I am thinking of a four digit positive integer with distinct digits.
Of course, there's a total of 4 ! − 1 = 2 3 ways to rearrange the digits to form a new 4 digit positive integer.
If the sum of these other 23 numbers is 157193, what is the number that I was thinking of?
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nice solution
Outstanding work, Garrett.
Wew... Best solution for this problem!
Where did 3875 and 2791 come from?
Very clever. I couldn't see any way to solve this so I wrote a Python program and found it. More than one way to skin a cat and more than one skill available!
Since all 24 permutations of a b c d would have each digit appear 6 times in each position, they add up to 6666(a+b+c+d). So a b c d bust be a multiple of 6666 minus 157193. The only candidates for a b c d are :
2 4 × 6 6 6 6 − 1 5 7 1 9 3 = 2 7 9 1 but 2 + 7 + 9 + 1 = 2 4
2 5 × 6 6 6 6 − 1 5 7 1 9 3 = 9 4 5 7 and 9 + 4 + 5 + 7 = 2 5
10^3a + 10^2b + 10c + d
including the original number = 6666 * (a + b + c + d)
157193 + 10^3a + 10^2b + 10c + d = 6666 * (a + b + c + d)
Easiest way to proceed is by trial and error:
Locate the next multiple of 6666 that is greater than 157193.
Check to see if the number 159984 satisfies the equation, which it does not.
6666 * 25 = 166650
166650 - 157193= 9457
9 + 4 + 5 + 7 = 25
Therefore, the answer is the number 9457 .
wolframalpha input solve mod(157193+n,6666)=0 result= 9457
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First of all, let's set our number n = 1 0 0 0 a + 1 0 0 b + 1 0 c + d , where a , b , c , and d are distinct integers in the interval [ 1 , 9 ] . Note that none of the integers can be 0 because we can't make the first digit 0 as the number would no longer have 4 digits, which would result in less than 2 3 possible permutations.
Let's assume that our number n is the 2 4 t h permutation. In counting these permutations, we notice that each digit must be in each spot an equal number of times. There are 4 variables ( a , b , c , d ) and 2 4 possible permutations, meaning that each variable must be in each spot 4 2 4 = 6 times.
If the sum of the 2 3 permutations is 1 5 7 1 9 3 , then 1 5 7 1 9 3 + n should be all 2 4 . Taking into account that every variable should be in each place 6 times:
1 5 7 1 9 3 + n = 6 6 6 6 ( a + b + c + d ) ⟹ 1 5 7 1 9 3 + n ≡ 0 (mod 6 6 6 6 )
1 5 7 1 9 3 + n ≡ 3 8 7 5 + n ≡ 0 (mod 6 6 6 6 ) ⟹ n ≡ 2 7 9 1 (mod 6 6 6 6 )
This means that n = 2 7 9 1 + 6 6 6 6 m . There are two integer solutions that result in 4 digit numbers: 2 7 9 1 and 9 4 5 7 . Let's plug them into our original equation to check which one works:
1 5 7 1 9 3 + 2 7 9 1 = 1 5 9 9 8 4 = 1 2 6 6 5 4 = 6 6 6 6 ( 2 + 7 + 9 + 1 )
1 5 7 1 9 3 + 9 4 5 7 = 1 6 6 6 5 0 = 6 6 6 6 ( 9 + 4 + 5 + 7 )
Therefore our answer must be 9 4 5 7 .