Discussion
Which is the least number that must be subtracted from 1936 so that the remainder when divided by 9, 10 and 15 will be 7 in each case?
Answer: B
First, find the smallest number that leaves a remainder of 7 when divided by 9, 10, and 15. This number is of the form \(k \times \text{LCM}(9,10,15) + 7\). LCM of 9, 10, 15 is 90. The numbers are 97, 187, 277, etc. We need to find a number of this form that is just less than 1936. Let's divide 1936 by 90. \(1936 = 21 \times 90 + 46\). The nearest multiple of 90 below 1936 is \(21 \times 90 = 1890\). The required number is \(1890+7=1897\). The number that must be subtracted from 1936 to get 1897 is \(1936 - 1897 = 39\).
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