Difference between revisions of "2011 AMC 10B Problems/Problem 10"
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− | == Problem | + | == Problem== |
Consider the set of numbers <math>\{1, 10, 10^2, 10^3, \ldots, 10^{10}\}</math>. The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer? | Consider the set of numbers <math>\{1, 10, 10^2, 10^3, \ldots, 10^{10}\}</math>. The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer? | ||
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<math> \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101 </math> | <math> \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101 </math> | ||
− | == Solution == | + | == Solution 1 == |
− | The | + | The requested ratio is <cmath>\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.</cmath> Using the formula for a geometric series, we have <cmath>10^9 + 10^8 + \ldots + 10 + 1 = \dfrac{10^{10} - 1}{10 - 1} = \dfrac{10^{10} - 1}{9},</cmath> which is very close to <math>\dfrac{10^{10}}{9},</math> so the ratio is very close to <math>\boxed{\mathrm{(B) \ } 9}.</math> |
+ | |||
+ | |||
+ | == Solution 2 == | ||
+ | The problem asks for the value of <cmath>\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.</cmath> Written in base 10, we can find the value of <math>10^9 + 10^8 + \ldots + 10 + 1</math> to be <math>1111111111.</math> Long division gives us the answer to be <math>\boxed{\mathrm{(B) \ } 9}.</math> | ||
+ | |||
+ | Alternate finish: multiply the denominator by 9 and notice that it is 1 less than <math>10^{10}</math>. So the answer is very very close to <math>\boxed{\mathrm{(B) } 9}</math>. | ||
+ | |||
+ | ~JH. L | ||
+ | |||
+ | == Solution 3 == | ||
+ | Let <math>f(n)=\dfrac{10^n}{1+10+10^2+10^3+\cdots+10^{n-1}}</math>. We are approximating <math>f(10)</math>. Trying several small values of <math>n</math> gives answers very close to <math>9</math>, so our answer is <math>\boxed{\textbf{(B)}~9}</math>. Note that <math>f(1)=10</math>, but <math>f(2)=\dfrac{100}{11}\approx9.09.</math> | ||
+ | ~Technodoggo | ||
== See Also== | == See Also== | ||
{{AMC10 box|year=2011|ab=B|num-b=9|num-a=11}} | {{AMC10 box|year=2011|ab=B|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} |
Latest revision as of 22:43, 31 August 2023
Problem
Consider the set of numbers . The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?
Solution 1
The requested ratio is Using the formula for a geometric series, we have
which is very close to
so the ratio is very close to
Solution 2
The problem asks for the value of Written in base 10, we can find the value of
to be
Long division gives us the answer to be
Alternate finish: multiply the denominator by 9 and notice that it is 1 less than . So the answer is very very close to
.
~JH. L
Solution 3
Let . We are approximating
. Trying several small values of
gives answers very close to
, so our answer is
. Note that
, but
~Technodoggo
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 | ||
All AMC 10 Problems and Solutions |
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