Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that there is not a single natural number $N$ with sum of digits equal to 15 that is the square of an integer.

share|cite|improve this question
Hint: Do you know anything else about properties of numbers with certain digit-sums? – MJD May 24 '12 at 13:55

Hint: If the sum of the digits is $15$, then $N$ is divisible by $3$, if $N$ is a square then it is also divisble by $9$, if $N$ is divisble by $9$, then the sum of digits...

share|cite|improve this answer

Hint: If the sum of the digits is $15$, then $N\equiv 15\equiv 6\mod 9$. Can you show that $6$ is not a square $\bmod 9$?

share|cite|improve this answer

Observe that the recursive sum of digit(R.D.) of any number of the form (9.a+b) where 0≤b<9 are same.

By recursive sum of digit, I mean continue taking the sum of digits until it becomes <10.

For example, $193^2 = 37249$ =>R.D. of $193^2$=R.D. of 37249 = R.D. of 25 = 7.

As 193≡4(mod 9) => $193^2≡4^2(mod\ 9)$ =>R.D. of $193^2$ is 7

Now R.D. of $(9a)^2$=9,

R.D. of $(9a±1)^2$=1,

R.D. of $(9a±2)^2$=4,

R.D. of $(9a±3)^2$=9 and

R.D. of $(9a±4)^2$=7.

But (R.D.) of 15 is 6.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.