2018 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 13 Medium

How many of the first 2018 numbers in the sequence 101, 1001, 10001, 100001, \cdots are divisible by 101? (2018 AMC 10B Problem, Question#13)

  • A.

    253

  • B.

    504

  • C.

    505

  • D.

    506

  • E.

    1009

Answer:C

The number 10^n+1 is divisible by 101 if and only if 10^n\equiv -1 (mod 101). We note that (10, 10^2, 10^3, 10^4)\equiv(10,-1, -10, 1) (mod 101), so the powers of 10 are 4-periodic mod 101. It follows that 10^n\equiv -1 (mod 101) if and only if  n\equiv 2 (mod 4). In the given list, 10^2+1, 10^3+1, 10^4+1, \cdots, 10^{2019}+1, the desired exponents are 2, 6, 10, \cdots, 2018, and there are \frac{2020}{4}=\rm(C)~505 numbers in that list.

Note that 10^{2k}+1 for some odd k will suffice mod 101 . Each 2k\in\left\{2, 6,10,\cdots,2018\right\}, so the answer is \rm (C)~505.

If we divide each number by 101, we see a pattern occuring in every 4 numbers. 101, 1000001, 10000000001, \cdots. We divide 2018 by 4 to get 504 with 2 left over. Looking at our pattern of four numbers from above, the first number is divisible by 101. This means that the first of the 2 left over will be divisible by 101, so our answer is \rm(C)~505.

Note that 909 is divisible by 101, and thus 9999 is too. We know that 101 is divisible and 1001 isn't so let us start from 10001. We subtract 9999 to get 2. Likewise from 100001 we subtract, but we instead subtract 9999 times 10 or 99990 to get 11. We do it again and multiply the \rm 9's by 10 to get 101. Following the same knowledge, we can use mod 101 to finish the problem. The sequence will just be subtracting 1, multiplying by 10, then adding 1. Thus the sequence is 0, -9, -99(2), 11, 0, \cdots. Thus it repeats every four. Consider the sequence after the \rm 1st term and we have 2017 numbers. Divide 2017 by four to get 504 remainder 1. Thus the answer is 504 plus the \rm 1st term or \rm (C)~505.

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