Can a cube end with exactly two zeroes (00)? Explain.

Question

Homi · Accepted Answer

A number ending in zeroes means it is a multiple of 10, so its prime factors must include both 2 and 5.

Step 1 — Understanding numbers ending in zeroes

Let us think about what it means for a number to end in zeroes. A number ending in one zero is a multiple of 10. $10 = 2 \times 5$ A number ending in two zeroes is a multiple of 100. $100 = 2^2 \times 5^2$ A number ending in three zeroes is a multiple of 1000. $1000 = 2^3 \times 5^3$ In general, if a number ends in exactly $k$ zeroes, its prime factorization must contain at least $k$ factors of 2 and at least $k$ factors of 5. More precisely, the lowest power of 2 or 5 in its prime factorization must be $k$ .

$\boxed{\text{A number ending in } k \text{ zeroes has } 2^k \text{ and } 5^k \text{ as factors.}}$

Step 2 — Prime factors of a cube

Let $N$ be any natural number. We want to find the prime factors of its cube, $N^3$ . Let us write $N$ using its prime factors. For example, if $N = 12$ , then $N = 2^2 \times 3^1$ . Then $N^3 = (2^2 \times 3^1)^3$ . $N^3 = 2^{2 \times 3} \times 3^{1 \times 3}$ $N^3 = 2^6 \times 3^3$ Notice that the exponents (powers) in the cube are all multiples of 3. In general, if $N = p_1^{a_1} \times p_2^{a_2} \times \dots$ , where $p_1, p_2, \dots$ are prime numbers and $a_1, a_2, \dots$ are their exponents. Then the cube of $N$ is $N^3$ . $N^3 = (p_1^{a_1} \times p_2^{a_2} \times \dots)^3$ $N^3 = p_1^{3a_1} \times p_2^{3a_2} \times \dots$ This means that in the prime factorization of any perfect cube, every exponent must be a multiple of 3.

$\boxed{\text{Exponents of prime factors in a cube are multiples of 3.}}$

Step 3 — Number of zeroes in a cube

Let us combine what we learned about zeroes and cubes. If a cube, $N^3$ , ends in exactly $k$ zeroes, then its prime factorization must contain $2^k$ and $5^k$ . From Step 2, we know that the exponent of any prime factor in $N^3$ must be a multiple of 3. So, the exponent of 2 in $N^3$ must be a multiple of 3. And the exponent of 5 in $N^3$ must also be a multiple of 3. This means that $k$ (the number of zeroes) must be a multiple of 3. So, a perfect cube can only end with 0, 3, 6, 9, etc., zeroes.

$\boxed{\text{The number of zeroes in a cube must be a multiple of 3.}}$

Step 4 — Conclusion

The question asks if a cube can end with exactly two zeroes. Here, the number of zeroes, $k$ , is 2. We found that $k$ must be a multiple of 3 for a number to be a perfect cube. Since 2 is not a multiple of 3, a cube cannot end with exactly two zeroes.

Answer

(i) No, a cube cannot end with exactly two zeroes. The number of zeroes at the end of a perfect cube must always be a multiple of 3.

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