Difference between revisions of "2018 AMC 10B Problems/Problem 13"
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\textbf{(E) }1009 \qquad | \textbf{(E) }1009 \qquad | ||
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+ | == Video Solution == | ||
+ | https://youtu.be/zfChnbMGLVQ?t=5122 | ||
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+ | ~ pi_is_3.14 | ||
== Solution 1 == | == Solution 1 == |
Latest revision as of 01:29, 17 January 2021
Problem
How many of the first numbers in the sequence are divisible by ?
Video Solution
https://youtu.be/zfChnbMGLVQ?t=5122
~ pi_is_3.14
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.
Minor fixes by fasterthanlight
More minor fixes by BakedPotato66
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. Looking at our pattern of four numbers from above, the first number is divisible by . This means that the first of the left over will be divisible by , 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 |
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