2022 AMC 10B Problem 8

Problem 8 of 25EasierNumber Theory

Consider the following

100

sets of

10

elements each:

\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?

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Solution

A block of ten consecutive integers contains exactly two multiples of

7

precisely when its first multiple of

7

is in one of the first three positions. Thus that multiple must have units digit

1,

2,

3.

The multiples of

7

with those units digits are, respectively,

\begin{gathered}21+70j,\\42+70j,\\63+70j.\end{gathered}

For each expression,

j=0,1,\ldots,13

gives a value at most

1000,

so each class contributes

14

blocks. Therefore, the total is

3\cdot14=42.

Thus, the answer is B .