1970 AMC 12 Problem 4

Question

Let

S

be the set of all numbers which are the sum of the squares of three consecutive integers. Then we can say that:

No member of

S

is divisible by

2

No member of

S

is divisible by

3

but some member is divisible by

11

No member of

S

is divisible by

3

or by

5

No member of

S

is divisible by

3

or by

7

None of these

Write the integers as

n-1,n,n+1

Their squared sum is

3n^2+2;

examine it modulo

3

and test a residue modulo

11

The sum is

\begin{gathered} (n-1)^2+n^2+(n+1)^2\\ =3n^2+2. \end{gathered}

It is always congruent to

2\pmod3,

so no member of

S

is divisible by

3.

On the other hand, taking

n=5

gives

3(25)+2=77,

which is divisible by

11.

Therefore, the correct answer is B.

Problem 4 in other years