# Thread: 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 $\displaystyle a^2 + b^2 + c^2 = \frac{N}{11}$

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)

2. Originally Posted by Deadstar
...and N/11 is equal to the sum of the squares of the digits of 11...
$\displaystyle 1^2+1^2=2$?

I'm sure this is a typo.

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

the actual value of $\displaystyle abc$ is

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

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

Let's check the first one :

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle c = 5-5 = 0$

Therefore , the required number is $\displaystyle 550$

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

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...

5. Originally Posted by Deadstar
@ 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...
Human being never thought about introducing complex numbers but luckily we have Gauss !