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Difference between revisions of "1992 AHSME Problems/Problem 18"

(Created page with "== Problem == The increasing sequence of positive integers <math>a_1,a_2,a_3,\cdots </math> has the property that <cmath>a_{n+2}=a_n+a_{n+1} \text{ for all } n\ge 1.</cmath> ...")
 
(Added a solution with explanation)
 
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== Solution ==
 
== Solution ==
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<math>\fbox{D}</math>
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<math>\fbox{D}</math> Let <math>a_{1} = a, a_{2} = b</math>, so <math>5a + 8b = 120</math>. Now <math>8b</math> and <math>120</math> are divisible by <math>8</math>, so <math>5a</math> is divisible by 8, so <math>a</math> is divisible by 8. It's now easy to try the multiples of <math>8</math> to get that <math>a = 8, b=10</math> (all the other possibilities violate the condition <math>a < b</math>, which comes from the fact that the sequence is increasing). Hence <math>a_8 = 8a + 13b = 8 \times 8 + 13 \times 10 = 194</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 03:02, 20 February 2018

Problem

The increasing sequence of positive integers $a_1,a_2,a_3,\cdots$ has the property that

\[a_{n+2}=a_n+a_{n+1} \text{ for all } n\ge 1.\]

If $a_7=120$, then $a_8$ is

$\text{(A) } 128\quad \text{(B) } 168\quad \text{(C) } 193\quad \text{(D) } 194\quad \text{(E) } 210$

Solution

$\fbox{D}$ Let $a_{1} = a, a_{2} = b$, so $5a + 8b = 120$. Now $8b$ and $120$ are divisible by $8$, so $5a$ is divisible by 8, so $a$ is divisible by 8. It's now easy to try the multiples of $8$ to get that $a = 8, b=10$ (all the other possibilities violate the condition $a < b$, which comes from the fact that the sequence is increasing). Hence $a_8 = 8a + 13b = 8 \times 8 + 13 \times 10 = 194$.

See also

1992 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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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