If N = 2^5 × 3^2 × 7 × □ and 100 divides evenly into N, which of the following numbers could be placed in the box?

Steps to Solve

1. Understand the prime factorization of 100

The first step is to find the prime factorization of 100.
We have:
$$ 100 = 10^2 = (2 \cdot 5)^2 = 2^2 \cdot 5^2 $$

2. Identify the current factorization of N

The expression for ( N ) is given as:
$$ N = 2^5 \times 3^2 \times 7 \times x $$
This means we already have ( 2^5 ) from ( N ).

3. Determine the requirements for divisibility by 100

For ( N ) to be divisible by 100, it must contain at least:

• ( 2^2 )
• ( 5^2 )

Since ( 2^5 ) already meets the requirement for the prime factor 2, we need to ensure that the factor ( x ) provides ( 5^2 ).

4. Examine the candidates for x

Now we check which of the options can provide ( 5^2 ):

• A. ( 5 = 5^1 )
• B. ( 20 = 2^2 \cdot 5^1 )
• C. ( 36 = 2^2 \cdot 3^2 )
• D. ( 75 = 3 \cdot 5^2 )

5. Evaluate each option

• A. ( 5 ): Only provides ( 5^1 ). Needs another ( 5 ) to fulfill ( 5^2 ).
• B. ( 20 ): Only provides ( 5^1 ) (same issue as A).
• C. ( 36 ): Provides no ( 5 ) factors.
• D. ( 75 ): Provides ( 5^2 ) (meets the requirement!).

6. Conclusion

Thus, the only value for ( x ) which fulfills the requirement that ( N ) is divisible by 100 is 75.