AMC 10 · 2013 · #4

Grade 8 algebra
0
Problem
A fraction has the same two huge powers on top and bottom, added and subtracted. Find its value.

Pick an answer.

(A)
-1
(B)
1
(C)
$\frac{5}{3}$
(D)
2013
(E)
$2^{4024}$
How to solve
Strategy Solve an Easier Related Problem

The exponents 2014 and 2012 look like the difficulty, but they are decoys: the only fact used anywhere is that they differ by 2. That points at Tool #9 (Solve an Easier Related Problem) — replace the unwritable powers by the same fraction built from one small block. Tool #4 (Introduce a Variable) does the replacing: name the shared power x=2²⁰¹², so the top and bottom become 4x+x and 4x-x. The move the whole problem turns on is cancelling the common factor x, and that step is legal only because x ≠ 0, so it deserves a check that does not use it at all: Tool #11 (Work Backwards) treats the value itself as the unknown t and recovers it from a linear equation, never factoring anything. Tool #3 (Eliminate Possibilities) then bounds the value between 1 and 2, which by itself leaves only one choice standing.

1STEP 1

The exponents differ by two

The exponents differ by only two.

2²⁰¹⁴=2²⁰¹²⁺²=2² · 2²⁰¹²=4 · 2²⁰¹²
2STEP 2

Name the shared power x

One letter names the shared power.

x=2²⁰¹² ＞ 0, (2²⁰¹⁴+2²⁰¹²)/(2²⁰¹⁴-2²⁰¹²)=(4x+x)/(4x-x)
3STEP 3

Factor out x and cancel it

Factoring it out lets it cancel.

(4x+x)/(4x-x)=5x/3x=5/3
4STEP 4

Confirm without cancelling anything

The value is 5/3, choice (B).

a/b=(t+1)/(t-1) and a/b=2²⁰¹⁴⁻²⁰¹²=4 → t+1=4t-4 → t=5/3
Answer
5/3
The answer choices can be narrowed by size alone, which is an independent confirmation rather than a feeling. With b=2²⁰¹² ＞ 0 the numerator is 5b and the denominator is 3b, both positive, so the value is positive and (A) -1 is out. Since 5b ＞ 3b, the value exceeds 1, so (B) 1 is out. Since 5b ＜ 6b=2 · 3b, the value is less than 2, so (D) 2013 and (E) 2⁴⁰²⁴ are out — both are far above 2. Exactly one choice, 5/3, lies strictly between 1 and 2, so the bounds alone force (C), and 5/3≈ 1.67 indeed sits there. A numerical spot check agrees: with exponents 4 and 2 in place of 2014 and 2012, (16+4)/(16-4)=20/12=5/3. That small case is a confirmation of the identity already proved, not the reason for believing it — the proof works for any exponents two apart. The bait choices are explained too: (E) is what wrongly adding the exponents would give, and (D) is just the year.
💡Key takeaway

When powers of the same base are added or subtracted, pull the smaller power out as a common factor: it cancels, and all that is left is the gap between the exponents.

  • The exponents differ by two
  • Name the shared power x
  • Factor out x and cancel it
  • Confirm without cancelling anything