AMC 10 · 2017 · #5

Grade 7 arithmetic
fraction-arithmeticratio-proportion symmetric-polynomialsconvert-to-algebra ↑ Prerequisites: fraction-arithmetic
📏 Short solution 💡 2 insights
Problem
Two nonzero real numbers have a sum that equals 4 times their product. Find the sum of the reciprocals of the two numbers.

Pick an answer.

(A)
1
(B)
2
(C)
4
(D)
8
(E)
12

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The two numbers are never told to us, so Tool #4 (Introduce a Variable) lets us name them a and b and write the single given fact as an equation. Tool #13 (Convert to Algebra) then turns the target — the sum of reciprocals — into one fraction (a+b)/ab, which lines up perfectly with the given a+b=4ab. Tool #6 (Guess and Check) confirms the result on a concrete pair of numbers.

1STEP 1

Name the numbers, write the fact

Name the numbers a and b; "their sum is 4 times their product" becomes the single equation a+b=4ab.

a + b = 4ab
2STEP 2

Write the target as one fraction

Add the reciprocal sum 1a\frac{1}{a}+1b\frac{1}{b} over the common denominator ab into the single fraction a+bab\frac{a+b}{ab}.

1a\frac{1}{a}+1b\frac{1}{b}=bab\frac{b}{ab}+aab\frac{a}{ab}=a+bab\frac{a+b}{ab}
3STEP 3

Substitute and simplify

Since a+b=4ab, the numerator of a+bab\frac{a+b}{ab} is 4ab; the nonzero ab cancels to leave 4 — the answer is (C).

a+bab\frac{a+b}{ab}=4abab\frac{4ab}{ab}=4 → (C)
Answer
4
Test a concrete pair. Take a=1; then 1+b=4b gives b=13\frac{1}{3}, and indeed the product 13\frac{1}{3} times 4 is 43\frac{4}{3}, the same as the sum 1+13\frac{1}{3}. The reciprocals are 1 and 3, summing to 4 — matching (C). The pair a=b=12\frac{1}{2} also works and gives 2+2=4.
💡Key takeaway

The sum of two reciprocals is always the sum over the product, so once the sum equals 4 times the product, the reciprocals add to 4.

  • Name the numbers, write the fact
  • Write the target as one fraction
  • Substitute and simplify