1
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ Hint: Do you know anything else about properties of numbers with certain digit-sums? $\endgroup$ – MJD May 24 '12 at 13:55
10
$\begingroup$

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...

$\endgroup$
8
$\begingroup$

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 $\bmod 9$?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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