Difference between revisions of "2014 AMC 12A Problems/Problem 23"
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==Problem== | ==Problem== | ||
− | |||
− | + | The fraction | |
− | |||
− | |||
− | |||
− | |||
− | = | + | <cmath>\dfrac1{99^2}=0.\overline{b_{n-1}b_{n-2}\ldots b_2b_1b_0},</cmath> |
− | the fraction <math>\dfrac{1}{99}</math> can be written as <cmath>\sum^{\infty}_{n=1}\dfrac{1}{10^{ | + | where <math>n</math> is the length of the period of the repeating decimal expansion. What is the sum <math>b_0+b_1+\cdots+b_{n-1}</math>? |
− | similarly the fraction <math>\dfrac{1}{99^2}</math> can be written as <math>\sum^{\infty}_{m=1}\dfrac{1}{10^{ | + | |
+ | <math>\textbf{(A) }874\qquad \textbf{(B) }883\qquad \textbf{(C) }887\qquad \textbf{(D) }891\qquad \textbf{(E) }892\qquad</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | the fraction <math>\dfrac{1}{99}</math> can be written as <cmath>\sum^{\infty}_{n=1}\dfrac{1}{10^{2n}}</cmath>. | ||
+ | similarly the fraction <math>\dfrac{1}{99^2}</math> can be written as <math>\sum^{\infty}_{m=1}\dfrac{1}{10^{2m}}\sum^{\infty}_{n=1}\dfrac{1}{10^{2n}}</math> which is equivalent to <cmath>\sum^{\infty}_{m=1}\sum^{\infty}_{n=1} \dfrac{1}{10^{2(m+n)}}</cmath> | ||
and we can see that for each <math>n+m=k</math> there are <math>k-1</math> <math>(n,m)</math> combinations so the above sum is equivalent to: | and we can see that for each <math>n+m=k</math> there are <math>k-1</math> <math>(n,m)</math> combinations so the above sum is equivalent to: | ||
− | <cmath>\sum^{\infty}_{k=2}\dfrac{k-1}{10^{ | + | <cmath>\sum^{\infty}_{k=2}\dfrac{k-1}{10^{2k}}</cmath> |
we note that the sequence starts repeating at <math>k = 102</math> | we note that the sequence starts repeating at <math>k = 102</math> | ||
− | yet consider <cmath>\sum^{101}_{k=99}\dfrac{k-1}{10^{ | + | yet consider <cmath>\sum^{101}_{k=99}\dfrac{k-1}{10^{2k}}=\dfrac{98}{{10^{198}}}+\dfrac{99}{{10^{200}}}+\dfrac{100}{10^{{202}}}=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{100}{10000})=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{1}{100})=\dfrac{1}{10^{198}}(98+\dfrac{100}{100})=\dfrac{1}{10^{198}}(99)</cmath> |
so the decimal will go from 1 to 99 skipping the number 98 | so the decimal will go from 1 to 99 skipping the number 98 | ||
− | and we can easily compute the sum of the digits from 0 to 99 to be <cmath>45 | + | and we can easily compute the sum of the digits from 0 to 99 to be <cmath>45\cdot10\cdot2=900</cmath> subtracting the sum of the digits of 98 which is 17 we get |
<cmath>900-17=883\textbf{(B) }\qquad</cmath> | <cmath>900-17=883\textbf{(B) }\qquad</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | <math>\frac{1}{99^2}\\\\ | ||
+ | =\frac{1}{99} \cdot \frac{1}{99}\\\\ | ||
+ | =\frac{0.\overline{01}}{99}\\\\ | ||
+ | =0.\overline{00010203...9799}</math> | ||
+ | |||
+ | So, the answer is <math>0+0+0+1+0+2+0+3+...+9+7+9+9=2\cdot10\cdot\frac{9\cdot10}{2}-(9+8)</math> or <math>\boxed{\textbf{(B)}\ 883}</math>. | ||
+ | |||
+ | There are two things to notice here. First, <math>\frac{1}{99}</math> has a very simple and unique decimal expansion, as shown. Second, for <math>\frac{0.\overline{01}}{99}</math> to itself produce a repeating decimal, <math>99</math> has to evenly divide a sufficiently extended number of the form <math>101010101..</math>. This number will have <math>99</math> ones (197 digits in total), as to be divisible by <math>9</math> and <math>11</math>. The enormity of this number forces us to look for a pattern, and so we divide out as shown. Indeed, upon division (seeing how the remainders always end in "501" or "601" or, at last, "9801"), we find the repeating part <math>.00010203040506070809101112131415....9799</math>. If we wanted to further check our pattern, we could count the total number of digits in our quotient (not counting the first three): 195. Since <math>99<100</math>, multiplying by it will produce either <math>1</math> or <math>2</math> extra digits, so our quotient passes the test. | ||
+ | |||
+ | Or note that the 1 has to be carried when you get to 100 so the 99 becomes 00 and the 1 gets carried again to 98 which becomes 99. | ||
+ | |||
+ | |||
+ | == Video Solution by Richard Rusczyk == | ||
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2014amc12a/382 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC12 box|year=2014|ab=A|num-b=22|num-a=24}} | ||
+ | {{MAA Notice}} |
Latest revision as of 09:35, 26 May 2023
Problem
The fraction
where is the length of the period of the repeating decimal expansion. What is the sum
?
Solution 1
the fraction can be written as
.
similarly the fraction
can be written as
which is equivalent to
and we can see that for each
there are
combinations so the above sum is equivalent to:
we note that the sequence starts repeating at
yet consider
so the decimal will go from 1 to 99 skipping the number 98
and we can easily compute the sum of the digits from 0 to 99 to be
subtracting the sum of the digits of 98 which is 17 we get
Solution 2
So, the answer is or
.
There are two things to notice here. First, has a very simple and unique decimal expansion, as shown. Second, for
to itself produce a repeating decimal,
has to evenly divide a sufficiently extended number of the form
. This number will have
ones (197 digits in total), as to be divisible by
and
. The enormity of this number forces us to look for a pattern, and so we divide out as shown. Indeed, upon division (seeing how the remainders always end in "501" or "601" or, at last, "9801"), we find the repeating part
. If we wanted to further check our pattern, we could count the total number of digits in our quotient (not counting the first three): 195. Since
, multiplying by it will produce either
or
extra digits, so our quotient passes the test.
Or note that the 1 has to be carried when you get to 100 so the 99 becomes 00 and the 1 gets carried again to 98 which becomes 99.
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2014amc12a/382
See Also
2014 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.