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2022 AMC 12B Problem 6

Problem 6 of 25EasierNumber TheoryCounting & Probability

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?

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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 100x100 - x contain one, then 2x+(100x)=142,2x + (100 - x) = 142, so x=42.x = 42. Thus, the correct answer is B.

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Concepts: multiple · counting integers in a range

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.