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Problem 638

AMC 10/12, early questions

Number theory Difficulty 3.8 Multiple choice CEMC Pascal · Canada · 2016

There are $n$ students in the math club at Scoins Secondary School. When Mrs. Fryer tries to put the $n$ students in groups of 4, there is one group with fewer than 4 students, but all of the other groups are complete. When she tries to put the $n$ students in groups of 3, there are 3 more complete groups than there were with groups of 4, and there is again exactly one group that is not complete. When she tries to put the $n$ students in groups of 2, there are 5 more complete groups than there were with groups of 3, and there is again exactly one group that is not complete. The sum of the digits of the integer equal to $n^2-n$ is

Pick one

Official solutions — 2

Solution 1

Suppose that, when the $n$ students are put in groups of 2, there are $g$ complete groups and 1 incomplete group.

Since the students are being put in groups of 2, an incomplete group must have exactly 1 student in it.

Therefore, $n = 2g+1$ .

Since the number of complete groups of 2 is 5 more than the number of complete groups of 3, then there were $g-5$ complete groups of 3.

Since there was still an incomplete group, this incomplete group must have had exactly 1 or 2 students in it.

Therefore, $n = 3(g-5)+1$ or $n = 3(g-5)+2$ .

If $n=2g+1$ and $n = 3(g-5) + 1$ , then $2g+1 = 3(g-5)+1$ or $2g + 1 = 3g - 14$ and so $g = 15$ .

In this case, $n = 2g+1=31$ and there were 15 complete groups of 2 and 10 complete groups of 3.

If $n=2g+1$ and $n = 3(g-5) + 2$ , then $2g+1 = 3(g-5)+2$ or $2g + 1 = 3g - 13$ and so $g = 14$ .

In this case, $n = 2g+1=29$ and there were 14 complete groups of 2 and 9 complete groups of 3.

If $n=31$ , dividing the students into groups of 4 would give 7 complete groups of 4 and 1 incomplete group.

If $n=29$ , dividing the students into groups of 4 would give 7 complete groups of 4 and 1 incomplete group.

Since the difference between the number of complete groups of 3 and the number of complete groups of 4 is given to be 3, then it must be the case that $n = 31$ .

In this case, $n^2 - n = 31^2 - 31 = 930$ ; the sum of the digits of $n^2-n$ is 12.

Solution 2

Since the $n$ students cannot be divided exactly into groups of 2, 3 or 4, then $n$ is not a multiple of 2, 3 or 4.

The first few integers larger than 1 that are not divisible by 2, 3 or 4 are 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, and 35.

In each case, we determine the number of complete groups of each size:

$n$
5
7
11
13
17
19
23
25
29
31
35

# of complete groups of 2
2
3
5
6
8
9
11
12
14
15
17

# of complete groups of 3
1
2
3
4
5
6
7
8
9
10
11

# of complete groups of 4
1
1
2
3
4
4
5
6
7
7
8

Since the number of complete groups of 2 is 5 more than the number of complete groups of 3 which is 3 more than the number of complete groups of 4, then of these possibilities, $n=31$ works.

In this case, $n^2 - n = 31^2 - 31 = 930$ ; the sum of the digits of $n^2-n$ is 12.

(Since the problem is a multiple choice problem and we have found a value of $n$ that satisfies the given conditions and for which an answer is present, then this answer must be correct. Solution 1 shows why $n=31$ is the only value of $n$ that satisfies the given conditions.)