AMC 10 · 2010 · #9

Grade 8 number-theory

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

prime-factorizationexponentsperfect-squareslcm convert-to-algebraextreme-principle ↑ Prerequisites: prime-factorization

📏 Long solution 💡 2 insights

Problem

A number is a multiple of twenty, its square is a cube, and its cube is a square. Count the digits of the smallest one.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Organize Information in More Ways

The three conditions talk about three different numbers (n, n², n³), so they look unrelated. Rewriting n by its prime factorization re-organizes all three into statements about one list of exponents, because squaring and cubing only multiply those exponents. Naming the exponents turns 'is a cube' and 'is a square' into plain divisibility rules. After that the smallest n is the one whose exponents are each pushed down to the smallest legal value, and a short list of small candidates confirms that nothing smaller slips past.

1STEP 1

Break n into prime powers

Prime powers turn the number into a list of exponents.

n=Π_i=1^k p_i^a_i, n²=Π_i=1^k p_i²a_i, n³=Π_i=1^k p_i³a_i

2STEP 2

What makes a cube or a square

Cubes and squares are just divisibility of exponents.

N=Π p^e_p is a cube⇔ 3 ∣ e_p ∀ p, N=(Π p^e_p/3)³

3STEP 3

Turn both conditions into exponent rules

Each condition becomes one plain exponent rule.

3 ∣ 2a_i → 3 ∣ a_i, 2 ∣ 3a_i → 2 ∣ a_i

4STEP 4

Both rules mean sixth power

Both rules together force a sixth power.

2 ∣ a_i and 3 ∣ a_i ⇔ 6 ∣ a_i ⇔ n=t⁶; n=t⁶→ n²=(t⁴)³, n³=(t⁹)²

5STEP 5

Feed in divisibility by 20

The multiple condition forces a factor of ten.

20=2² · 5; 6c ≥ 2, 6d ≥ 1 ⇔ c ≥ 1, d ≥ 1 ⇔ 10 ∣ t

6STEP 6

Take the smallest one

The smallest such number is ten to the sixth.

{ n }={ t⁶ : 10 ∣ t }, t_min=10 → n=10⁶

7STEP 7

Count the digits

It has 7 digits, choice (D).

10⁶ = 1 000 000 → 7 digits

Answer

Check n = 1000000 against all three demands directly. It is 20 times 50000, so 20 divides it. Its square is 10¹2 = (10⁴)³, a perfect cube. Its cube is 10¹8 = (10⁹)², a perfect square. So it qualifies, and it has 7 digits, matching choice (E). Minimality can be confirmed without any theory by a finite check: a number with 6 or fewer digits is below 1000000, and the only sixth powers below 1000000 are 1, 64, 729, 4096, 15625, 46656, 117649, 262144 and 531441. The only one of those divisible by 5 is 15625 = 5⁶, and that one is odd, so it is not divisible by 20. So no qualifying number has 6 digits or fewer, and the answer cannot be (A) through (D).

💡Key takeaway

Squaring and cubing only stretch the prime exponents, so asking for a cube and a square at the same time forces every exponent to be a multiple of 6.

Break n into prime powers
What makes a cube or a square
Turn both conditions into exponent rules
Both rules mean sixth power
Feed in divisibility by 20
Take the smallest one
Count the digits