Algebra
posted by Anonymous .
a and b are digits such that \overline{120ab} is a multiple of 3, 5 and 7. What is the value of \overline{ab} ?

ab = 75
Archaic solution: only possibilities for ab are 01 through 99. Eliminate all of those not divisible by 5, which leaves: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55. 60, 65, 70, 75, 80, 85, 90, and 95. Eliminate those not divisible by 3 (divisibility rule for 3) , which leaves 15, 30, 45, 60, 75, and 90. Now eliminate those not divisible by 7, which leaves ab = 75. Divisibility rule for 7: double the last digit and subtract from the remaining digits. If the ramaining digits are divisible by 7, then the original number is divisible by 7. So, our number is 12075. 5 + 5 =10. Subtracting: 1207  10 = 1197. 1197mod7 has no remainder, thus our solution.
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