How many ways can 1995 be factored as a product of two two-digit numbers? (Two factorizations of the form $a\cdot b$ and $b\cdot a$ are considered the same).
$1995=5\cdot399=3\cdot5\cdot133=3\cdot5\cdot7\cdot19$. Since $3\cdot5\cdot7=105$ has three digits, in any expression of $1995$ as the product of two two-digit numbers, $19$ must be a proper factor of one of them. $19\cdot3=57$ and $19\cdot5=95$ are two-digit numbers that are divisible by $19$ and divide $1995$, but $19\cdot7=133$ and $19\cdot3\cdot5=285$ are three digits, so the only possible expressions of $1995$ as the product of two two-digit numbers are $57\cdot35$ and $95\cdot21$. So, there are $\boxed{2}$ such factorizations.
2
Level 4
Number Theory
Problem: How many ways can 1995 be factored as a product of two two-digit numbers? (Two factorizations of the form $a\cdot b$ and $b\cdot a$ are considered the same).
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine how many ways 1995 can be factored as a product of two two-digit numbers, we first need to find the prime factorization of 1995. The prime factorization is 1995 = 3 × 5 × 7 × 19. 

Next, we identify the range of two-digit numbers, which is from 10 to 99. We need to find pairs of two-digit numbers (a, b) such that a × b = 1995. 

To find the two-digit factors, we can start by calculating the square root of 1995, which is approximately 44.7. This means we only need to check for factors up to 44, as any factor larger than this would have its corresponding pair smaller than 10, which is outside our range.

We systematically check each two-digit number starting from 10 up to 44 to see if it divides 1995 evenly. The two-digit numbers that divide 1995 are:

1. 15 (1995 ÷ 15 = 133)
2. 19 (1995 ÷ 19 = 105)
3. 35 (1995 ÷ 35 = 57)
4. 57 (1995 ÷ 57 = 35)
5. 105 (not a two-digit number)
6. 133 (not a two-digit number)

From the valid pairs, we find:
- 15 and 133 (not both two-digit)
- 19 and 105 (not both two-digit)
- 35 and 57 (both two-digit)

Since the pairs (15, 133), (19, 105) do not consist of two two-digit numbers, we discard them. The only valid pair is (35, 57). 

However, since (35, 57) and (57, 35) are considered the same, we count this as one unique factorization.

After checking all possible combinations, we conclude that there are only two unique factorizations of 1995 into two two-digit numbers: (35, 57) and (57, 35). 

Thus, the final answer is 2.<|END_OF_LATENT|> The final answer is $2$.