Difference between revisions of "2018 AMC 10B Problems/Problem 13"
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==Solution 3== | ==Solution 3== | ||
− | If we divide each number by <math>101</math>, we see a pattern occuring in every 4 numbers. <math>101, 1000001, 10000000001, \dots</math>. We divide <math>2018</math> by <math>4</math> to get <math>504</math> with <math>2</math> left over. One divisible number will be in the <math>2</math> left over, so | + | If we divide each number by <math>101</math>, we see a pattern occuring in every 4 numbers. <math>101, 1000001, 10000000001, \dots</math>. We divide <math>2018</math> by <math>4</math> to get <math>504</math> with <math>2</math> left over. One divisible number will be in the <math>2</math> left over, so our answer is <math>\boxed{\textbf{(C) } 505}</math>. |
==Solution 4== | ==Solution 4== |
Revision as of 15:44, 9 February 2019
Problem
How many of the first numbers in the sequence are divisible by ?
Solution 1
The number is divisible by 101 if and only if . We note that , so the powers of 10 are 4-periodic mod 101. It follows that if and only if .
In the given list, , the desired exponents are , and there are numbers in that list.
Solution 2
Note that for some odd will suffice . Each , so the answer is (AOPS12142015)
Solution 3
If we divide each number by , we see a pattern occuring in every 4 numbers. . We divide by to get with left over. One divisible number will be in the left over, so our answer is .
Solution 4
Note that is divisible by , and thus is too. We know that is divisible and isn't so let us start from . We subtract to get 2. Likewise from we subtract, but we instead subtract times or to get . We do it again and multiply the 9's by to get . Following the same knowledge, we can use mod to finish the problem. The sequence will just be subtracting 1, multiplying by 10, then adding 1. Thus the sequence is . Thus it repeats every four. Consider the sequence after the 1st term and we have 2017 numbers. Divide by four to get remainder . Thus the answer is plus the 1st term or .
-googleghosh
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.