2022 AMC 12B Problem 6

Attempt Problem 6 of the 2022 AMC 12B 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 12B solutions, or check the answer key.

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

6.

Consider the following 100100 sets of 1010 elements each:
{1,2,3,…,10},\{1,2,3,\ldots,10\}, {11,12,13,…,20},\{11,12,13,\ldots,20\}, {21,22,23,…,30},\{21,22,23,\ldots,30\}, ⋮\vdots {991,992,993,…,1000}.\{991,992,993,\ldots,1000\}.
How many of these sets contain exactly two multiples of 7?7?
4040
4242
4343
4949
5050
Answer: B
Concepts:multiplecounting integers in a range
Small hint:
A block of 1010 consecutive integers contains either one or two multiples of 77
Big hint:
Count the total multiples of 77 in {1,…,1000},\{1,\ldots,1000\}, then let xx blocks have two and 100−x100-x have one
Solution:
Among 11 to 10001000 there are ⌊10007⌋=142\left\lfloor \tfrac{1000}{7} \right\rfloor = 142 multiples of 7.7. Because 10>7,10 \gt 7, each block of 1010 consecutive integers contains one or two multiples of 7.7.
If xx blocks contain two and the remaining 100−x100 - x contain one, then 2x+(100−x)=142,2x + (100 - x) = 142, so x=42.x = 42.
Thus, the correct answer is B.
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