AMC 10 · 2014 · #17
Grade 8 number-theoryPick an answer.
AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The numbers are astronomically large, so Tool #7 (Identify Subproblems) is the spine: first pull out the obvious shared power of 2, then the whole job shrinks to counting how many extra 2s hide inside a single leftover factor 5¹⁰⁰² - 1. Tool #5 (Look for a Pattern) reads the powers of 5 modulo 4 to see which pieces are divisible by 4 and which carry only one 2. Tool #9 (Solve an Easier Related Problem) replaces the giant 5¹⁰⁰²-1 with a tiny factored form (4 × odd) that is easy to count.
Write both terms as powers of 2
Since 10 = 2 · 5, 10¹⁰⁰² = 2¹⁰⁰² · 5¹⁰⁰²; since 4 = 2², 4⁵⁰¹ = (2²)⁵⁰¹. So both terms openly carry 2¹⁰⁰².
Putting both numbers over the same base 2 makes the shared factors of 2 visible instead of hidden.
8.EE.A.1Identify SubproblemsFactor out the shared 2¹002
Pull the shared factor out: 10¹⁰⁰² - 4⁵⁰¹ = 2¹⁰⁰²(5¹⁰⁰² - 1). As 5¹⁰⁰² is odd, the leftover is even — more 2s hide in it.
Pulling out the common factor turns one scary number into a known power of 2 times a smaller mystery.
6.EE.A.3Identify SubproblemsSplit with difference of squares
With 5¹⁰⁰² = (5⁵⁰¹)², the leftover splits: 5¹⁰⁰² - 1 = (5⁵⁰¹ - 1)(5⁵⁰¹ + 1) — two consecutive even numbers.
Two even numbers in a row always sit right next to each other, and only one of them can be a multiple of 4.
Splitting with a difference of squares gives two even numbers two apart, and only one can carry an extra factor of two.
▸ Why?
A difference of two squares is the two numbers added multiplied by the two subtracted.
▸ Why?
Two even numbers two apart cannot both be multiples of four, so one of them is only barely even.
Read the factors modulo 4
Since 5 ≡ 1 (mod 4), also 5⁵⁰¹ ≡ 1, so 5⁵⁰¹ + 1 ≡ 2 (mod 4) carries exactly one factor of 2, while 5⁵⁰¹ - 1 is a multiple of 4.
Every power of 5 is one more than a multiple of 4, so subtracting 1 lands on a multiple of 4 and adding 1 lands two past it.
4.OA.B.4Look For A PatternPin down 5⁵01 - 1 exactly
Factor 5⁵⁰¹ - 1 = 4 · S, where S sums 501 odd powers of 5 and is therefore odd: two 2s here, plus one before, gives 3 in all.
The leftover sum is odd, so the only 2s come from the clean factor of 4 out front.
8.EE.A.1Solve An Easier Related ProblemCombine the powers of 2
Stack the two counts: 2¹⁰⁰² · 2³ = 2¹⁰⁰⁵, and no larger power of 2 divides the number — choice (D).
Adding exponents when multiplying powers of the same base stacks all the 2s into one count.
8.EE.A.1Identify SubproblemsTo find the biggest power of 2 hiding in a difference, pull out the shared power first, then count the leftover 2s one factor at a time.
- Write both terms as powers of 2
- Factor out the shared 2¹002
- Split with difference of squares
- Read the factors modulo 4
- Pin down 5⁵01 - 1 exactly
- Combine the powers of 2