Let's break the problem down step by step.

1. Finding $N$: the least 2025-digit number whose product of digits is 2025.

We are tasked with finding the least 2025-digit number, $N$, whose product of digits is 2025.

• First, let's factor 2025:
  $2025 = 5^2 \times 3^4.$ To construct a number whose digits multiply to 2025, we should choose digits whose product gives us this factorization. The digits that could be used must lie between 1 and 9 (since these are the possible digits for a number in decimal notation).

• We want to minimize the number, so we should use as many 1's as possible, since they don't affect the product.

• The prime factorization of 2025 suggests that we should use digits like 5 and 3. We need to determine how many of each digit we need.

Constructing the Number:

• We need two 5's to account for $5^2$.
• We need four 3's to account for $3^4$.

Thus, we need a number that contains:

• Two 5's,
• Four 3's,
• The rest of the digits should be 1's.

The smallest 2025-digit number that satisfies this condition will be one where we place the two 5's and four 3's in the smallest possible positions (i.e., at the end), and fill the rest with 1's. This gives us:

$N = \underbrace{111\cdots1}_{2021 \text{ ones}} 3355.$

Thus, the least 2025-digit number whose product of digits is 2025 is $N = 111\cdots 1 3355$, where there are 2021 ones followed by the digits 3, 3, 5, 5.

2. Finding $2025N \mod 10^{2025}$

Now, we need to find the quotient and remainder when $2025N$ is divided by $10^{2025}$.

Let $N = 111\cdots1 3355$ (as discussed). When multiplying $N$ by 2025, we have:

$2025N = 2025 \times (111\cdots1 3355).$

Since $10^{2025}$ is a large number with 2025 zeros, when we compute $2025N \mod 10^{2025}$, the result will be the last 2025 digits of $2025N$.

• The number $N$ ends in "3355". When multiplied by 2025, the last 4 digits of $2025N$ will be the product of the last 4 digits of $N$ and 2025.

So, we need to compute the product of the last digits: $2025 \times 3355 = 2025 \times (3000 + 355) = 2025 \times 3000 + 2025 \times 355.$

Breaking it down: $2025 \times 3000 = 6075000, \quad 2025 \times 355 = 718875.$ Thus: $2025 \times 3355 = 6075000 + 718875 = 6793875.$

Now, we focus only on the last 4 digits of $6793875$, which are 3875. Hence, the last 4 digits of $2025N$ will be 3875.

Therefore, $2025N \mod 10^{2025}$ is the number formed by placing 3875 at the end of a number of 2021 ones:

$2025N \mod 10^{2025} = \underbrace{111\cdots1}_{2021 \text{ ones}} 3875.$

Final Answer:

• Quotient: The quotient when $2025N$ is divided by $10^{2025}$ is simply the first part of $2025N$, i.e., $2025N$ without the last 2025 digits. This is a large number, but we can summarize it as a number with 2021 ones followed by 6793.

• Remainder: The remainder is the last 2025 digits of $2025N$, which are $\underbrace{111\cdots1}_{2021 \text{ ones}} 3875$.

Do you need more details or have further questions on this? Here are some related questions:

1. What is the total number of digits in $2025N$?
2. How does the product of digits affect the choice of digits in such problems?
3. What are some techniques to efficiently compute large modular operations like $2025N \mod 10^{2025}$?
4. Can the pattern of 1's in $N$ be generalized to other cases with different products of digits?
5. How would the answer change if we had a different number of digits in $N$?

Tip: Breaking a large number into parts (as done here with the multiplication of the last digits) helps in handling large computations more efficiently.