Number System Question 1 - CAT 2020
How many 3-digit numbers are there for which the product of their digits is more than 2 but less than 7?
Step-by-Step Explanation
Let the 3-digit number be abc. The problem states that the product of the digits must satisfy 2 < a Ă— b Ă— c < 7. This means the possible integer values for the product are 3, 4, 5, or 6.
Let's find the combinations of single-digit numbers that give these products:
- Product = 3: The only possible digits are 1,1,3.
Number of arrangements = (3!)/(2!) = 3 ways.
- Product = 4: This can happen in two ways:
- Using digits 1,1,4 ⇒ (3!)/(2!) = 3 ways.
- Using digits 1,2,2 ⇒ (3!)/(2!) = 3 ways.
- Product = 5: The only possible digits are 1,1,5.
Number of arrangements = (3!)/(2!) = 3 ways.
- Product = 6: This can also happen in two ways:
- Using digits 1,1,6 ⇒ (3!)/(2!) = 3 ways.
- Using digits 1,2,3 ⇒ 3! = 6 ways.
Now, add up all the possible arrangements:
Total Numbers = 3 + 3 + 3 + 3 + 3 + 6 = 21
đź’ˇ FundaMakers Tip:
Instead of picking numbers randomly, list the target products first (3, 4, 5, 6) and find their factor sets. Use the unique permutation formula (3!)/(2!) for cases with repeating digits to quickly find total arrangements without missing any combinations.