2022 AMC 10B Problem 8

Attempt Problem 8 of the 2022 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

8.

Consider the following 100100 sets of 1010 elements each: {1,2,3,…,10},{11,12,13,…,20},{21,22,23,…,30},⋮{991,992,993,…,1000}.\begin{gathered} \{1,2,3,\ldots,10\},\\ \{11,12,13,\ldots,20\},\\ \{21,22,23,\ldots,30\},\\ \vdots\\ \{991,992,993,\ldots,1000\}. \end{gathered} How many of these sets contain exactly two multiples of 7?7?
 40 \ 40
 42 \ 42
 43 \ 43
 49 \ 49
 50 \ 50
Answer: B
Concepts:multiplemodular arithmeticcasework
Small hint:
A block has two multiples exactly when its first multiple of 77 ends in 1,1, 2,2, or 33
Big hint:
Track the units digits of the multiples of 77
Solution:
A block of ten consecutive integers contains exactly two multiples of 77 precisely when its first multiple of 77 is in one of the first three positions. Thus that multiple must have units digit 1,1, 2,2, or 3.3.
The multiples of 77 with those units digits are, respectively, 21+70j,42+70j,63+70j.\begin{gathered}21+70j,\\42+70j,\\63+70j.\end{gathered} For each expression, j=0,1,…,13j=0,1,\ldots,13 gives a value at most 1000,1000, so each class contributes 1414 blocks. Therefore, the total is 3⋅14=42.3\cdot14=42.
Thus, the answer is B .
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