IMO 1960 Problem 1
We seek all three-digit integers whose quotient upon division by $11$ equals the sum of the squares of their digits.
Proposed by: -
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 27m41s
Problem
Find all three-digit numbers for which one obtains, when dividing the number by 11, the sum of the squares of the digits of the initial number.
Problem Understanding
We seek all three-digit integers whose quotient upon division by $11$ equals the sum of the squares of their digits.
Let the number be
$$N=100a+10b+c,$$
where
$$a\in{1,2,\dots,9},\qquad b,c\in{0,1,\dots,9}.$$
The condition of the problem is therefore
$$\frac{N}{11}=a^2+b^2+c^2,$$
or equivalently
$$100a+10b+c=11(a^2+b^2+c^2).$$
The task is to determine all digit triples $(a,b,c)$ satisfying this equation and to prove that no others exist.
Key Observations
Since the right-hand side is divisible by $11$, the number $N$ itself must be divisible by $11$. Hence
$$100a+10b+c\equiv0\pmod{11}.$$
Using
$$100\equiv1\pmod{11},\qquad 10\equiv-1\pmod{11},$$
we obtain
$$a-b+c\equiv0\pmod{11}.$$
Now
$$1-9+0=-8, \qquad 9-0+9=18,$$
so the integer $a-b+c$ lies between $-8$ and $18$. The only multiples of $11$ in this interval are $0$ and $11$. Therefore the only possibilities are
$$a-b+c=0 \qquad\text{or}\qquad a-b+c=11.$$
These two cases exhaust all possibilities.
Solution
Let
$$N=100a+10b+c,$$
with
$$a\in{1,\dots,9},\qquad b,c\in{0,\dots,9}.$$
The defining equation is
$$100a+10b+c=11(a^2+b^2+c^2).$$
As shown above, either
$$a-b+c=11$$
or
$$a-b+c=0.$$
We examine the two cases separately.
Case 1
Assume
$$a-b+c=11.$$
Then
$$b=a+c-11.$$
Since $b$ is a digit, we must have
$$0\le a+c-11\le9,$$
so in particular
$$a+c\ge11.$$
Substituting $b=a+c-11$ into the defining equation gives
$$100a+10(a+c-11)+c = 11\bigl(a^2+(a+c-11)^2+c^2\bigr).$$
The left-hand side simplifies to
$$110a+11c-110.$$
Also,
$$(a+c-11)^2 = a^2+2ac+c^2-22a-22c+121.$$
Hence
$$110a+11c-110 = 11(2a^2+2ac+2c^2-22a-22c+121).$$
Dividing by $11$ gives
$$10a+c-10 = 2a^2+2ac+2c^2-22a-22c+121.$$
Rearranging,
$$2a^2+2ac+2c^2-32a-23c+131=0.$$
Define
$$F(a,c)=2a^2+2ac+2c^2-32a-23c+131.$$
Because $a+c\ge11$, write
$$c=11-a+t,$$
where $t\ge0$. Substituting into $F(a,c)$ gives
$$\begin{aligned} F(a,11-a+t) &= 2a^2+2a(11-a+t)+2(11-a+t)^2 \ &\qquad -32a-23(11-a+t)+131. \end{aligned}$$
Expanding carefully,
$$\begin{aligned} 2a^2+2a(11-a+t) &= 2a^2+22a-2a^2+2at = 22a+2at, \end{aligned}$$
and
$$\begin{aligned} 2(11-a+t)^2 &= 2(121+a^2+t^2-22a+22t-2at) \ &= 242+2a^2+2t^2-44a+44t-4at. \end{aligned}$$
Therefore
$$\begin{aligned} F(a,11-a+t) &= 22a+2at +242+2a^2+2t^2-44a+44t-4at \ &\qquad -32a-253+23a-23t+131. \end{aligned}$$
Collecting like terms yields
$$\begin{aligned} F(a,11-a+t) &= 2t^2+(21-2a)t+2a^2-31a+120. \end{aligned}$$
Now complete the square in the $a$-terms:
$$\begin{aligned} 2a^2-31a+120 &= 2\left(a-\frac{31}{4}\right)^2-\frac{961}{8}+120 \ &= 2\left(a-\frac{31}{4}\right)^2-\frac1{8}. \end{aligned}$$
Since $a$ is an integer between $1$ and $9$, we check directly:
$$\begin{array}{c|ccccccccc} a & 1&2&3&4&5&6&7&8&9\ \hline 2a^2-31a+120 & 91&66&45&28&15&6&1&0&3 \end{array}$$
Thus
$$2a^2-31a+120\ge0$$
for all admissible $a$, with equality only at $a=8$.
Also,
$$21-2a\ge3$$
for every digit $a\le9$. Since $t\ge0$, every term in
$$F(a,11-a+t)=2t^2+(21-2a)t+(2a^2-31a+120)$$
is nonnegative. Hence
$$F(a,c)\ge0.$$
To satisfy the defining equation we would need $F(a,c)=0$. Therefore equality must hold in every nonnegative term above. In particular,
$$2a^2-31a+120=0.$$
From the table, this occurs only when $a=8$. Then
$$21-2a=5,$$
so
$$F(8,3+t)=2t^2+5t.$$
Since $t\ge0$, this vanishes only for $t=0$. Hence
$$a=8,\qquad c=3.$$
But then
$$b=a+c-11=8+3-11=0,$$
and substituting into the original equation gives
$$803\ne11(64+0+9)=803\ne803?$$
Indeed,
$$64+0+9=73, \qquad 11\cdot73=803.$$
Thus we obtain the valid equality
$$803=11(64+0+9).$$
However,
$$803$$
is not divisible by $11$, because
$$8-0+3=11,$$
so actually it is divisible by $11$, and
$$803/11=73.$$
Hence $803$ is in fact a valid solution.
Therefore the branch $a-b+c=11$ yields the solution
$$803.$$
Case 2
Assume
$$a-b+c=0.$$
Then
$$b=a+c.$$
Since $b$ is a digit,
$$a+c\le9.$$
Substituting into the defining equation gives
$$100a+10(a+c)+c = 11(a^2+(a+c)^2+c^2).$$
Thus
$$110a+11c = 11(2a^2+2ac+2c^2).$$
Dividing by $11$,
$$10a+c = 2a^2+2ac+2c^2.$$
Rearranging,
$$2c^2+(2a-1)c+(2a^2-10a)=0.$$
Treat this as a quadratic equation in $c$. Its discriminant is
$$\Delta=(2a-1)^2-8(2a^2-10a).$$
Expanding,
$$\Delta = 4a^2-4a+1-16a^2+80a = -12a^2+76a+1.$$
We now test all possible values of the digit $a$.
For $a=1$,
$$\Delta=65,$$
not a square.
For $a=2$,
$$\Delta=105,$$
not a square.
For $a=3$,
$$\Delta=121.$$
The roots are
$$c=\frac{-5\pm11}{4},$$
namely
$$c=\frac32,,-4,$$
not digits.
For $a=4$,
$$\Delta=113,$$
not a square.
For $a=5$,
$$\Delta=81.$$
The equation becomes
$$2c^2+9c=0,$$
so
$$c=0.$$
Then
$$b=a+c=5,$$
giving
$$N=550.$$
For $a=6$,
$$\Delta=25.$$
The roots are
$$c=\frac{-11\pm5}{4},$$
namely
$$-\frac32,,-4,$$
not digits.
For $a=7$,
$$\Delta=-55<0.$$
For $a=8$,
$$\Delta=-159<0.$$
For $a=9$,
$$\Delta=-287<0.$$
Hence the only solution in this branch is
$$550.$$
Finally,
$$\frac{550}{11}=50,$$
and
$$5^2+5^2+0^2=25+25=50.$$
Also,
$$\frac{803}{11}=73,$$
and
$$8^2+0^2+3^2=64+9=73.$$
Therefore the complete set of three-digit numbers satisfying the condition is
$$\boxed{550,\ 803}.$$
Verification of Key Steps
The congruence argument is exhaustive because
$$-8\le a-b+c\le18,$$
and the only multiples of $11$ in this interval are
$$0,\ 11.$$
In the branch
$$a-b+c=11,$$
the critical computation is
$$2a^2+2ac+2c^2-32a-23c+131=0.$$
Writing
$$c=11-a+t, \qquad t\ge0,$$
transforms this into
$$2t^2+(21-2a)t+(2a^2-31a+120)=0.$$
The table
$$\begin{array}{c|ccccccccc} a & 1&2&3&4&5&6&7&8&9\ \hline 2a^2-31a+120 & 91&66&45&28&15&6&1&0&3 \end{array}$$
shows explicitly that this expression is always nonnegative. Equality can occur only when $a=8$. Substituting $a=8$ forces $t=0$, giving $(a,b,c)=(8,0,3)$.
In the branch
$$a-b+c=0,$$
the discriminant
$$\Delta=-12a^2+76a+1$$
was checked for every digit $a=1,\dots,9$, which is exhaustive.
Alternative Approaches
After deriving
$$a-b+c=0 \qquad\text{or}\qquad a-b+c=11,$$
one may proceed by direct finite checking.
In the branch
$$a-b+c=11,$$
the condition $b=a+c-11$ leaves only finitely many possibilities:
$$(a,c)\in{(2,9),(3,8),\dots,(9,9)}.$$
Substituting each pair into the defining equation quickly isolates the solution $(8,0,3)$.
In the branch
$$a-b+c=0,$$
the equation
$$2c^2+(2a-1)c+(2a^2-10a)=0$$
may also be handled modulo $2$. Since
$$(2a-1)c$$
has the parity of $c$, the equation implies that $c$ is even. Testing
$$c=0,2,4,6,8$$
reduces the search immediately, and only
$$(a,c)=(5,0)$$
works.