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>2^4 = 16</math>. Let <math>a = 16k</math>. |
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. | 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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== See also == | == See also == | ||
− | {{AHSME box|year=1967|num-b=35|num-a=37}} | + | {{AHSME 40p box|year=1967|num-b=35|num-a=37}} |
[[Category: Intermediate Algebra Problems]] | [[Category: Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:40, 16 August 2023
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 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
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