Difference between revisions of "2002 AMC 12A Problems/Problem 21"
(New page: == Problem == Consider the sequence of numbers: <math>4,7,1,8,9,7,6,\dots</math> For <math>n>2</math>, the <math>n</math>-th term of the sequence is the units digit of the sum of the two ...) |
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We have <math>\lfloor 10000/60 \rfloor = 166</math> and <math>166\cdot 12 = 1992</math>, therefore <math>S_{1992} = 60\cdot 166 = 9960</math>. | We have <math>\lfloor 10000/60 \rfloor = 166</math> and <math>166\cdot 12 = 1992</math>, therefore <math>S_{1992} = 60\cdot 166 = 9960</math>. | ||
− | The rest can now be computed by hand, we get <math>S_{1998} = 9960+4+7+1+8+9+7= 9996</math>, and <math>S_{1999}=9996 + 6 = 10002</math>, thus the answer is <math>\boxed{1999}</math>. | + | The rest can now be computed by hand, we get <math>S_{1998} = 9960+4+7+1+8+9+7= 9996</math>, and <math>S_{1999}=9996 + 6 = 10002</math>, thus the answer is <math>\boxed{(B)1999}</math>. |
== See Also == | == See Also == | ||
{{AMC12 box|year=2002|ab=A|num-b=20|num-a=22}} | {{AMC12 box|year=2002|ab=A|num-b=20|num-a=22}} | ||
+ | {{MAA Notice}} |
Latest revision as of 09:13, 4 July 2013
Problem
Consider the sequence of numbers: For , the -th term of the sequence is the units digit of the sum of the two previous terms. Let denote the sum of the first terms of this sequence. The smallest value of for which is:
Solution
The sequence is infinite. As there are only pairs of digits, sooner or later a pair of consecutive digits will occur for the second time. As each next digit only depends on the previous two, from this point on the sequence will be periodic.
(Additionally, as every two consecutive digits uniquely determine the previous one as well, the first pair of digits that will occur twice must be the first pair .)
Hence it is a good idea to find the period. Writing down more terms of the sequence, we get:
and we found the period. The length of the period is , and its sum is . Hence for each we have .
We have and , therefore . The rest can now be computed by hand, we get , and , thus the answer is .
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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 |
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