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Difference between revisions of "1997 AHSME Problems/Problem 29"
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==Solution==
 
==Solution==
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Define a super-special number to be a number whose decimal expansion only consists of <math>0</math>'s and <math>1</math>'s. The problem is equivalent to finding the number of super-special numbers necessary to add up to <math>\frac{1}{7}=0.142857142857\hdots</math>. This can be done in <math>8</math> numbers if we take
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<cmath>
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0.111111\hdots, 0.011111\hdots, 0.010111\hdots, 0.010111\hdots, 0.000111\hdots, 0.000101\hdots, 0.000101\hdots, 0.000100\hdots
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</cmath>
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Now assume for sake of contradiction that we can do this with strictly less than <math>8</math> super-special numbers (in particular, less than <math>10</math>.) Then the result of the addition won't have any carry over, so each digit is simply the number of super-special numbers which had a <math>1</math> in that place. This means that in order to obtain the <math>8</math> in <math>0.1428\hdots</math>, there must be <math>8</math> super-special numbers, so the answer is <math>\boxed{\textbf{(B)}\ 8}</math>.
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1997|num-b=28|num-a=30}}
 
{{AHSME box|year=1997|num-b=28|num-a=30}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:11, 9 August 2013

Problem

Call a positive real number special if it has a decimal representation that consists entirely of digits $0$ and $7$. For example, $\frac{700}{99}= 7.\overline{07}= 7.070707\cdots$ and $77.007$ are special numbers. What is the smallest $n$ such that $1$ can be written as a sum of $n$ special numbers?

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\\ \textbf{(E)}\ \text{The number 1 cannot be represented as a sum of finitely many special numbers.}$

Solution

Define a super-special number to be a number whose decimal expansion only consists of $0$'s and $1$'s. The problem is equivalent to finding the number of super-special numbers necessary to add up to $\frac{1}{7}=0.142857142857\hdots$. This can be done in $8$ numbers if we take \[0.111111\hdots, 0.011111\hdots, 0.010111\hdots, 0.010111\hdots, 0.000111\hdots, 0.000101\hdots, 0.000101\hdots, 0.000100\hdots\] Now assume for sake of contradiction that we can do this with strictly less than $8$ super-special numbers (in particular, less than $10$.) Then the result of the addition won't have any carry over, so each digit is simply the number of super-special numbers which had a $1$ in that place. This means that in order to obtain the $8$ in $0.1428\hdots$, there must be $8$ super-special numbers, so the answer is $\boxed{\textbf{(B)}\ 8}$.

See also

1997 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
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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