Understanding the Problem: Finding the Largest Power of 10

The question asks for the largest value of $n$ such that $10^n$ divides the given product. To find this, we need to determine how many times the factor 10 appears in the prime factorization of the product. Since $10 = 2 \times 5$, the number of factors of 10 in a product is limited by the minimum count of its prime factors, 2 and 5.

Therefore, the strategy is to find the total exponent of 2 and the total exponent of 5 in the prime factorization of the entire product. The smallest of these two exponents will be the largest possible value of $n$ such that $10^n$ divides the product.

Prime Factorization of Each Term

Let's break down each term in the product into its prime factors and determine the contribution of 2 and 5 from each term:

• $2^5$: Contains $2^5$. No factors of 5.
• $3^3$: Contains $3^3$. No factors of 2 or 5.
• $4^8 = (2^2)^8 = 2^{16}$: Contains $2^{16}$. No factors of 5.
• $5^3$: Contains $5^3$. No factors of 2.
• $6^7 = (2 \times 3)^7 = 2^7 \times 3^7$: Contains $2^7$. No factors of 5.
• $7^6$: Contains $7^6$. No factors of 2 or 5.
• $8^{12} = (2^3)^{12} = 2^{36}$: Contains $2^{36}$. No factors of 5.
• $9^9 = (3^2)^9 = 3^{18}$: Contains $3^{18}$. No factors of 2 or 5.
• $10^6 = (2 \times 5)^6 = 2^6 \times 5^6$: Contains $2^6$ and $5^6$.
• $15^{12} = (3 \times 5)^{12} = 3^{12} \times 5^{12}$: Contains $5^{12}$. No factors of 2.
• $20^{14} = (2^2 \times 5)^{14} = (2^2)^{14} \times 5^{14} = 2^{28} \times 5^{14}$: Contains $2^{28}$ and $5^{14}$.
• $22^{11} = (2 \times 11)^{11} = 2^{11} \times 11^{11}$: Contains $2^{11}$. No factors of 5.
• $25^{15} = (5^2)^{15} = 5^{30}$: Contains $5^{30}$. No factors of 2.

Calculating the Total Power of 2

To find the total power of 2 in the product, we sum the exponents of 2 from all terms that contribute a factor of 2:

Terms contributing factors of 2: $2^5, 4^8, 6^7, 8^{12}, 10^6, 20^{14}, 22^{11}$.

The exponents of 2 are: $5$ (from $2^5$), $16$ (from $4^8$), $7$ (from $6^7$), $36$ (from $8^{12}$), $6$ (from $10^6$), $28$ (from $20^{14}$), $11$ (from $22^{11}$).

Total exponent of 2 $= 5 + 16 + 7 + 36 + 6 + 28 + 11$

$= 21 + 7 + 36 + 6 + 28 + 11$

$= 28 + 36 + 6 + 28 + 11$

$= 64 + 6 + 28 + 11$

$= 70 + 28 + 11$

$= 98 + 11 = 109$

So, the total power of 2 in the product is $2^{109}$.

Calculating the Total Power of 5

To find the total power of 5 in the product, we sum the exponents of 5 from all terms that contribute a factor of 5:

Terms contributing factors of 5: $5^3, 10^6, 15^{12}, 20^{14}, 25^{15}$.

The exponents of 5 are: $3$ (from $5^3$), $6$ (from $10^6$), $12$ (from $15^{12}$), $14$ (from $20^{14}$), $30$ (from $25^{15}$).

Total exponent of 5 $= 3 + 6 + 12 + 14 + 30$

$= 9 + 12 + 14 + 30$

$= 21 + 14 + 30$

$= 35 + 30 = 65$

So, the total power of 5 in the product is $5^{65}$.

Determining the Largest Value of n

The product can be written in its prime factorization form as $2^{109} \times 5^{65} \times (\text{other prime factors})$.

We are looking for the largest value of $n$ such that $10^n$ divides this product. Since $10^n = (2 \times 5)^n = 2^n \times 5^n$, for $2^n \times 5^n$ to divide the product, the power of 2 must be at least $n$, and the power of 5 must also be at least $n$.

From our calculations, we have $2^{109}$ and $5^{65}$ in the product.

• We need $n \le 109$ (from the power of 2).
• We need $n \le 65$ (from the power of 5).

To satisfy both conditions, $n$ must be less than or equal to the minimum of the two exponents.

$n_{max} = \min(\text{Total power of 2}, \text{Total power of 5})$

$n_{max} = \min(109, 65)$

$n_{max} = 65$

Therefore, the largest value of $n$ such that $10^n$ divides the given product is 65.

Since the minimum exponent is 65 (for the prime factor 5), the largest power of 10 that divides the product is $10^{65}$.

Revision Table: Key Exponents

The largest power of 10 is determined by the minimum of these total exponents, which is 65.

Additional Information: Trailing Zeros

This problem is closely related to finding the number of trailing zeros in a number, especially factorials. The number of trailing zeros in an integer is equal to the exponent of the largest power of 10 that divides it. This, in turn, is equal to the exponent of 5 in its prime factorization, because factors of 2 are always more abundant than factors of 5 in products involving many numbers (like factorials).

For a general product, as seen in this problem, we must calculate both the total power of 2 and the total power of 5 and take the minimum, as either factor can be the limiting one.