Difference between revisions of "1995 AJHSME Problems/Problem 15"

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<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8</math>
 
<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8</math>
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==Solution==
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<math>\frac{4}{37}=\frac{12}{111}=\frac{108}{999}=0.108108108...</math>
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Since this repeats every three digits, digit number x = digit number (x mod 3), and the 100th digit = (100 mod 3)th digit = 1st digit = <math>\boxed{\text{(B)}\ 1}</math>
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==See Also==
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{{AJHSME box|year=1995|num-b=14|num-a=16}}
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{{MAA Notice}}

Latest revision as of 00:23, 5 July 2013

Problem

What is the $100^\text{th}$ digit to the right of the decimal point in the decimal form of $4/37$?

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$

Solution

$\frac{4}{37}=\frac{12}{111}=\frac{108}{999}=0.108108108...$

Since this repeats every three digits, digit number x = digit number (x mod 3), and the 100th digit = (100 mod 3)th digit = 1st digit = $\boxed{\text{(B)}\ 1}$


See Also

1995 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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 AJHSME/AMC 8 Problems and Solutions

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