1970 AHSME Problems/Problem 4

Revision as of 19:09, 6 January 2018 by Miziziriz (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

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

$\text{(A) No member of S is divisible by } 2\quad\\ \text{(B) No member of S is divisible by } 3 \text{ but some member is divisible by } 11\quad\\ \text{(C) No member of S is divisible by } 3 \text{ or } 5\quad\\ \text{(D) No member of S is divisible by } 3 \text{ or } 7 \quad\\ \text{(E) None of these}$

Solution

Consider $3$ consecutive integers $a, b,$ and $c$. Exactly one of these integers must be divisible by 3; WLOG, suppose $a$ is divisible by 3. Then $a \equiv 0 \pmod {3},  b \equiv 1 \pmod{3},$ and $c \equiv 2 \pmod{3}$. Squaring, we have that $a^{2} \equiv 0 \pmod{3}, b^{2} \equiv 1 \pmod{3},$ and $c^{2} \equiv 1 \pmod{3}$, so $a^{2} + b^{2} + c^{2} \equiv 2 \pmod{3}$. Therefore, no member of $S$ is divisible by 3.

Now consider $3$ more consecutive integers $a, b,$ and $c$, which we will consider mod 11. We will assign $k$ such that $a \equiv k \pmod{11}, b \equiv k + 1 \pmod{11},$ and $c \equiv k + 2 \pmod{11}$. Some experimentation shows that when $k = 4, a \equiv 4 \pmod{11}$ so $a^{2} \equiv 5 \pmod{11}$. Similarly, $b \equiv 5 \pmod{11}$ so $b^{2} \equiv 3 \pmod{11}$, and $c \equiv 6 \pmod{11}$ so $c^{2} \equiv 3 \pmod{11}$. Therefore, $a^{2} + b^{2} + c^{2} \equiv 0 \pmod{11}$, so there is at least one member of $S$ which is divisible by 11. Thus, $\fbox{B}$ is correct.

See also

1970 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS