How many three-digit integers are multiples of both 4 and 6 and have a unit digit of 2?
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3-digit numbers range from 100 to 999 (900 numbers). 4 and 6 have an LCM of 12, so we need multiples of 12 that end in 2. The multipliers must end in 1 or 6. If we divide 12 into 100 and 999, we get about 9, and 83 (multiples are 108 and 996). So the multipliers ending in 1 or 6 are: 11, 16, ..., 81. Subtract 1 from each of these and divide by 5: 10/5=2, 15/5=3, ..., 80/5=16; so there are 15 multiples of 12 ending in 1 or 6. These multiples are 11×12=132, 16×12=192, 21×12=252, ..., 76×12=912, 81×12=972.

We can list them all and check that there are only 15:

132, 192, 252, 312, 372, 432, 492, 552, 612, 672, 732, 792, 852, 912, 972.

4 and 6 go into all these numbers, and they all have 2 in the ones (unit) place.

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