Inspired By Sandeep Sir

Number Theory Level 3

How many of 10-digit integers can be formed by using all the digits 0 to 9 (inclusive) such that it is a prime number?

Inspiration , not Quite related, but I got the idea from here.

The answer is 0.

1 solution

Mehul Arora
May 7, 2015

Each of the Numbers Formed using digits From 0-9 Will Be Divisible By 3 Because, 0+1+2+3+4+5+6+7+8+9=45 which is Divisible by 3.

Hence, The answer is 0.

Cheers! :D

Moderator note:

Good job identifying divisibility rule as the key. Bonus question: How many 45-digit integers can be formed using one 1, two 2's, three 3's, and so on till nine 9's such that it is a prime number?

Really nice problem.Cheers!

Nihar Mahajan - 6 years, 1 month ago

Cheers! Thanks! ^_^

Mehul Arora - 6 years, 1 month ago

[Response to Challenge Master Note]

@Calvin Lin , there's a typo in the note. It should be " $\color{#D61F06}{45}$ -digit integers".

The total digit sum is an invariant property, i.e, it doesn't change on permutation of the digits of the number. The invariant digit sum for the stated $45$ -digit integers is given by,

$\sum_{i=1}^9\sum_{j=1}^ii=\sum_{i=1}^9i^2=\frac{9\times 10\times 19}{6}=285$

By the divisibility rule of $3$ , we get that $285\equiv 0\pmod3$ . As such, there can be no such integer which is prime.

Prasun Biswas - 6 years, 1 month ago

K fixed. Thanks!

Calvin Lin Staff - 6 years, 1 month ago

You know, you sounded a bit like Sreejato (a.k.a. "the bunny") in that reply! :P