When a positive three-digit dividend with a units digit of 2 is divided by a positive one-digit divisor, the result is a wholenumber quotient with a remainder of 1. How many distinct values are possible for this three-digit dividend?
+18 Having a units digit of 2 and a remainder of 1 means the units digit of the quotient is 1. So the problem is asking "How many 3 digit integers have a units digit of 1 and are divisible by a number 2-9?"
Because 2, 4, 5, 6, 8 dont have any multiples that end in 1, we can exclude those
Because 9 is a multiple of 3, counting the 3's already includes 9's so we can exclude 9 too
To find the number of 3 digit integers that end in 1 and are divisible by 3 and/or 7, we can see that every 3 terms in the sequence 101, 111, 121, 131, ... is divisible by 3, so that is simply 90 (The number of terms in the sequence)/3 = 30.
To find the number of 3 digit integers that are divisible by 7, we can find that, extending the sequence to 1001, there are 91/7 = 13 terms, and subtracting the 1001 gives 12.
This logic counts integers that are divisible by 21 twice, so subtracting ⌊90/21⌋ = 4 would even it out.
Finally, the sum turns out to be
30(Multiples of 3) + 12 (Multiples of 7) - 4 (Multiples of 21) = 38 3 digit numbers
