# Problem of the Day

A woman who has two sons enrolled in an elementary school brings a batch of cookies to the school. Her youngest son is in a class with 10 other students and her oldest son is in a class with 7 other students. If she divides the cookies evenly among the students in her youngest son’s class there will be 5 left over. If she divides the cookies evenly among the students in her oldest son’s class there will be 7 left over. If she has more than 200 cookies, which of the following is the sum of the digits of the smallest possible value of ?

A. 9

B. 11

C. 13

D. 17

E. 23

### Solution

[latexpage]

We know that $n$ is 5 more than a multiple of 11 and 7 more than a multiple of 8. That is, $n=11a+5$ and $n=8b+7$ for some positive integers $a,b$. Start by making a list of numbers that satisfies the first equation and stop when you get to one that also satisfies the 2nd. To generate this list, start with the remainder, 5, and add the divisor, 11: $5,16,27,38,49,60,71$. 71 is the first number that is also 7 more than a multiple of 8. Thus, 71 is the smallest positive integer that satisfies both equations. We can obtain further solutions by adding the least common multiple of 11 and 8: LCM(11,8)=88. That gives us $71, 159, 247$. 247 is thus the smallest possible value of $n$ that is greater than 200. The sum of its digits is $2+4+7=13$.

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