AMC 10 · 2010 · #24
Grade 8 number-theoryPick an answer.
AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #9 (Solve an Easier Related Problem): the full number 90! is impossible to compute, so replace it with a much smaller question about remainders. The last two nonzero digits are N mod 100, where N is 90! with its trailing zeros removed. Tool #7 (Identify Subproblems): 100 = 4 × 25 with 4 and 25 sharing no factors, so split the one hard remainder into an easy N mod 4 and a harder N mod 25, then reassemble. Tool #5 (Look for a Pattern): mod 25, each clean block of 25 consecutive numbers leaves the same footprint, and the powers of 2 repeat in a short cycle — patterns that collapse the huge product to a few small steps.
Count and strip the trailing zeros
Factors of 5 are scarcer than 2s, so the zero count is ⌊90/5⌋+⌊90/25⌋ = 21. Strip them off: N = 90!/10²¹, and we want N mod 100.
Every trailing zero is one 2 paired with one 5, and the 5s are the scarce partner, so counting 5s counts the zeros.
Every trailing zero is one two paired with one five, and the fives are the scarce partner.
▸ Why?
Every number has exactly one prime recipe, so the twos and fives can be counted separately.
▸ Why?
Ending in zeros means being a multiple of ten that many times, which the last places record.
Split the target into a 4-clock and a 25-clock
Since 100 = 4 × 25 with no shared factor, do mod 4 and mod 25 apart. 90! has 86 twos and only 21 leave with the zeros, so N ≡ 0 (mod 4).
Two coprime clocks — a 4-clock and a 25-clock — together give every value 0 to 99 a unique reading, so their two answers rebuild the last two digits.
6.NS.B.4Identify SubproblemsStrip the fives block by block, mod 25
In each run of 25, the numbers coprime to 5 multiply to -1 (mod 25). There are four such runs here, so together they give (-1)⁴ ≡ 1.
Each clean block of 25 leaves the identical footprint -1 mod 25, so you only have to count how many blocks there are.
4.OA.C.5Look For A PatternPeel the fives off the multiples of 5
Peel one 5 off each of 5 · 1…5 · 18, then one more off 5,10,15 inside 18!. What survives is 4 and 6, so A ≡ 4 · 6 = 24 ≡ -1 (mod 25).
Peeling one 5 off every multiple of 5 turns the scary product into a small factorial you can finish by hand.
4.OA.B.4Identify SubproblemsDivide out the 2s using cycling powers
N = A/2²¹ and 2¹⁰ ≡ -1 makes 2²¹ ≡ 2 (mod 25), whose inverse is 13, so N ≡ (-1) · 13 ≡ 12 (mod 25).
Powers of 2 loop every 20 steps mod 25, so a giant exponent collapses to a tiny one you can read off.
8.EE.A.1Look For A PatternGlue the two clocks together
Below 100 only one value is both ≡ 0 (mod 4) and ≡ 12 (mod 25), so n = 12, choice (A).
Among 0–99, only 12 reads 0 on the 4-clock and 12 on the 25-clock at the same time.
6.NS.B.4Identify SubproblemsTo find the last nonzero digits of a huge factorial, strip the zeros, then track the number on a 4-clock and a 25-clock and glue the two readings back together.
- Count and strip the trailing zeros
- Split the target into a 4-clock and a 25-clock
- Strip the fives block by block, mod 25
- Peel the fives off the multiples of 5
- Divide out the 2s using cycling powers
- Glue the two clocks together