AMC 10 · 2007 · #14

Grade 7 number-theoryalgebra

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

prime-factorizationfactorssystematic-enumeration bound-inequality-then-enumeratecaseworkwork-backwards ↑ Prerequisites: prime-factorization

📏 Long solution 💡 3 insights

Problem

Five different integers give a product of 45 when each is subtracted from six. Find the sum of the five integers.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Extreme Principle

The question is about the letters, but everything the problem tells you is about the five factors, so Tool #4 (Introduce a Variable) renames them x_i=6-a_i and converts the target sum into a sum of factors. From there the whole problem is one counting question: how large can a factor afford to be? Tool #14 (Extreme Principle) is the spine — five factors must all be different, only two integers (1 and -1) have absolute value 1, and every other factor of 45 costs at least a factor of 3. Since 3⁴=81 already overshoots 45, the budget is what pins the list down. Tool #3 (Eliminate Possibilities) narrows the candidates to the twelve signed divisors of 45 and then settles the last sign; Tool #2 (Make a Systematic List) exhausts the few ways three numbers of size ≥ 3 can multiply to 45. Tool #11 (Work Backwards) does the step the fast solutions skip: turning the forced factor list back into actual values of a,…,e and checking they really are five distinct integers.

1STEP 1

Rename the five factors

Renaming turns the problem into five factors of one product.

x_i=6-a_i, Σ_i=1⁵x_i=30-(a+b+c+d+e), Π_i=1⁵x_i=45

2STEP 2

Every factor divides 45

Each factor divides the product, so the options are few.

45=3² · 5; x_i∈{± 1,± 3,± 5,± 9,± 15,± 45}

3STEP 3

Only two factors can be cheap

Without both size-one factors the product would be far too big.

at most one size-1 → |{Π} x_i| ≥ 3⁴=81 ＞ 45; hence {1,-1}⊆{x₁,…,x₅}

4STEP 4

List the ways the big three multiply

Only one way splits the rest into three factors.

45=3 · 3 · 5 is the only factorization into three factors each ≥ 3; so {x_i}⊇{1,-1,3,-3} and the fifth has size 5

5STEP 5

The sign of the last factor

The sign of the last factor is forced by the product.

1·(-1) · 3·(-3)=9, 9x₅=45→ x₅=5; {x_i}={-3,-1,1,3,5}, Σ x_i=5

6STEP 6

Turn the factors back into letters

Turning them back gives the sum 25, choice (C).

a_i=6-x_i: {9,7,5,3,1}; (-3)(-1)(1)(3)(5)=45; 9+7+5+3+1=25=30-5 → (C)

Answer

The final quintuple survives every condition in the problem read literally: 9,7,5,3,1 are integers, they are pairwise distinct, and (-3)(-1)(1)(3)(5)=45 exactly. The sum 25 is choice (C). The wrong choices are readable as specific slips, which is a good sign the reasoning is aimed at the right target. (A) 5 is the sum of the factors, -3-1+1+3+5, reported without the final step a+b+c+d+e=30-5. (B) 17 is what you get if you forget the word "distinct": the factors 1,1,3,3,5 also multiply to 45 and sum to 13, giving 30-13=17. (E) 30 is the belief that the factors cancel in pairs and sum to 0 — tempting because -3,-1,1,3 do cancel, but the leftover 5 breaks the symmetry. A last independent check on size: 30-Σ x_i must be 25 only if Σ x_i=5, and any legal factor list contains 1,-1,3,-3, so its sum is whatever the fifth factor is — 5. Consistent.

💡Key takeaway

When different numbers have to multiply to something small, count how many of them can be tiny: only two can be ± 1, every other one costs at least a factor of 3, and that budget alone can force the whole list.

Rename the five factors
Every factor divides 45
Only two factors can be cheap
List the ways the big three multiply
The sign of the last factor
Turn the factors back into letters