2018 AMC 10B Problems/Problem 13
How many of the first numbers in the sequence are divisible by ?
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.
Note that for some odd will suffice . Each , so the answer is
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 .
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 .
Note that and , where . We have that must have a remainder of . By the remainder theorem, the roots of must also be roots of . Plugging in to yields that . Because the sequence starts with , the answer is
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