There are 150 pages in a book determine the total number of pages in such 1652 books if the number of pages in each book is the same

Problem:
The product of two digits number is 1998 if the product of their units digits is 28 and that of tens digits is 15. Find the numbers.

Solution:

To find the two-digit numbers, we need to consider their tens and units digits. Let's assume the tens digit of the first number is "a" and the units digit is "b". Similarly, the tens digit of the second number is "c" and the units digit is "d".

Step 1: Identify the given information:
- The product of the units digits is 28, so we have the equation b * d = 28.
- The product of the tens digits is 15, so we have the equation a * c = 15.
- The product of the two numbers is 1998, so we have the equation (10a + b)(10c + d) = 1998.

Step 2: Simplify the equations:
- From the equation b * d = 28, we can list the possible combinations of b and d that multiply to 28: (1, 28), (2, 14), (4, 7), (7, 4), (14, 2), (28, 1).
- From the equation a * c = 15, we can list the possible combinations of a and c that multiply to 15: (1, 15), (3, 5), (5, 3), (15, 1).

Step 3: Find the two-digit numbers:
- We need to find the combination of (a, c, b, d) that satisfies all the given equations.
- Let's try the different combinations of (a, c, b, d) and check if they satisfy the equation (10a + b)(10c + d) = 1998.

Possible combinations:
1. (1, 15, 1, 28): (10 * 1 + 1)(10 * 15 + 28) = 1999 * 178 = 355222 (Not equal to 1998)
2. (1, 15, 28, 1): (10 * 1 + 28)(10 * 15 + 1) = 281 * 151 = 42431 (Not equal to 1998)
3. (3, 5, 1, 28): (10 * 3 + 1)(10 * 5 + 28) = 31 * 78 = 2418 (Not equal to 1998)
4. (3, 5, 28, 1): (10 * 3 + 28)(10 * 5 + 1) = 58 * 51 = 2958 (Not equal to 1998)
5. (5, 3, 1, 28): (10 * 5 + 1)(10 * 3 + 28) = 51 * 58 = 2958 (Not equal to 1998)
6. (5, 3, 28, 1): (10 * 5 + 28)(10 * 3 + 1) = 78 * 31 = 2418 (Not equal to 1998)