Competition · AMC preparation · step 4 of 4
AMC 8 · 2001 · #18
Grade 7 probabilitycountingPick an answer.
AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
First reduce 'product is a multiple of 5' to the cleaner event 'at least one die shows a 5' — true because 5 is prime and 5 is the only multiple of 5 on a die. The phrase 'at least one' is the classic flag for Tool #16 (Count the Complement): instead of summing 'exactly one 5' and 'exactly two 5s' separately, count the easier opposite event 'neither die shows a 5' and subtract from 1. Tool #2 (Make a Systematic List) supplies the 6 × 6 = 36 equally likely ordered outcomes that anchor the probability fractions.
Rewrite the event
Because 5 is prime and 5 is the only multiple of 5 on a die, the event becomes 'at least one die shows a 5'.
Replacing a hard event with a simpler equivalent event is the Grade 6 'factor / prime' move applied to probability.
The product of the two dice is a multiple of 5 exactly when at least one die shows a 5.
▸ Why?
'Exactly when' means the two things happen in the very same rolls, so we check both directions: a rolled 5 forces the product to be a multiple of 5, and a product that is a multiple of 5 forces one of the dice to have shown a 5.
▸ Why?
If one die shows a 5, the product is 5 multiplied by the other face, which is that many 5s added together — and a whole number of 5s added up is exactly what 'multiple of 5' means.
▸ Why?
If the product is a multiple of 5, then 5 divides the product, so 5 must divide one of the two faces, and on a die the only face 5 divides is 5 itself.
▸ Why?
Because 5 is prime, a 5 in the product's prime factorization has to come from the prime factorization of one of the two faces — multiplying cannot invent a prime factor that neither face already had.
▸ Why?
Each whole number breaks into primes in only one way, and the product's single prime breakdown is just the two faces' prime factors pooled together, so every prime in the product — including 5 — already sits inside one of the faces.
▸ Why?
Checking the faces 1,2,3,4,5,6, only 5 divides by 5 with nothing left over; 1,2,3,4 are too small and 6 leaves a remainder of 1.
Count all outcomes
The two dice are independent, so the equally likely ordered pairs fill a 6 × 6 = 36 grid.
Listing the ordered pairs (a,b) in a 6 × 6 grid is the standard Grade 7 set-up for two independent trials.
7.SP.C.8Make A Systematic ListFind the chance of no 5
Count the easy opposite instead: neither die shows a 5, with probability · = .
'At least one' is almost always easier as 1 - P(none) — that is exactly Tool #16's complement trick.
7.SP.C.8Change Focus Count The ComplementSubtract from 1
Subtract from 1: the probability is 1 - = .
Every outcome either has a 5 or has no 5, so the two probabilities add to 1.
7.SP.C.7Change Focus Count The ComplementA product is a multiple of 5 only when at least one die rolls a 5. Count the easy opposite — neither die is a 5, · = — and the answer is 1 - = .
- Rewrite the event
- Count all outcomes
- Find the chance of no 5
- Subtract from 1
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