Four Digit Number - Can You Find It?

I am thinking of a four digit positive integer with distinct digits.

Of course, there's a total of 4 ! 1 = 23 4!-1=23 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?


The answer is 9457.

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3 solutions

Garrett Clarke
Jun 29, 2015

First of all, let's set our number n = 1000 a + 100 b + 10 c + d n=1000a+100b+10c+d , where a a , b b , c c , and d d are distinct integers in the interval [ 1 , 9 ] [1,9] . Note that none of the integers can be 0 0 because we can't make the first digit 0 0 as the number would no longer have 4 4 digits, which would result in less than 23 23 possible permutations.

Let's assume that our number n n is the 24 t h 24th permutation. In counting these permutations, we notice that each digit must be in each spot an equal number of times. There are 4 4 variables ( a , b , c , d ) (a,b,c,d) and 24 24 possible permutations, meaning that each variable must be in each spot 24 4 = 6 \frac{24}{4}=6 times.

If the sum of the 23 23 permutations is 157193 157193 , then 157193 + n 157193+n should be all 24 24 . Taking into account that every variable should be in each place 6 6 times:

157193 + n = 6666 ( a + b + c + d ) 157193 + n 0 157193+n=6666(a+b+c+d) \Longrightarrow 157193+n \equiv 0 (mod 6666 6666 )

157193 + n 3875 + n 0 157193+n \equiv 3875+n \equiv 0 (mod 6666 6666 ) n 2791 \Longrightarrow n \equiv 2791 (mod 6666 6666 )

This means that n = 2791 + 6666 m n=2791+6666m . There are two integer solutions that result in 4 4 digit numbers: 2791 2791 and 9457 9457 . Let's plug them into our original equation to check which one works:

157193 + 2791 = 159984 126654 = 6666 ( 2 + 7 + 9 + 1 ) 157193+2791=159984\neq126654=6666(2+7+9+1)

157193 + 9457 = 166650 = 6666 ( 9 + 4 + 5 + 7 ) 157193+9457=166650=6666(9+4+5+7)

Therefore our answer must be 9457 \boxed{9457} .

nice solution

Vighnesh Raut - 5 years, 11 months ago

Outstanding work, Garrett.

Lew Sterling Jr - 5 years, 11 months ago

Wew... Best solution for this problem!

hanif adzkiya - 5 years, 3 months ago

Where did 3875 and 2791 come from?

Greg Grapsas - 5 years ago

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Calculator or Euclidean division, I suppose.

Daan Smets - 2 years, 6 months ago

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!

Paul Koegler - 2 years, 2 months ago
K T
Feb 26, 2019

Since all 24 permutations of a b c d \overline{abcd} would have each digit appear 6 times in each position, they add up to 6666(a+b+c+d). So a b c d \overline{abcd} bust be a multiple of 6666 minus 157193. The only candidates for a b c d \overline{abcd} are :

24 × 6666 157193 = 2791 24×6666-157193=2791 but 2 + 7 + 9 + 1 24 2+7+9+1 \neq 24

25 × 6666 157193 = 9457 25×6666-157193=9457 and 9 + 4 + 5 + 7 = 25 9+4+5+7=25

Lew Sterling Jr
Jul 15, 2015

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:

  1. Locate the next multiple of 6666 that is greater than 157193.

  2. 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

Harout G. Vartanian - 4 years, 1 month ago

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