1, 10, 100, 1000, 10000, 100000

Find the smallest positive integer k k such that there exists a positive integer n n whose sum of divisors is equal to 1 0 k 10^k .


The answer is 4.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Mark Hennings
Apr 8, 2017

Since 7 , 1249 7,1249 are prime and therefore σ 1 ( 8743 ) = σ 1 ( 7 ) σ 1 ( 1249 ) = 8 × 1250 = 10000 \sigma_1(8743) = \sigma_1(7)\sigma_1(1249) = 8\times1250 = 10000 , we see that k 4 k \le 4 . It is simple enough to check the values of σ 1 ( n ) \sigma_1(n) for 1 n 1000 1 \le n \le 1000 to see that it is not possible for σ 1 ( n ) \sigma_1(n) to be equal to 10 10 , 100 100 or 1000 1000 . Thus k = 4 k = \boxed{4} .

For interest, the other number n n with σ 1 ( n ) = 10000 \sigma_1(n) = 10000 is 9481 = 19 × 499 9481 = 19 \times 499 .

Giorgos K.
Mar 16, 2018

Using Mathematica we can find the first few n numbers

Rest@Select[Range@100000, IntegerQ[Log10@DivisorSigma[1, #]] &]

which returns
{8743, 9481, 71193, 94981, 98671, 99301}

so the k numbers we are looking for are
4, 4, 5, 5, 5, 5

so the smallest k is 4

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...