Discussion
Find the least number which when divided by 12, 15 and 20 leaves a remainder of 5 in each case.
Answer: A
Correct answer: 65
A number leaving the same remainder 5 on division by each divisor is 5 more than a common multiple of them all. The smallest such multiple is the LCM: \( 12 = 2^{2} \times 3 \), \( 15 = 3 \times 5 \), \( 20 = 2^{2} \times 5 \), so \[ \text{LCM} = 2^{2} \times 3 \times 5 = 60. \] The required number is therefore \[ 60 + 5 = 65. \] Checking: 65 divided by 12 gives remainder 5, by 15 gives 5, and by 20 gives 5. The general form is \( \text{LCM} \times n + r \), so the next such numbers are 125, 185 and so on. A variation asks for a number leaving different remainders whose difference from the divisor is constant - say remainders 7, 10 and 15 for divisors 12, 15 and 20, each 5 short. The answer is then \( \text{LCM} - 5 = 55 \), and recognising which form is being asked is the whole difficulty.
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