Difference between revisions of "1965 AHSME Problems/Problem 24"
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| − | + | == Problem == | |
| + | |||
| + | Given the sequence <math>10^{\frac {1}{11}},10^{\frac {2}{11}},10^{\frac {3}{11}},\ldots,10^{\frac {n}{11}}</math>, | ||
| + | the smallest value of n such that the product of the first <math>n</math> members of this sequence exceeds <math>100000</math> is: | ||
| + | |||
| + | <math>\textbf{(A)}\ 7 \qquad | ||
| + | \textbf{(B) }\ 8 \qquad | ||
| + | \textbf{(C) }\ 9 \qquad | ||
| + | \textbf{(D) }\ 10 \qquad | ||
| + | \textbf{(E) }\ 11 </math> | ||
| + | |||
| + | == Solution == | ||
| + | |||
| + | Note that the given sequence is a [[geometric sequence]] with a common ratio <math>10^{\frac{1}{11}}</math>. Let the product of the first <math>n</math> terms of the sequence be denoted <math>P_n</math>. It is a consequence of the laws of exponents that <math>P_1=(10^{\frac{1}{11}})^1</math>, <math>P_2=(10^{\frac{1}{11}})^{1+2}</math>, and, in general, <math>P_n=(10^{\frac{1}{11}})^{T_n}</math>, where <math>T_n</math> denotes the <math>n</math>th [[triangular number]]. Setting <math>P_n</math> equal to <math>100,000</math>, we see that: | ||
| + | \begin{align*} \\ | ||
| + | (10^{\frac{1}{11}})^{\frac{n(n+1)}{2}}&=10^5 \\ | ||
| + | \frac{1}{11}*\frac{n(n+1)}{2}&=5 \\ | ||
| + | n(n+1)&=110 \\ | ||
| + | n^2+n-110&=0 \\ | ||
| + | (n-10)(n+11)&=0 \\ | ||
| + | \end{align*} | ||
| + | Because <math>n</math> must be positive, we are left with <math>n=10</math>. Given this information, choice (D) may seem appealing. Do not be fooled. The product of the first <math>n</math> terms is ''exactly'' <math>100,000</math>, but the problem asks for the smallest <math>n</math> such that <math>P_n</math> ''exceeds'' 100,000. Thus, the minimum value of <math>n</math> which satsifies the problem is <math>\boxed{\textbf{(E) }11}</math>. | ||
| + | |||
| + | == See Also == | ||
| + | {{AHSME 40p box|year=1965|num-b=23|num-a=25}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:Intermediate Algebra Problems]] | ||
Latest revision as of 18:25, 18 July 2024
Problem
Given the sequence 
,
the smallest value of n such that the product of the first 
members of this sequence exceeds 
is:
Solution
Note that the given sequence is a geometric sequence with a common ratio 
. Let the product of the first terms of the sequence be denoted 
. It is a consequence of the laws of exponents that 
, 
, and, in general, 
, where 
denotes the th triangular number. Setting equal to 
, we see that:
\begin{align*} \\
(10^{\frac{1}{11}})^{\frac{n(n+1)}{2}}&=10^5 \\
\frac{1}{11}*\frac{n(n+1)}{2}&=5 \\
n(n+1)&=110 \\
n^2+n-110&=0 \\
(n-10)(n+11)&=0 \\
\end{align*}
Because must be positive, we are left with 
. Given this information, choice (D) may seem appealing. Do not be fooled. The product of the first terms is exactly , but the problem asks for the smallest such that exceeds 100,000. Thus, the minimum value of which satsifies the problem is 
.
See Also
| 1965 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
