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Number theory Difficulty 3.1 AMC 10/12 Prove it Canada

IMG0 The positive divisors of 6 are 1, 2, 3,
and 6. What is the sum of the positive divisors of 64?
Fionn wrote 4 consecutive integers on a
whiteboard. Lexi came along and erased one of the integers. Fionn
noticed that the sum of the remaining integers was 847. What integer did
Lexi erase?
An arithmetic sequence with 7 terms has
first term $d^2$ and common difference $d$ . The sum of the 7 terms in the sequence is 756. Determine all possible values of $d$ . (An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant, called the common difference. For example, $3, 5, 7, 9$ are the first four terms of an
arithmetic sequence.)

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Solution

Since $64 = 2^6$ , its positive divisors are 1, 2, 4, 8, 16, 32, and 64. The sum of these divisors is $1 + 2 + 4 + 8 +
16 + 32 + 64 = 127. Suppose that the four consecutive integers that Fionn originally wrote on the blackboard were

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Figure for this problem x $,$ x+1 $,$ x+2 $, and$ x+3. When Lexi erases one of these integers, the sum of the remaining three integers is equal to one of the following:

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Figure for this problem $\text{(x+1) + (x+2) + (x+3) = 3x + 6 x + (x+2) + (x+3) = 3x + 5 x + (x+1) + (x+3) = 3x + 4 x + (x+1) + (x+2) = 3x + 3}$ We are told that the sum of these integers is 847. We note that

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Figure for this problem 847 = 3 $\cdot$ 282 + 1 $, which is one more than a multiple of 3. Since$ 3x+3 $and$ 3x+6 $are always multiples of 3 and$ 3x+5 is 2 more than a multiple of 3, then we must have

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Figure for this problem 3x + 4 = 847 $and so$ 3x = 843 $or$ x = 281. (Alternatively, we could have set each of the four sums above equal to 847 to determine in which case or cases we obtained an integer solution for

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Figure for this problem x.) Therefore, the original integers were 281, 282, 283, 284 and Lexi erased

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Figure for this problem x + 2 = 283. From the given information, the 7 terms in the arithmetic sequence are

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Figure for this problem $d^2,~~d^2 + d,~~d^2 + 2d,~~d^2 + 3d,~~d^2 + 4d,~~d^2 + 5d,~~d^2 + 6d$ Since the sum of these 7 terms is 756, we obtain the following equivalent equations:

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Figure for this problem $\text{d 2 + (d 2 + d) + (d 2 + 2d) + (d 2 + 3d) + (d 2 + 4d) + (d 2 + 5d) + (d 2 + 6d) = 756 7d 2 + 21d = 756 d 2 + 3d = 108 d 2 + 3d - 108 = 0 (d + 12)(d - 9) = 0}$ $and so$ d = -12 $or$ d = 9$.

The corresponding arithmetic sequences are $144, 132, 120, 108, 96, 84, 72 ~~\text{ and } ~~ 81, 90, 99, 108, 117, 126, 135$

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.