AMC 10 · 2018 · #14

Grade 8 arithmetic
exponentsweighted-averagefloor-functionfraction-arithmetic bound-inequality-then-enumerateestimation ↑ Prerequisites: exponents
📏 Medium solution 💡 2 insights
Problem
Find the greatest integer that is less than or equal to the fraction (3¹⁰⁰+2¹⁰⁰)/(3⁹⁶+2⁹⁶). In other words, compute ⌊ (3¹⁰⁰+2¹⁰⁰)/(3⁹⁶+2⁹⁶) ⌋, the floor of that value.

Pick an answer.

(A)
80
(B)
81
(C)
96
(D)
97
(E)
625

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

Tool #4 (Introduce a Variable): the exponents only differ by 4, so pulling out the common powers and naming a=3⁹⁶ and b=2⁹⁶ collapses the monster fraction into the tidy expression (81a+16b)/(a+b) that I can actually manipulate. Tool #9 (Solve an Easier Related Problem): I never need the true gigantic value — I only need to trap it between two integers, so I replace exact arithmetic with a simple bound on one small leftover term. Tool #3 (Eliminate Possibilities): once I show the value sits strictly between 80 and 81, four of the five choices are impossible and only one survives.

1STEP 1

Pull out the common powers

Peel off 3⁹⁶ and 2⁹⁶ and name a=3⁹⁶, b=2⁹⁶; since 3¹⁰⁰=81· 3⁹⁶ and 2¹⁰⁰=16· 2⁹⁶, the fraction becomes (81a+16b)/(a+b).

(3¹⁰⁰+2¹⁰⁰)/(3⁹⁶+2⁹⁶)=(81· 3⁹⁶+16· 2⁹⁶)/(3⁹⁶+2⁹⁶)=(81a+16b)/(a+b), a=3⁹⁶, b=2⁹⁶
2STEP 2

Rewrite to expose 81

Compare the top to 81(a+b): it falls short by 65b, so 81a+16b=81(a+b)-65b and the fraction splits into 81-65b/(a+b).

81a+16b=81(a+b)-65b → (81a+16b)/(a+b)=81-65b/(a+b)
3STEP 3

The value is below 81

Both a and b are positive, so 65b/(a+b) is a positive amount subtracted from 81, forcing the value strictly below 81.

65b/(a+b) > 0 → 81-65b/(a+b) < 81
4STEP 4

The leftover is tiny

Since a+b > a, 65b/(a+b) < 65(2/3)⁹⁶, and 3⁹⁶ dwarfs 65· 2⁹⁶ so this is less than 1, keeping the value above 80.

65b/(a+b) < 65b/a=65(2/3)⁹⁶ < 1 → 81-65b/(a+b) > 80
5STEP 5

Trap between 80 and 81

The value is under 81 but over 80, so it sits between consecutive integers, making its floor 80 — the answer is (A).

80 < (3¹⁰⁰+2¹⁰⁰)/(3⁹⁶+2⁹⁶) < 81 → ⌊ · ⌋=80 → (A)
Answer
80
The value is a weighted average of 81 (weight a=3⁹⁶) and 16 (weight b=2⁹⁶), because (81a+16b)/(a+b) mixes 81 and 16. Since 3⁹⁶ utterly dwarfs 2⁹⁶, the average is pulled almost all the way to 81, landing just a hair under it — exactly matching 'between 80 and 81.' The big choices make no sense: 96, 97, and especially 625 exceed even the larger ingredient 81, which a true average can never do, and 81 itself is ruled out because a positive amount was subtracted. Only (A) 80 fits.
💡Key takeaway

Pull out the shared powers so the fraction becomes 81 minus a tiny leftover; the leftover is less than 1, so the value is squeezed between 80 and 81 and its floor is 80.

  • Pull out the common powers
  • Rewrite to expose 81
  • The value is below 81
  • The leftover is tiny
  • Trap between 80 and 81