The number of two-digit positive integers which have at least one 7 as a digit is

Solution

The numbers 70 to 79 have 7 in the tens place, and that is 10 numbers. The two-digit numbers ending in 7 are 17, 27, 37, 47, 57, 67, 87 and 97, which is 8 more, because 77 has already been counted. So the total is $$10 + 8 = 18$$.

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