Difference between revisions of "1967 AHSME Problems/Problem 36"

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== Solution ==
 
== Solution ==
<math>\fbox{C}</math>
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Let the first term be <math>a</math> and the common ratio be <math>r</math>, and WLOG let <math>r \ge 1</math>.  The five terms are <math>a, ar, ar^2, ar^3, ar^4</math>, and the sum is <math>a(1 + r + r^2 + r^3 + r^4)</math>.  Clearly <math>r</math> must be rational for all terms to be integers.  If <math>r</math> were an integer, it could not be <math>1</math>, since <math>a</math> would equal <math>\frac{211}{1 + r + r^2 + r^3 + r^4</math>, which is not an integer.  In fact, quickly testing <math>r = 2, 3, 4</math> shows that <math>r</math> cannot be an integer.
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We now consider non-integers.  If <math>r = \frac{x}{y}</math> and <math>\gcd(x, y) = 1</math>, then <math>a</math> would have to be divisible by <math>y^4</math>, since <math>ar^4 = a\frac{x^4}{y^4}</math> is an integer.  If <math>y = 3</math>, then <math>a</math> would have to be a multiple of <math>81</math>, which would make the five terms sum to at least <math>5 \cdot 81</math>.  It only gets worse if <math>y > 3</math>.  Thus, <math>y = 2</math>, and <math>a</math> is a multiple of <math>16</math>.  Let <math>a = 16k</math>.
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We now look at the last term, <math>ar^4 = 16k\frac{x^4}{16} = kx^4</math>.  The smallest allowable values for <math>x</math>, given that <math>x</math> cannot be even, are <math>3</math> and <math>5</math>.  If <math>x = 5</math>, the last term will be way too big.  Thus, <math>x = 3</math> is the only possibility.
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We now have a sum of <math>a(1 + r + r^2 + r^3 + r^4) = 211</math>, and we know that <math>r = \frac{3}{2}</math> is the only possibility.  The terms in the parentheses happen to equal <math>\frac{211}{16}</math> when you plug them in, so <math>a = 16</math>.
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Thus, the terms are <math>16, 24, 36, 54, 81</math>, and the first, third, and fifth terms are squares, with a sum of <math>133</math>, which is answer <math>\fbox{C}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 13:45, 12 July 2019

Problem

Given a geometric progression of five terms, each a positive integer less than $100$. The sum of the five terms is $211$. If $S$ is the sum of those terms in the progression which are squares of integers, then $S$ is:

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 91\qquad \textbf{(C)}\ 133\qquad \textbf{(D)}\ 195\qquad \textbf{(E)}\ 211$

Solution

Let the first term be $a$ and the common ratio be $r$, and WLOG let $r \ge 1$. The five terms are $a, ar, ar^2, ar^3, ar^4$, and the sum is $a(1 + r + r^2 + r^3 + r^4)$. Clearly $r$ must be rational for all terms to be integers. If $r$ were an integer, it could not be $1$, since $a$ would equal $\frac{211}{1 + r + r^2 + r^3 + r^4$ (Error making remote request. No response to HTTP request), which is not an integer. In fact, quickly testing $r = 2, 3, 4$ shows that $r$ cannot be an integer.

We now consider non-integers. If $r = \frac{x}{y}$ and $\gcd(x, y) = 1$, then $a$ would have to be divisible by $y^4$, since $ar^4 = a\frac{x^4}{y^4}$ is an integer. If $y = 3$, then $a$ would have to be a multiple of $81$, which would make the five terms sum to at least $5 \cdot 81$. It only gets worse if $y > 3$. Thus, $y = 2$, and $a$ is a multiple of $16$. Let $a = 16k$.

We now look at the last term, $ar^4 = 16k\frac{x^4}{16} = kx^4$. The smallest allowable values for $x$, given that $x$ cannot be even, are $3$ and $5$. If $x = 5$, the last term will be way too big. Thus, $x = 3$ is the only possibility.

We now have a sum of $a(1 + r + r^2 + r^3 + r^4) = 211$, and we know that $r = \frac{3}{2}$ is the only possibility. The terms in the parentheses happen to equal $\frac{211}{16}$ when you plug them in, so $a = 16$.

Thus, the terms are $16, 24, 36, 54, 81$, and the first, third, and fifth terms are squares, with a sum of $133$, which is answer $\fbox{C}$.

See also

1967 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 35
Followed by
Problem 37
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