AMC 10 · 2004 · #4

Grade 6 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

divisibility-rulesprime-factorizationfactors extremal-constructionsystematic-enumeration ↑ Prerequisites: factors

📏 Medium solution 💡 2 insights

Problem

A standard six-sided die is rolled, and $P$ is the product of the five numbers that are visible. What is the largest number that is certain to divide $P$ ?

Pick an answer.

(A)

(B)

(C)

(D)

144

(E)

720

AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: A standard die (faces $1,2,3,4,5,6$) is rolled so exactly one face rests on the bottom and is hidden. $P$ is the product of the five visible faces. Find the largest number that divides $P$ no matter which face ends up hidden.

Givens: The six faces are the numbers $1,2,3,4,5,6$; Exactly one face is hidden on the bottom; the other five are visible; $P$ is the product of the five visible numbers; Answer choices: (A) $6$, (B) $12$, (C) $24$, (D) $144$, (E) $720$

Unknowns: The largest number guaranteed to divide $P$ for every possible hidden face

Understand

Plan

Primary tool: #14 Extreme Principle

Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities

"Certain to divide" is a guarantee: the divisor has to survive the worst possible roll, so this is an Extreme Principle (#14) question — for each prime factor, find the roll that leaves the fewest of them. Make a Systematic List (#2) turns the vague "any hidden face" into six concrete products so the worst case is visible, and Eliminate Possibilities (#3) matches the surviving number to a choice while ruling out the larger traps that only work sometimes.

Execute — Answer: B

#2 Make a Systematic List 5.NBT.B.5 Step 1

Fix the total product

All six faces multiply to a single fixed number: $1\cdot2\cdot3\cdot4\cdot5\cdot6=720$.
Hiding one face just removes that face from the product, so the five visible numbers multiply to $720$ divided by the hidden face.
That means $P$ is always $720$ split up by whichever number is on the bottom.

$$1\cdot2\cdot3\cdot4\cdot5\cdot6=720,\qquad P=\dfrac{720}{\text{hidden face}}$$

💡 The whole product never changes; hiding a face just divides $720$ by that one number.

#14 Extreme Principle 4.OA.B.4 Step 2

Turn "certain" into worst case

A number is "certain" to divide $P$ only if it divides $P$ for every hidden face.
So the answer is the largest number that fits inside the smallest amount of each prime factor $P$ can have.
Break $720$ into primes: $720=2^4\cdot3^2\cdot5$.
Now track, prime by prime, the fewest copies that can survive when a face is hidden.

$$720=2^4\cdot3^2\cdot5$$

💡 A guaranteed divisor must fit the poorest roll, so we hunt for the minimum of each prime.

#2 Make a Systematic List 4.OA.B.4 Step 3

List the six possible products

Write $P=720\div h$ for each hidden face $h$ and factor each into primes.
This makes the worst case for every prime visible at a glance.

$$\begin{aligned}h=1&:720=2^4\cdot3^2\cdot5\\ h=2&:360=2^3\cdot3^2\cdot5\\ h=3&:240=2^4\cdot3\cdot5\\ h=4&:180=2^2\cdot3^2\cdot5\\ h=5&:144=2^4\cdot3^2\\ h=6&:120=2^3\cdot3\cdot5\end{aligned}$$

💡 Seeing all six factorizations at once exposes the smallest supply of each prime.

#14 Extreme Principle 6.NS.B.4 Step 4

Take the smallest power of each prime

The guaranteed divisor is the greatest common divisor of all six products: for each prime, keep only as many copies as the poorest roll has.
The fewest $2$s is $2^2$ (when $4$ is hidden, the visible evens are just $2$ and $6$).
The fewest $3$s is $3^1$ (when $3$ or $6$ is hidden).
The fewest $5$s is none at all (when $5$ is hidden, $P=144$ has no factor of $5$).
Multiplying the survivors gives $2^2\cdot3=12$.

$$\gcd=2^{\min(4,3,4,2,4,3)}\cdot3^{\min(2,2,1,2,2,1)}\cdot5^{\min(1,1,1,1,0,1)}=2^2\cdot3^1\cdot5^0=12$$

💡 The greatest common divisor keeps each prime only as many times as the weakest case allows.

#3 Eliminate Possibilities 6.NS.B.4 Step 5

Match to a choice and rule out the traps

$12$ divides all six products, so it is certain.
The bigger choices each fail on some roll: $24=2^3\cdot3$ needs three $2$s but $180$ has only two; $144$ and $720$ both demand a factor of $5$ that vanishes when $5$ is hidden; $6$ is guaranteed but not the largest.
So the largest certain divisor is $12$, choice (B).

$$12\mid P\text{ for all }h,\quad 24\nmid180,\quad 5\nmid144\;\Rightarrow\;\textbf{(B)}$$

💡 Any divisor bigger than $12$ leans on a prime that some hidden face takes away.

[1] #2 5.NBT.B.5 All six faces multiply to a single fixed number: $1\cdot2\cdot3\cdot4\cdot5\cdot

[2] #14 4.OA.B.4 A number is "certain" to divide $P$ only if it divides $P$ for every hidden face

[3] #2 4.OA.B.4 Write $P=720\div h$ for each hidden face $h$ and factor each into primes. This m

[4] #14 6.NS.B.4 The guaranteed divisor is the greatest common divisor of all six products: for e

[5] #3 6.NS.B.4 $12$ divides all six products, so it is certain. The bigger choices each fail on

Review

Reasonableness: Check $12$ against the extremes of the list: $180$ (its smallest value) gives $180\div12=15$, and $144$ gives $144\div12=12$ — both whole, so $12$ really does divide every case. Pushing higher breaks: $24$ fails on $180$ (only two $2$s available) and anything with a factor of $5$ fails on $144$. So $12$ is exactly the ceiling, which lands on (B) and rules out the larger (C), (D), (E) while beating (A).

Alternative: Skip the full list and reason one prime at a time. Three of the faces ($2,4,6$) are even, so at least two even numbers are always visible, guaranteeing $2^2=4$; among $3$ and $6$ at least one is always visible, guaranteeing one factor of $3$; but the single face $5$ can be hidden, so no factor of $5$ is guaranteed. Multiplying the guarantees gives $4\cdot3=12$ directly.

CCSS standards used (min grade 6)

5.NBT.B.5 Fluently multiply multi-digit whole numbers (Computing the fixed total $1\cdot2\cdot3\cdot4\cdot5\cdot6=720$ and dividing it by each hidden face to get $P$.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Breaking $720$ and each of the six products into prime factors to see the supply of each prime.)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Taking the greatest common divisor of the six products by keeping the smallest power of each prime, giving $12$.)

⭐ "Certain to divide" means it has to survive the worst roll, so for each prime keep only as many as the poorest case leaves — here that is $2\cdot2\cdot3=12$.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Fix the total product

Turn "certain" into worst case

List the six possible products

Take the smallest power of each prime

Match to a choice and rule out the traps

Review

CCSS standards used (min grade 6)

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