Question

Consider the following $100$ sets of $10$ elements each: $\begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*}$ How many of these sets contain exactly two multiples of $7?$

$\ 40$

$\ 42$

$\ 43$

$\ 49$

$\ 50$

Solution:

We can analyze the units digit of the first multiple of $7$ in each set.

If the last digit is $4,5,6,7,$ then adding $7$ yields numbers outside the set, so the sets with a multiple of $7$ such that its units digit is $4,5,6,7$ would have only one multiple.

If the last digit is $1,2,3$ then adding $7$ would yield a units digit of $8,9,0$ in the same set.

Out of the first $98$ sets, there are an equal number of occurrences of sets such that the first multiple of $7$ has each of the units digit of $1,2,3,4,5,6,7$ since they cycle every $7$ sets. These yield $42$ sets whose first multiple of $7$ contains two multiples of $7.$

The $99$ th and $100$ th set wouldn't work since they yield a set whose first multiples are $987$ and $994$ respectively, which can't have $2$ multiples of $7.$ This means we have $42$ sets that work.

Thus, the answer is B .

2022 AMC 10B Problem 8

Solution:

Problem 8 in Other Years