Digit sum of square root

Number Theory Level 2

A 5-digit number has digits k , k + 1 , k + 2 , 3 k , k + 3 k, k + 1, k + 2, 3k, k + 3 in order, from left to right.

If this number is m 2 m^2 for some natural number m m , find the sum of the digits of m m .


The answer is 15.

Shubhrajit Sadhukhan by Shubhrajit Sadhukhan , 16, India

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.

1 solution

Hongqi Wang
Mar 13, 2021

0 < 3 k ≤ 9 ⟹ 1 ≤ k ≤ 3 0 < 3k \leq 9 \implies 1 \leq k \leq 3

  • if k = 1 k = 1 , the last 2 digits are 34, it is not a square number
  • if k = 2 k = 2 , the last 2 digits are 65, not a square number either.
  • if k = 3 k = 3 , the number is 3 ? 596 3?596 . The only square number is 18 6 2 = 34596 186^2 = 34596
1 Helpful 1 Interesting 1 Brilliant 0 Confused

You're answer is incorrect. The digit sum of 34,596 = 27.

tom engelsman - 3 months ago

Log in to reply

the question is the digits sum of m m , but not the number 34596.

Hongqi Wang - 3 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...