1970 AHSME Problems/Problem 4: Difference between revisions

Latest revision as of 19:09, 6 January 2018

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.

1970 AHSC (Problems • Answer Key • Resources)
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

@@ Line 10: / Line 10: @@
 == Solution ==
-<math>\fbox{B}</math>
+Consider <math>3</math> consecutive integers <math>a, b,</math> and <math>c</math>.  Exactly one of these integers must be divisible by 3; WLOG, suppose <math>a</math> is divisible by 3.  Then <math>a \equiv 0 \pmod {3},  b \equiv 1 \pmod{3},</math> and <math>c \equiv 2 \pmod{3}</math>.  Squaring, we have that <math>a^{2} \equiv 0 \pmod{3}, b^{2} \equiv 1 \pmod{3},</math> and <math>c^{2} \equiv 1 \pmod{3}</math>, so <math>a^{2} + b^{2} + c^{2} \equiv 2 \pmod{3}</math>.  Therefore, no member of <math>S</math> is divisible by 3.
+Now consider <math>3</math> more consecutive integers <math>a, b,</math> and <math>c</math>, which we will consider mod 11.  We will assign <math>k</math> such that <math>a \equiv k \pmod{11}, b \equiv k + 1 \pmod{11},</math> and <math>c \equiv k + 2 \pmod{11}</math>.  Some experimentation shows that when <math>k = 4, a \equiv 4 \pmod{11}</math> so <math>a^{2} \equiv 5 \pmod{11}</math>.  Similarly, <math>b \equiv 5 \pmod{11}</math> so <math>b^{2} \equiv 3 \pmod{11}</math>, and <math>c \equiv 6 \pmod{11}</math> so <math>c^{2} \equiv 3 \pmod{11}</math>.  Therefore, <math>a^{2} + b^{2} + c^{2} \equiv 0 \pmod{11}</math>, so there is at least one member of <math>S</math> which is divisible by 11.  Thus, <math>\fbox{B}</math> is correct.
 == See also ==
-{{AHSME box|year=1970|num-b=3|num-a=5}}
+{{AHSME 35p box|year=1970|num-b=3|num-a=5}}
 [[Category: Introductory Algebra Problems]]
 {{MAA Notice}}

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1970 AHSME Problems/Problem 4: Difference between revisions

Latest revision as of 19:09, 6 January 2018

Problem

Solution

See also