Difference between revisions of "1960 IMO Problems/Problem 1"

Revision as of 22:27, 28 May 2018

Problem

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

Solutions

Solution 1

Let $N = 100a + 10b+c$ for some digits $a,b,$ and $c$ . Then $100a + 10b+c = 11m$ for some $m$ . We also have $m=a^2+b^2+c^2$ . Substituting this into the first equation and simplification, we get $100a+10b+c = 11a^2 +11b^2 +11c^2$ For an integer divisible by $11$ , the the sum of digits in the odd positions minus the sum of digits in the even positions is divisible by $11$ . Thus we get: $b = a + c$ or $b = a + c - 11$ .

Case $1$ : Let $b=a+c$ . We get $100a+c+10a+10c = 11a^2 +11c^2+11(a+c)^2$ $10a+c = 2a^2+2ac+2c^2$ Since the right side is even, the left side must also be even. Let $c=2q$ for some $q = 0,1,2,3,4$ . Then $10a+2q=2a^2+4aq+8q^2$ $5a+q=a^2+2aq+4q^2$ Substitute $q=0,1,2,3,4$ into the last equation and then solve for $a$ .

When $q=0$ , we get $a=5$ . Thus $c=0$ and $b=5$ . We get that $N=550$ which works.

When $q=1$ , we get that $a$ is not an integer. There is no $N$ for this case.

When $q=2$ , we get that $a$ is not an integer. There is no $N$ for this case.

When $q=3$ , we get that $a$ is not an integer. There is no $N$ for this case.

When $q=4$ , we get that $a$ is not an integer. There is no $N$ for this case.

Case $2$ : Let $b = a + c - 11$ . We get $100a+c+10a+10c -110= 11(a^2+(a+c)^2-22(a+c)+c^2+121)$ $10a+c=2a^2+2c^2+2ac-22a-22c+131$ $2(a-8)^2+2(c-\frac{23}{4})^2+2ac-\frac{505}{8}=0$ Now we test all $c=0\rightarrow10$ . When $c=0,1,2,4,5,6,7,8,9$ , we get no integer solution to $a$ . Thus, for these values of $c$ , there is no valid $N$ . However, when $c=3$ , we get $2(a-8)^2+2(3-\frac{23}{4})^2+6a-\frac{505}{8}=0$ $2(a-8)^2+6a-48 = 0$ We get that $a=8$ is a valid solution. For this case, we get $a=8,b=0,c=3$ , so $N=803$ , and this is a valid value. Thus, the answers are $\boxed{N=550,803}$ .

Solution 2

Define a ten to be all ten positive integers which begin with a fixed tens digit.

We can make a systematic approach to this:

By inspection, $\dfrac{N}{11}$ must be between 10 and 90 inclusive. That gives us 8 tens to check, and 90 as well.

For a given ten, the sum of the squares of the digits of $N$ increases faster than $\dfrac{N}{11}$ , so we can have at most one number in every ten that works.

We check the first ten:

$11*11=121$

$1^2+2^2+1^2=4$

$12*11=132$

$1^2+3^2+2^2=14$

11 is too small and 12 is too large, so all numbers below 11 will be too small and all numbers above 12 will be too large, so no numbers in the first ten work.

We try the second ten:

$21*11=231$

$2^2+3^2+1^2=14$

$22*11=242$

$2^2+4^2+2^2=24$

Therefore, no numbers in the second ten work.

We continue, to find out that 50 and 73 are the only ones that works.

$N=50*11=550$ , $N=73*11=803$ so there are two $N$ that works.

Solution 3

Note that there are only 900 three-digit numbers. Of these, only 81 are divisible by 11. This brings to mind the great strategy of listing and bashing. Being organized while listing will get you the answer. There is plenty of time on the IMO, and listing is highly unlikely to miss cases.

@@ Line 65: / Line 65: @@
 ===Solution 3===
-Note that there are only 900 three-digit numbers. This brings to mind the great strategy of listing and bashing. Being organized while listing will get you the answer.
+Note that there are only 900 three-digit numbers. Of these, only 81 are divisible by 11. This brings to mind the great strategy of listing and bashing. Being organized while listing will get you the answer. There is plenty of time on the IMO, and listing is highly unlikely to miss cases.
 ==See Also==
 {{IMO7 box|year=1960|before=First Question|num-a=2}}

1960 IMO (Problems)
Preceded by First Question	1 • 2 • 3 • 4 • 5 • 6 • 7	Followed by Problem 2

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