# Math Help - IMO 1960 problem 1

1. ## IMO 1960 problem 1

Determine all three-digit numbers N having the property that N is divisible by 11, and N/11 is equal to the sum of the squares of the digits of 11.

Anyone know how to solve this one. We write N as abc then...

Since N is divisible by 11 we know that a - b + c is divisible by 11. So a-b+c = 0 or 11.

Then we also know that $a^2 + b^2 + c^2 = \frac{N}{11}$

I can't get any sort of 'nice' solution from this and just end up with having to check 18 cases. (a+c=b, sub into above than a=0,1,2,... etc)

...and N/11 is equal to the sum of the squares of the digits of 11...
$1^2+1^2=2$?

I'm sure this is a typo.

3. Maybe there is another method which is better ...

the actual value of $abc$ is

$100a + 10b + c = (99a + 11b) + (a-b+c)$

which leads to two cases $a-b+c = 0$ and $a-b+c = 11$

Let's check the first one :

We have $a^2 + b^2 + c^2 = 9a+b$

Since $c = b-a$ substitute it we obtain

$a^2 + b^2 + (b-a)^2 = 9a + b$

$2a^2 - a(2b+9) + 2b^2 - b = 0$ which is a quadratic equation of $a$ , check its discriminant and it should be a square $k^2$ ie:

$(2b+9)^2 - 8(2b^2 - b) = k^2$

$-12b^2 + 44b + 81 = k^2$

$-(6b-11)^2 + 364 = 3k^2$

$(6b-11)^2 + 3k^2 = 364 = 4 \times 91$

We have
$4 = 1^2 + 3(1)^2 = (1 + \sqrt{3} i )(1 - \sqrt{3} i )$
$91 = 4^2 + 3(5)^2 = (4 + 5\sqrt{3} i )(4 - 5\sqrt{3} i )$

( Note we neglect the case $4 = 2^2 + 3(0)^2$ because later we will find that $b$ must not be an integer . )

$(1 + \sqrt{3} i )( 4 + 5\sqrt{3} i ) = -11 + 9\sqrt{3}i$
$(1 + \sqrt{3} i )( 4 + 5\sqrt{3} i ) = 19 - \sqrt{3} i$

Therefore , the two possible ways to write $364$ in the form $x^2 + 3y^2$ are :

$11^2 + 3(9)^2$ and
$19^2 + 3(1)^2$

and we have $|6b-11| = 11$ or $|6b-11| = 19$ the only solution is $b=5$ and $|k| = 1$ ( corresponding to $b$ )

so $a = \frac{2b+9 + 1}{4}$ or $\frac{2b+9 - 1}{4}$ which gives $a = 5$

$c = 5-5 = 0$

Therefore , the required number is $550$

Use the similar method to see if there is another solution in the second case . It should be $803$ .

4. @ Chip yes should have been N.

@ Simple pendulum nice solution! I don't have the answers so can't confirm but I never thought about introducing complex numbers...