Some integers m with 1 < m < 100000 have the property that the product of the digits of m is equal to 200. If N is the number of such integers m, what is the integer formed by the rightmost two digits of N?

1. Factorising the target product

The product condition is

$200 = 2^3 \times 5^2$

So each valid digit‐set must contain three 2’s and two 5’s in some form.

2. Converting prime factors into single digits

Single digits we can legally use (0–9, but 0 is useless because it makes the product 0) are

• $2,4,8$ for the powers of 2 (because $4 = 2^2$ and $8 = 2^3$)
• $5$ for the power of 5
• $1$ for “doing nothing” (product stays the same)

There is no digit that contains $5^2$ inside it, so we must use two separate 5’s.

3. Creating all multisets of digits whose product is 200

Work with the powers of 2:

• Option A $\{2,2,2\}$
• Option B $\{2,4\}$
• Option C $\{8\}$

Add the compulsory two 5’s to each option. Thus the basic digit packs are

1. $\{2,2,2,5,5\}$ (5 digits already)
2. $\{2,4,5,5\}$ (4 digits)
3. $\{8,5,5\}$ (3 digits)

Digit 1 may be appended because it does not change the product. We may append just enough 1’s so the total length never exceeds 5 digits (largest allowed number).

4. Counting permutations of each pack

Because numbers are just different orders of the same digits, use

\text{# permutations} = \frac{n!}{(\text{repeats of each digit})!}

Count case by case:

5. Adding everything together

What does the question say?

Imagine you have lots of 1-, 2-, 3-, …, 9-shaped stickers.
If you stick some of them next to each other you make a number.
The rule is: when you multiply the stickers (digits) you used, the answer must be 200.

Your job is to count how many different numbers (up to five stickers long, because the biggest allowed number is 99 999) obey that rule.
Finally, look at that big count and tell only its last two digits.

How could you think about it?

1. Break 200 into small pieces (prime factors). 200 = 2×2×2×5×5.
2. Match those pieces with single digits like 2, 4 (=2×2), 8 (=2×2×2), 5, 1, … (1 does nothing in multiplication).
3. List every collection of digits whose multiplication gives 200.
4. Mix them in every possible order (because 25245 and 54252 are different numbers) but avoid repeats coming from identical digits.
5. Keep numbers to at most five digits (because 100 000 is not allowed).

The real solution does exactly this counting, one neat case at a time.

Step-by-step counting

1. Prime break-up of 200
   $200 = 2^3 \times 5^2$

2. Translate primes to possible digits

   • Powers of 2: digits $2,4,8$
   • Powers of 5: digit $5$ (must appear exactly twice)
   • Digit $1$ (optional, neutral)

3. Construct all digit packs (must exactly give $2^3 \times 5^2$):

   • Case 1 $\{2,2,2,5,5\}$
   • Case 2 $\{2,4,5,5\}$ (+ optional one 1)
   • Case 3 $\{8,5,5\}$ (+ optional one or two 1’s)

4. Count permutations inside each case (use $\frac{n!}{(\text{repeat factorials})}$):

   • Case 1: $10$
   • Case 2: $12$ (no 1) + $60$ (one 1) = $72$
   • Case 3: $3$ (no 1) + $12$ (one 1) + $30$ (two 1’s) = $45$

5. Total numbers

   $N = 10 + 72 + 45 = 127$

6. Rightmost two digits of $127$ are 27.

[\boxed{27}]