Multiple choice

Find the smallest number which leaves the remainders 13, 41 and 29 at the end when divided by 20, 48 and 36 respectively?

  1. 187

  2. 713

  3. 720

  4. 725

  5. 727

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The number N satisfies N = 20k1 + 13, N = 48k2 + 41, N = 36k3 + 29. Note that the difference between divisor and remainder is 7 in all cases (20-13=7, 48-41=7, 36-29=7). Thus, N = LCM(20, 48, 36) - 7. LCM(20, 48, 36) = 720. N = 720 - 7 = 713.