setlocal enabledelayedexpansion
for /l %%a in (0 1 9) do (
   for /l %%b in (0 1 9) do (
     for /l %%c in (0 1 9) do (
       set/a Fnum=2%%a%%b%%c5
       set Bnum=!Fnum:5=A!
       set Bnum=!Bnum:2=5!
       set/a Bnum=!Bnum:A=2!
       set/a Fnum_=2*!Fnum!+2
       if !Bnum! EQU !Fnum_! echo !Fnum!
       )))
echo Checking completed!
pause>nul

#07 ！
The sum of five consecutive natural numbers can be divisible by 2, 3, 4, 5, and 6 respectively. Find the smallest set of numbers that meets this condition.

【Problem-solving idea】
There is no particular skill. Use the arithmetic sequence formula, first calculate the sum of 5 numbers, and then check whether it can be divisible by 4, 5, 6.


@echo off
echo Checking data....
setlocal enabledelayedexpansion
for /l %%i in (1 1 10000) do (
   set /a flag=0
   set /a num=5*%%i+10
   set /a num1=!num!%%4
   set /a num2=!num!%%5
   set /a num3=!num!%%6
   for %%a in (!num1! !num2! !num3!) do (
      if %%a NEQ 0 set /a flag=1 
    )
   if !flag! EQU 0 set /a num=%%i & goto :exit
)
:exit
for /l %%i in (0 1 4) do (
   set /a num%%i=%num%+%%i
)
echo The required five consecutive natural numbers are: %num0% %num1% %num2% %num3% %num4%
pause>nul

#08 ！
I am a three-digit number. There is one digit that is "3", another digit is "1", and the other digit is unknown. If "3" is changed to "4" and "1" is changed to "3", then the original me will be 9 less than half of the assumed me. Do you know what the original number is?

【Problem-solving idea】
Using the method of proof by contradiction, it can be known that the first digit must be 1, so the arrangement of the number may be: 1*3 or 13*, and then check.


@echo off
rem After deduction, the first digit must be 1, (unless the unknown is 1, and the situation of 1 is also processed in the first for)
setlocal enabledelayedexpansion
for /l %%i in (0 1 9) do (
   set /a Fnum=1%%i3
   set /a Bnum=3%%i4
   set /a num=!Bnum!/2-9
   if !Fnum! EQU !num! echo !Fnum!)

for /l %%i in (0 1 9) do (
   set /a Fnum=13%%i
   set /a Bnum=34%%i
   set /a num=!Bnum!/2-9
   if !Fnum! EQU !num! echo !Fnum!)
pause>nul

#09 ！
Farmer Jones said to his wife: "Hey, Maria, if we sell 75 chicks according to my method, then our chicken feed can last for 20 days. However, if we follow your suggestion and buy 100 more chicks, then the chicken feed will only last for 15 days."
"Ah, dear," she replied, "then how many chicks do we have now?"
The problem is here. How many chicks do they have exactly?

【Problem-solving idea】
Throughout the process, the amount of feed the chicks eat is unchanged. Suppose there are X chicks, then there should be: (x-75)*20=(x+100)*15, and then write a loop to see which number is appropriate?!


@echo off
rem The number of chicks is at least 76:

setlocal enabledelayedexpansion
for /l %%i in (76 1 10000) do (
   set /a Fnum=%%i*20-75*20
   set /a Bnum=%%i*15+100*15
  if !Fnum! EQU !Bnum! echo The number of chicks is: %%i & goto :exit)
:exit
pause>nul

#10 ！
Among all five-digit numbers, how many have exactly two 3s?

【Problem-solving idea】
The following idea is relatively novel. Its principle is: replace all 3s in the number with empty, and the detection method is illustrated by an example:
For example: 75332 becomes 752 after replacement, which is a number with less than 4 digits. Just check its size to meet the requirements of the question.


@echo off
echo Checking data........
setlocal enabledelayedexpansion
set /a flag=0
for /l %%i in (10000,1,99999) do (
   set num=%%i
rem Adding a 1 in front is to prevent special cases like 30820.
   set /a num=1!num:3=!
   if !num! lss 2000 if !num! gtr 200 set /a flag+=1)
echo %flag%
echo Checking completed!
pause>nul

#11 ！
Divide 17 into the sum of several natural numbers, and find the maximum product of these natural numbers?

【Problem-solving idea】
Using high school mathematics knowledge, it can be proved that any number greater than or equal to 4 can be split into two numbers: their sum is equal to the number, and the product is greater than or equal to the number. According to this inference, all numbers will eventually be split into such forms: A*A*A*A....A is 2 or 3 (because only 2 and 3 cannot be split, and the remaining must be only 2 and 3).
For example: 8 can be split like this (without 3):
num1=2*2*2*2 or can be split into (with 2 threes): num2=2*3*3, then just compare the sizes of the two nums!


@echo off
setlocal enabledelayedexpansion
set num=17
set /a Cnum=%num%/3
set /a Rnum=%num%%%2
set /a Tnum=0
   for /l %%i in (%Rnum% 2 %Cnum%) do (
      set num_=1
      set /a num_tem=%num%/2-%%i*3/2
      for /l %%a in (1 1 %%i) do (
          set /a num_*=3) 
      for /l %%b in (1 1 !num_tem!) do (
          set /a num_*=2)
      if !num_! gtr !Tnum! set /a Tnum=!num_!
)
echo !Tnum!
pause>nul

#12 ！
Multiply natural numbers 2, 3...... together. The last 6 digits of their product are exactly all 0s. What is the minimum possible value of the last natural number?


@echo off
echo Checking data........
setlocal enabledelayedexpansion
set /a num=1
set /a flag=0
for /l %%i in (2 1 10000) do (
   call :test %%i
   if !flag! equ 5 set /a num=%%i & goto :exit)

:exit
echo.
echo The minimum natural number is:!num!
echo.
echo Checking completed!
pause>nul

:test 
   set /a num=!num!*%1
   for /l %%i in (1 1 5) do (
      if !num:~-1! EQU 0 (set /a flag+=1 & set /a num=!num:~0,-1!) else (
          set /a num=!num:~-1! & goto :eof))

#13 ！
The sum of the dividend, divisor, and quotient is 181, and the quotient is 12. Find the dividend.


@echo off
echo Checking data........
setlocal enabledelayedexpansion

for /l %%a in (90 1 180) do (
     for /l %%b in (1 1 %%a) do (
        set /a num=%%a %% %%b
        if !num! EQU 0 (
           set /a num=%%a/%%b
           set /a num_=!num!+%%a+%%b
           if !num_! EQU 181 echo Such numbers exist. Dividend: %%a Divisor: %%b Quotient:!num! 
         )
     )
)
echo.
echo Checking completed!
pause>nul

#14 ！
There are six boxes of goods in the store, weighing 15, 16, 18, 19, 20, and 31 kilograms respectively. Two customers bought five of them. It is known that the weight of the goods bought by one customer is twice that of the other customer. Then, what is the weight of the remaining box of goods in the store?

【Problem-solving idea】
A known condition that can be quickly inferred is: the total mass of the two bags of goods bought by the two customers can be divisible by 3, and the other customer's can be divisible by 2.......


@echo off
set /a num=15+16+18+19+20+31
set Tnum=15 16 18 19 20 31 
setlocal enabledelayedexpansion
rem ==========================
rem Question 14:
for %%i in (15 16 18 19 20 31) do call :test %%i
echo The mass of the remaining bag is:!result1!
echo The masses of the two bags with less mass are:!result2! !result3!
pause>nul

:test
   set /a num_tmp=%num%-%1
   set /a Rnum=!num_tmp! %% 3
   if !Rnum! NEQ 0 goto :eof
   set Tnum_=!Tnum:%1 =!
   set /a Snum=!num_tmp!/3
   for %%a in (!Tnum_!) do (
       set /a Rt=!Snum!-%%a
       echo !Tnum_! | find "!Rt!" >nul 2>nul && (
          set result3=%%a
          set /a result2=!Rt!
          set /a result1=%1))

#15 ！
A number has a remainder of 2 when divided by 3 and a remainder of 1 when divided by 5. What is the remainder when this number is divided by 15?


@echo off
setlocal enabledelayedexpansion
echo Within 1~10000, such numbers and their remainders when divided by 15 are respectively:
for /l %%i in (1 1 10000) do (
   set /a num1=%%i %% 3
   set /a num2=%%i %% 5
   if !num1! EQU 2 if !num2! EQU 1 (
       set /a Result=%%i
       set /a num=!Result! %% 15
       echo Number:!Result!       Remainder:!num!
   )
)

pause>nul

#16 ！
1. p is a prime number, and p×p+1 is also a prime number. Find 2006×p.
2. What is the remainder when the product of 2006 2s is divided by 7?

【Problem-solving idea】
1> PxP+1 is a prime number, so PxP+1 is odd (except 2), then P×P is even, so P must be even, then P=2
2> 2^2006 can be written in the form: 2^2006=8^668×2×2=(7+1)^668×4. According to the binomial theorem, the remainder of 2^2006 divided by 7 is: 4


@echo off
setlocal enabledelayedexpansion
for %%i in (2 3 5 6 7 9 10 15 17 31 33 63 65) do (
   set /a tmp=1
   set /a Res=1
   call :test %%i 
   echo The remainder of 2^^^^2006 divided by %%i is:!Res!)
pause>nul

::Process even numbers
:test
  if %1 GTR 2 (
    set /a var=%1 %% 2
    if !var! EQU 0  (
        set /a num=%1/2
        call :test !num!) else call :test1 %1) else set Res=0
    goto :eof

:test1 
    for /l %%i in (1 1 10) do (
       set /a tmp*=2
       set /a Rnum=%1 %% 4
       if !tmp! GEQ %1 (
          if !Rnum! EQU 3 ( 
             set /a Inum=%%i 
             set /a Inum_=2006 %% !Inum!
             for /l %%a in (0 1 !Inum_!) do set /a Res*=2
             set /a Res=!Res!/2
             goto :eof)
          set /a Inum=%%i-1
          set /a Inum_=2006 %% !Inum!
          for /l %%a in (0 1 !Inum_!) do set /a Res*=2
          set /a Res=%1-!Res!/2   
          goto :eof
          ))

#17 ！
It is said that in a holy temple in India, there is a brass plate with three gem needles inserted on it. On the first gem needle, from bottom to top, there are 64 golden plates with holes in the center, arranged from largest to smallest. The monks in the temple move the plates according to the following rules: only one plate can be moved at a time, and the small plate must always be placed on the large plate. At that time, it was said that when all 64 golden plates were moved to another gem needle, the world would be destroyed in a thunderclap. How many times does it take to move 64 golden plates to another gem needle? This is a very large number!
Answer: 18446744073709551615

【Problem-solving idea】According to relevant knowledge such as geometric sequences, it is known that the total number of times required is: 2^64-1. This number is extremely large, and direct calculation is definitely not feasible. Then we have to adopt the method of "segmented calculation in a clock-like manner". If it is not original, please forgive my rudeness!

The following code divides the data into 6 segments, each with 6 digits to save respectively. Its basic principle is: when the last segment of data is multiplied by 2, if its value is greater than 1000000, carry over, add the carried data to the previous segment of data, and so on. This segment of code currently supports the continuous multiplication of 200 2s. Of course, you can "modify" it to calculate more massive numbers.


@echo off
setlocal enabledelayedexpansion

::Initialize each segment of data;
for /l %%i in (1 1 5) do set /a num%%i=0
set /a num6=1

for /l %%i in (1 1 64) do (
   rem ================================================
   rem Initialize the carry data and the calculation formula of the segmented data (that is, multiply by 2)

   for /l %%i in (1 1 5) do (
     set /a num%%i_=0
     set /a num%%i*=2)
   set /a num6*=2
   rem ================================================

  for /l %%a in (6 -1 1) do ( 
     rem Each segment of data saves a 6-digit number;  
     if !num%%a! GTR 1000000 (
        set /a Inum=%%a-1
        set /a tmp=!num%%a!
        rem ========================================================================
        rem The 1 in front is "borrowed" from the previous segment of data; adding a 1 in front is to prevent the occurrence of set /a test=0003
        rem Similar situations!

        set /a num%%a=1!tmp:~-6!
        rem ========================================================================
        
        rem The previous segment of data plus the carry of the current segment of data minus 1, because a 1 was "borrowed"!
        set /a num!Inum!+=!tmp:~0,-6!-1
        )
     )
)
   set /a num6-=1
   rem ===================================================================
   rem Tailoring work, because the previous step used the method of "borrowing" data for debugging, and now restore it!

   for /l %%a in (6 -1 4) do (
        set /a Inum=%%a-1
        set /a tmp=!num%%a!
        set num%%a=!tmp:~1!
        set /a num!Inum!+=!tmp:~0,1!)
    rem ===================================================================
     for %%i in (%num1% %num2% %num3% %num4% %num5% %num6%) do (
        if %%i neq 0 set Result=!Result!%%i)
    echo !Result!
   
pause>nul

#18 ！
There are ten RMB notes with denominations of 1 cent, 2 cents, 5 cents, 1 jiao, 2 jiao, 5 jiao, 1 yuan, 2 yuan, 5 yuan, and 10 yuan. How many different denominations can be formed?

【Problem-solving idea】According to the relevant knowledge of permutations and combinations, it is known that the total number of combinations may be: num=C(10,1)+C(10,2)+..+C(10,10)


@echo off
setlocal enabledelayedexpansion

set /a num=1
set /a Result=0

for /l %%i in (1 1 10) do (
    call :test %%i 10
    set /a Result+=!num! & set /a num=1)
echo The number of denominations is:!Result!
pause>nul

::Find C(n,r);
:test start_num end_num
    set /a tmp=%2-%1+1
    for /l %%i in (%tmp% 1 %2) do set /a num*=%%i
    for /l %%i in (1 1 %1) do set /a num/=%%i

Floor46　youxi01	Posted 2006-11-17 02:32
高级用户　Posts 247　Credits 846　From 湖南==》广东
#Serial Number ！Solved ？Unsolved #01 ！(Code provided by lxmxn) Find narcissistic numbers? @echo off&&setlocal ENABLEDELAYEDEXPANSION for %%a in (1 2 3 4 5 6 7 8 9) do ( for %%b in (0 1 2 3 4 5 6 7 8 9) do ( for %%c in (0 1 2 3 4 5 6 7 8 9) do ( set/a result=%%a%%a%%a+%%b%%b%%b+%%c%%c%%c if "!result!"=="%%a%%b%%c" ( echo %%a%%b%%c is a narcissistic number! ) ) ) ) pause Similar type of problem: Find Pythagorean triples? (Code provided by namejm) @echo off echo. echo Pythagorean triples within 100 are as follows: echo. setlocal enabledelayedexpansion for /l %%i in (1,1,100) do ( for /l %%j in (1,1,100) do ( for /l %%k in (1,1,100) do ( set /a a=%%i%%i set /a b=%%j%%j set /a c=%%k%%k set /a sum=!a!+!b! if !sum! equ !c! echo %%i %%j %%k ) ) ) pause #02 ！(Code provided by zouzhxi) There are four numbers. When any three numbers are added together, the resulting sums are 84, 88, 99, and 110 respectively. Find these four numbers? @ECHO OFF echo. setlocal enabledelayedexpansion SET A=84 SET B=88 SET C=99 SET D=110 SET /A ABCD=A+B+C+D SET /A SUM=ABCD/3 SET /A NO1=SUM-A SET /A NO2=SUM-B SET /A NO3=SUM-C SET /A NO4=SUM-D ECHO. ECHO.NO1^=%NO1% ECHO.NO2^=%NO2% ECHO.NO3^=%NO3% ECHO.NO4^=%NO4% PAUSE #03 ！ Miss Zhao's age has the following characteristics: 1. Its cube is a four-digit number, and its fourth power is a six-digit number; 2. These four-digit number and six-digit number are exactly composed of the ten digits from 0 to 9. Ask, what number should this be? 【Problem-solving idea】 Take an unknown number, first obtain the cube and fourth power of this number respectively; then divide the ten digits respectively and define them, and finally check whether all from 0 to 9 have been defined. If all have been defined, then this number meets the requirements; otherwise, there are repeated digits and it does not meet the requirements. @echo off setlocal enabledelayedexpansion for /l %%i in (10 1 30) do ( ::Clear variables set flag= for /l %%a in (0 1 9) do set %%a= ::Obtain cube and fourth power set /a cube=%%i%%i%%i set /a s=!cube!%%i if !cube! geq 1000 if !cube! lss 10000 ( if !s! geq 100000 if !s! lss 1000000 ( set num=!s!!cube! call :test !num! if not defined flag echo %%i !num! ) ) ) pause>nul :test for /l %%a in (0 1 9) do ( set var=%1 ::Assign the first digit of the parameter to var_; check whether the variable value !var_! has been defined as a variable. set var_=!var:~%%a,1! if defined !var_! set flag=1 & goto :eof set !var_!=A ) #04 ！ A total of 4889 digits were used to number the pages of a dictionary. How many pages does this dictionary have? Answer: 1499 【Problem-solving idea】 The following method is relatively direct. Directly check the size of the number. If it is less than 10, it is a 1-digit number, 10~99 is 2 digits,.... Then add up all the digits. @echo off set /a t_num=0 setlocal enabledelayedexpansion echo Checking data..... for /l %%i in (1 1 4889) do ( if !t_num! lss 4889 ( if %%i lss 10 set /a t_num+=1 if %%i geq 10 if %%i lss 100 set /a t_num+=2 if %%i geq 100 if %%i lss 1000 set /a t_num+=3 if %%i geq 1000 set /a t_num+=4 ) else set /a num=%%i-1 & goto :exit ) :exit echo The required number is: %num% pause>nul #05 ！ Ah Cong said that he saw a group of camels in the northwest this time. There are 23 humps and 60 feet. How many single-humped and double-humped camels are there respectively? 【Problem-solving idea】 First determine the total number of camels from the number of feet of camels (divided by 4), and then successively check between 1 and 15 for data that meets the requirements; @echo off set /a F_num=23 set /a J_num=60/4 setlocal enabledelayedexpansion for /l %%i in (1 1 %J_num%) do ( set /a S_num=%%i set /a B_num=%J_num%-%%i set /a num=!S_num!+2!B_num! call :test !num! !S_num! !B_num! ) pause>nul :test if %1 EQU %F_num% echo Possible combination: Number of single-humped camels=%2 Number of double-humped camels=%3 goto :eof #06 ！ There is a five-digit odd number. Replace all 2s with 5s and all 5s with 2s in this five-digit odd number, and other numbers remain unchanged, to get a new five-digit number. If half of the new five-digit number is still 1 larger than the original five-digit number, what is the original five-digit number? 【Problem-solving idea】 The clear idea is: the first digit must be 2, and the last digit must be 5 (because it was originally odd, but later can have "half", indicating it is even, so it can be determined that the last digit is 5, which was later replaced with 2. @echo off & echo Checking data... setlocal enabledelayedexpansion for /l %%a in (0 1 9) do ( for /l %%b in (0 1 9) do ( for /l %%c in (0 1 9) do ( set/a Fnum=2%%a%%b%%c5 set Bnum=!Fnum:5=A! set Bnum=!Bnum:2=5! set/a Bnum=!Bnum:A=2! set/a Fnum_=2!Fnum!+2 if !Bnum! EQU !Fnum_! echo !Fnum! ))) echo Checking completed! pause>nul #07 ！ The sum of five consecutive natural numbers can be divisible by 2, 3, 4, 5, and 6 respectively. Find the smallest set of numbers that meets this condition. 【Problem-solving idea】 There is no particular skill. Use the arithmetic sequence formula, first calculate the sum of 5 numbers, and then check whether it can be divisible by 4, 5, 6. @echo off echo Checking data.... setlocal enabledelayedexpansion for /l %%i in (1 1 10000) do ( set /a flag=0 set /a num=5%%i+10 set /a num1=!num!%%4 set /a num2=!num!%%5 set /a num3=!num!%%6 for %%a in (!num1! !num2! !num3!) do ( if %%a NEQ 0 set /a flag=1 ) if !flag! EQU 0 set /a num=%%i & goto :exit ) :exit for /l %%i in (0 1 4) do ( set /a num%%i=%num%+%%i ) echo The required five consecutive natural numbers are: %num0% %num1% %num2% %num3% %num4% pause>nul #08 ！ I am a three-digit number. There is one digit that is "3", another digit is "1", and the other digit is unknown. If "3" is changed to "4" and "1" is changed to "3", then the original me will be 9 less than half of the assumed me. Do you know what the original number is? 【Problem-solving idea】 Using the method of proof by contradiction, it can be known that the first digit must be 1, so the arrangement of the number may be: 13 or 13, and then check. @echo off rem After deduction, the first digit must be 1, (unless the unknown is 1, and the situation of 1 is also processed in the first for) setlocal enabledelayedexpansion for /l %%i in (0 1 9) do ( set /a Fnum=1%%i3 set /a Bnum=3%%i4 set /a num=!Bnum!/2-9 if !Fnum! EQU !num! echo !Fnum!) for /l %%i in (0 1 9) do ( set /a Fnum=13%%i set /a Bnum=34%%i set /a num=!Bnum!/2-9 if !Fnum! EQU !num! echo !Fnum!) pause>nul #09 ！ Farmer Jones said to his wife: "Hey, Maria, if we sell 75 chicks according to my method, then our chicken feed can last for 20 days. However, if we follow your suggestion and buy 100 more chicks, then the chicken feed will only last for 15 days." "Ah, dear," she replied, "then how many chicks do we have now?" The problem is here. How many chicks do they have exactly? 【Problem-solving idea】 Throughout the process, the amount of feed the chicks eat is unchanged. Suppose there are X chicks, then there should be: (x-75)20=(x+100)15, and then write a loop to see which number is appropriate?! @echo off rem The number of chicks is at least 76: setlocal enabledelayedexpansion for /l %%i in (76 1 10000) do ( set /a Fnum=%%i20-7520 set /a Bnum=%%i15+10015 if !Fnum! EQU !Bnum! echo The number of chicks is: %%i & goto :exit) :exit pause>nul #10 ！ Among all five-digit numbers, how many have exactly two 3s? 【Problem-solving idea】 The following idea is relatively novel. Its principle is: replace all 3s in the number with empty, and the detection method is illustrated by an example: For example: 75332 becomes 752 after replacement, which is a number with less than 4 digits. Just check its size to meet the requirements of the question. @echo off echo Checking data........ setlocal enabledelayedexpansion set /a flag=0 for /l %%i in (10000,1,99999) do ( set num=%%i rem Adding a 1 in front is to prevent special cases like 30820. set /a num=1!num:3=! if !num! lss 2000 if !num! gtr 200 set /a flag+=1) echo %flag% echo Checking completed! pause>nul #11 ！ Divide 17 into the sum of several natural numbers, and find the maximum product of these natural numbers? 【Problem-solving idea】 Using high school mathematics knowledge, it can be proved that any number greater than or equal to 4 can be split into two numbers: their sum is equal to the number, and the product is greater than or equal to the number. According to this inference, all numbers will eventually be split into such forms: AAAA....A is 2 or 3 (because only 2 and 3 cannot be split, and the remaining must be only 2 and 3). For example: 8 can be split like this (without 3): num1=2222 or can be split into (with 2 threes): num2=233, then just compare the sizes of the two nums! @echo off setlocal enabledelayedexpansion set num=17 set /a Cnum=%num%/3 set /a Rnum=%num%%%2 set /a Tnum=0 for /l %%i in (%Rnum% 2 %Cnum%) do ( set num_=1 set /a num_tem=%num%/2-%%i3/2 for /l %%a in (1 1 %%i) do ( set /a num_=3) for /l %%b in (1 1 !num_tem!) do ( set /a num_=2) if !num_! gtr !Tnum! set /a Tnum=!num_! ) echo !Tnum! pause>nul #12 ！ Multiply natural numbers 2, 3...... together. The last 6 digits of their product are exactly all 0s. What is the minimum possible value of the last natural number? @echo off echo Checking data........ setlocal enabledelayedexpansion set /a num=1 set /a flag=0 for /l %%i in (2 1 10000) do ( call :test %%i if !flag! equ 5 set /a num=%%i & goto :exit) :exit echo. echo The minimum natural number is:!num! echo. echo Checking completed! pause>nul :test set /a num=!num!%1 for /l %%i in (1 1 5) do ( if !num:~-1! EQU 0 (set /a flag+=1 & set /a num=!num:~0,-1!) else ( set /a num=!num:~-1! & goto :eof)) #13 ！ The sum of the dividend, divisor, and quotient is 181, and the quotient is 12. Find the dividend. @echo off echo Checking data........ setlocal enabledelayedexpansion for /l %%a in (90 1 180) do ( for /l %%b in (1 1 %%a) do ( set /a num=%%a %% %%b if !num! EQU 0 ( set /a num=%%a/%%b set /a num_=!num!+%%a+%%b if !num_! EQU 181 echo Such numbers exist. Dividend: %%a Divisor: %%b Quotient:!num! ) ) ) echo. echo Checking completed! pause>nul #14 ！ There are six boxes of goods in the store, weighing 15, 16, 18, 19, 20, and 31 kilograms respectively. Two customers bought five of them. It is known that the weight of the goods bought by one customer is twice that of the other customer. Then, what is the weight of the remaining box of goods in the store? 【Problem-solving idea】 A known condition that can be quickly inferred is: the total mass of the two bags of goods bought by the two customers can be divisible by 3, and the other customer's can be divisible by 2....... @echo off set /a num=15+16+18+19+20+31 set Tnum=15 16 18 19 20 31 setlocal enabledelayedexpansion rem ========================== rem Question 14: for %%i in (15 16 18 19 20 31) do call :test %%i echo The mass of the remaining bag is:!result1! echo The masses of the two bags with less mass are:!result2! !result3! pause>nul :test set /a num_tmp=%num%-%1 set /a Rnum=!num_tmp! %% 3 if !Rnum! NEQ 0 goto :eof set Tnum_=!Tnum:%1 =! set /a Snum=!num_tmp!/3 for %%a in (!Tnum_!) do ( set /a Rt=!Snum!-%%a echo !Tnum_! \| find "!Rt!" >nul 2>nul && ( set result3=%%a set /a result2=!Rt! set /a result1=%1)) #15 ！ A number has a remainder of 2 when divided by 3 and a remainder of 1 when divided by 5. What is the remainder when this number is divided by 15? @echo off setlocal enabledelayedexpansion echo Within 1~10000, such numbers and their remainders when divided by 15 are respectively: for /l %%i in (1 1 10000) do ( set /a num1=%%i %% 3 set /a num2=%%i %% 5 if !num1! EQU 2 if !num2! EQU 1 ( set /a Result=%%i set /a num=!Result! %% 15 echo Number:!Result! Remainder:!num! ) ) pause>nul #16 ！ 1. p is a prime number, and p×p+1 is also a prime number. Find 2006×p. 2. What is the remainder when the product of 2006 2s is divided by 7? 【Problem-solving idea】 1> PxP+1 is a prime number, so PxP+1 is odd (except 2), then P×P is even, so P must be even, then P=2 2> 2^2006 can be written in the form: 2^2006=8^668×2×2=(7+1)^668×4. According to the binomial theorem, the remainder of 2^2006 divided by 7 is: 4 @echo off setlocal enabledelayedexpansion for %%i in (2 3 5 6 7 9 10 15 17 31 33 63 65) do ( set /a tmp=1 set /a Res=1 call :test %%i echo The remainder of 2^^^^2006 divided by %%i is:!Res!) pause>nul ::Process even numbers :test if %1 GTR 2 ( set /a var=%1 %% 2 if !var! EQU 0 ( set /a num=%1/2 call :test !num!) else call :test1 %1) else set Res=0 goto :eof :test1 for /l %%i in (1 1 10) do ( set /a tmp=2 set /a Rnum=%1 %% 4 if !tmp! GEQ %1 ( if !Rnum! EQU 3 ( set /a Inum=%%i set /a Inum_=2006 %% !Inum! for /l %%a in (0 1 !Inum_!) do set /a Res=2 set /a Res=!Res!/2 goto :eof) set /a Inum=%%i-1 set /a Inum_=2006 %% !Inum! for /l %%a in (0 1 !Inum_!) do set /a Res=2 set /a Res=%1-!Res!/2 goto :eof )) #17 ！ It is said that in a holy temple in India, there is a brass plate with three gem needles inserted on it. On the first gem needle, from bottom to top, there are 64 golden plates with holes in the center, arranged from largest to smallest. The monks in the temple move the plates according to the following rules: only one plate can be moved at a time, and the small plate must always be placed on the large plate. At that time, it was said that when all 64 golden plates were moved to another gem needle, the world would be destroyed in a thunderclap. How many times does it take to move 64 golden plates to another gem needle? This is a very large number! Answer: 18446744073709551615 【Problem-solving idea】According to relevant knowledge such as geometric sequences, it is known that the total number of times required is: 2^64-1. This number is extremely large, and direct calculation is definitely not feasible. Then we have to adopt the method of "segmented calculation in a clock-like manner". If it is not original, please forgive my rudeness! The following code divides the data into 6 segments, each with 6 digits to save respectively. Its basic principle is: when the last segment of data is multiplied by 2, if its value is greater than 1000000, carry over, add the carried data to the previous segment of data, and so on. This segment of code currently supports the continuous multiplication of 200 2s. Of course, you can "modify" it to calculate more massive numbers. @echo off setlocal enabledelayedexpansion ::Initialize each segment of data; for /l %%i in (1 1 5) do set /a num%%i=0 set /a num6=1 for /l %%i in (1 1 64) do ( rem ================================================ rem Initialize the carry data and the calculation formula of the segmented data (that is, multiply by 2) for /l %%i in (1 1 5) do ( set /a num%%i_=0 set /a num%%i=2) set /a num6=2 rem ================================================ for /l %%a in (6 -1 1) do ( rem Each segment of data saves a 6-digit number; if !num%%a! GTR 1000000 ( set /a Inum=%%a-1 set /a tmp=!num%%a! rem ======================================================================== rem The 1 in front is "borrowed" from the previous segment of data; adding a 1 in front is to prevent the occurrence of set /a test=0003 rem Similar situations! set /a num%%a=1!tmp:~-6! rem ======================================================================== rem The previous segment of data plus the carry of the current segment of data minus 1, because a 1 was "borrowed"! set /a num!Inum!+=!tmp:~0,-6!-1 ) ) ) set /a num6-=1 rem =================================================================== rem Tailoring work, because the previous step used the method of "borrowing" data for debugging, and now restore it! for /l %%a in (6 -1 4) do ( set /a Inum=%%a-1 set /a tmp=!num%%a! set num%%a=!tmp:~1! set /a num!Inum!+=!tmp:~0,1!) rem =================================================================== for %%i in (%num1% %num2% %num3% %num4% %num5% %num6%) do ( if %%i neq 0 set Result=!Result!%%i) echo !Result! pause>nul #18 ！ There are ten RMB notes with denominations of 1 cent, 2 cents, 5 cents, 1 jiao, 2 jiao, 5 jiao, 1 yuan, 2 yuan, 5 yuan, and 10 yuan. How many different denominations can be formed? 【Problem-solving idea】According to the relevant knowledge of permutations and combinations, it is known that the total number of combinations may be: num=C(10,1)+C(10,2)+..+C(10,10) @echo off setlocal enabledelayedexpansion set /a num=1 set /a Result=0 for /l %%i in (1 1 10) do ( call :test %%i 10 set /a Result+=!num! & set /a num=1) echo The number of denominations is:!Result! pause>nul ::Find C(n,r); :test start_num end_num set /a tmp=%2-%1+1 for /l %%i in (%tmp% 1 %2) do set /a num*=%%i for /l %%i in (1 1 %1) do set /a num/=%%i

Floor47　namejm	Posted 2006-11-17 02:45
荣誉版主　Posts 1,737　Credits 5,226　From 成都
Since there are already codes in the previous post, there should be no need to start a new post to summarize, because this will be suspected of content duplication and will also occupy too much forum resources. Moreover, new questions may be raised at any time in the top post of that thread. It can be considered to merge with the post opened by zouzhxi. zouzhxi is responsible for collecting all the codes and placing them as an attachment package in the top post of the thread for everyone to download. At the same time, the links to the codes that solve each problem should also be posted in the top post of the thread. What do you think, Brother youxi01?

Floor48　youxi01	Posted 2006-11-17 02:53
高级用户　Posts 247　Credits 846　From 湖南==》广东
No problem, it's all for the convenience of everyone!~

Floor49　NaturalJ0	Posted 2006-11-17 03:01
银牌会员　Posts 533　Credits 1,181
I missed out = =b On the first day I came in and took a look, and found only narcissistic numbers and the Pythagorean theorem, so I didn't come back in. Today I thought it was strange: how could narcissistic numbers and the Pythagorean theorem be discussed for three pages? So I came in again to take a look, and found there were actually so many fun problems. Looking at these problems reminded me of one I did when I was little. There are two basins of water, one cold and one hot. There is a thermometer in the cold-water basin. Use a small cup to get a cup of hot water and pour it into the cold water, and the temperature rises by 5 degrees. Pour in another cup of hot water, and it rises by another 3 degrees. If you pour in one more cup, how many more degrees will it rise?

Floor50　zouzhxi	Posted 2006-11-17 03:59
中级用户　Posts 177　Credits 430　From 广东深圳
Re 46F: I put your topic on the top floor...

Floor51　zouzhxi	Posted 2006-11-17 04:12
中级用户　Posts 177　Credits 430　From 广东深圳
Okay, I'll collect the code issues and put them up later...

Floor52　youxi01	Posted 2006-11-17 06:12
高级用户　Posts 247　Credits 846　From 湖南==》广东
Moderator namejm, should you delete this question?

Floor53　不得不爱	Posted 2006-11-17 09:32
超级版主　Posts 2,044　Credits 5,310　From 四川南充
Should this topic be deleted or merged? If merging is needed, I can handle it!

Floor54　youxi01	Posted 2006-11-17 10:30
高级用户　Posts 247　Credits 846　From 湖南==》广东
Merge them, thanks~

Floor55　namejm	Posted 2006-11-17 11:32
荣誉版主　Posts 1,737　Credits 5,226　From 成都
Actually, Brother 3742668's essence-addition operation should be based on certain rules. When willsort was in office before, relevant rules had already been formulated. Please see here. It's actually me who is online almost every day but forgot to add essence, which is really dereliction of duty, sweat~. To zouzhxi: It feels that the post at the top of the building by Brother has some places that are not very convenient for retrieval in terms of typesetting, and posting my plan seems very wordy, and it may not be able to be accurately conveyed. So I want to directly modify the post at the top of your building to facilitate future retrieval. I want to ask for your opinion.

Floor56　zouzhxi	Posted 2006-11-17 21:10
中级用户　Posts 177　Credits 430　From 广东深圳
Okay, as long as everyone finds it convenient, that's fine... Don't change it too drastically either. Hehe.

Floor57　NaturalJ0	Posted 2006-11-17 22:22
银牌会员　Posts 533　Credits 1,181
It's a very interesting set of questions. I carefully looked through all the questions today. ^_^ PS: Question 20 seems to be to find the greatest common divisor of three numbers

Floor58　不得不爱	Posted 2006-11-17 23:55
超级版主　Posts 2,044　Credits 5,310　From 四川南充
——————————————————Moderation Records—————————————————— Executor: Must Love Operation: Merge topics into posts 46, 47, 48, 51, 52, 53 Explanation: The original topic "{ Use Batch Scripts to Make Fun Math Problems (Summary)}" and the topic "{ Use Batch Scripts to Make Fun Math Problems}" have a contextual relationship ————————————————————————————————————————

Floor59　NaturalJ0	Posted 2006-11-18 05:16
银牌会员　Posts 533　Credits 1,181
It seems that the essence post has been merged?

Floor60　namejm	Posted 2006-11-18 05:27
荣誉版主　Posts 1,737　Credits 5,226　From 成都
Probably a certain moderator made a misoperation and removed the essence, and now it is re-added as essence. However, there is a slight regret that the original poster of this post was zouzhxi, but when returning to the post list interface, it has become youxi01. Hehe, hope the two don't mind.