1992 AHSME Problems/Problem 18: Difference between revisions
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 == | ||
<math>\fbox{D}</math> | <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 02:02, 20 February 2018
Problem
The increasing sequence of positive integers 
has the property that
![\[a_{n+2}=a_n+a_{n+1} \text{ for all } n\ge 1.\]](http://latex.artofproblemsolving.com/7/2/e/72e5280db01a8da1c1a59c1a08713b9d4a3a8d25.png)
If 
, then 
is
Solution
Let 
, so 
. Now 
and 
are divisible by 
, so 
is divisible by 8, so 
is divisible by 8. It's now easy to try the multiples of to get that 
(all the other possibilities violate the condition 
, which comes from the fact that the sequence is increasing). Hence 
.
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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