$N$ is a positive integer without $0$ as a digit. There are $559$ other positive integers that share exactly the same digits as $N$ (and having the correct number of each). What is the smallest possible value of $N?$

Here's what I have (which was wrong):

559+1=660. So there are 660 total numbers that can be made with these unknown digits. The way to solve 660 is a!/(b!*c!*...) where a is how many digits there are and b, c, etc. is how many sets of repeating digits there are.

Ex. With the digits 444566 I can make 6!/(3!*2!) = 60 different numbers.

The prime factorization of 660 is 2^2*3*5*11. Which means that n! must include 11, so it has to be 11! or a greater factorial. 11! divided by 660 is 2^6*3^3*5*7.

The closest factorials that this is a multiple of is 9!(equialent to 2^6*3^3*5*7 times 6). 11! needs another factor of 6, so I need to up that to 12!. 12!/9!*2! is what I get. So the smallest number is in the form abbccccccccc (9 c's), or 100222222222.

But it's wrong. It's probably too big. Can someone help?

(To anyone who answers this question, please, PLEASE, just provide a little explanation. Even just your calculations would be fine. An answer out of the blue isn't going to help anyone. It also will stop others from clicking on this post because there is already an answer, however useless it may be.)

Thanks!

Guest Mar 31, 2021

#1**0 **

First of all, \(559+1=560\).

The prime factorization of 560 is \(2^4\cdot5\cdot7\), which means that N must be at least 7 digits.

If we test \(\frac{7!}{560}\), we get 9, which is not the product of factorials of integers, so it must be more than 7 digits.

If we test \(\frac{8!}{560}\), we get \(2!\cdot3!\cdot3!\), which is a product of factorials, so the smallest possible value of N is an 8 digit number.

This means that we will get a number in the form of aaabbbcc, where a, b, and c are digits.

Since 0 cannot be a digit, the smallest possible value is \(\boxed{11122233}\)

textot Mar 31, 2021