Difference between revisions of "1967 AHSME Problems/Problem 36"
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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. | + | 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. |
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>. | 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>. |
Revision as of 13:46, 12 July 2019
Problem
Given a geometric progression of five terms, each a positive integer less than . The sum of the five terms is . If is the sum of those terms in the progression which are squares of integers, then is:
Solution
Let the first term be and the common ratio be , and WLOG let . The five terms are , and the sum is . Clearly must be rational for all terms to be integers. If were an integer, it could not be , since would equal , which is not an integer. In fact, quickly testing shows that cannot be an integer.
We now consider non-integers. If and , then would have to be divisible by , since is an integer. If , then would have to be a multiple of , which would make the five terms sum to at least . It only gets worse if . Thus, , and is a multiple of . Let .
We now look at the last term, . The smallest allowable values for , given that cannot be even, are and . If , the last term will be way too big. Thus, is the only possibility.
We now have a sum of , and we know that is the only possibility. The terms in the parentheses happen to equal when you plug them in, so .
Thus, the terms are , and the first, third, and fifth terms are squares, with a sum of , which is answer .
See also
1967 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 35 |
Followed by Problem 37 | |
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 | ||
All AHSME Problems and Solutions |
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