Question

How many factors of $19!$ are there, whose unit digit is 5?

• A. \(1298 \)
• B. \(1296 \)
• C. \(1295 \)
• D. \(1297 \)
• E. \(1291 \)

Hint:

For unit digit problems: - Ending in 0 → both 2 and 5 present - Ending in 5 → only factor 5 present (no 2)

Answer 1

The Correct Option is B

Concept: A number ends with digit 5 if:
• It contains at least one factor of 5
• It does NOT contain factor 2 (otherwise it ends with 0)
Step 1: Prime factorization of $19!$.
\[ 19! = 2^{16} \cdot 3^{8} \cdot 5^{3} \cdot 7^{2} \cdot 11 \cdot 13 \cdot 17 \cdot 19 \]

Step 2: Conditions for unit digit 5.

• Power of 2 must be 0
• Power of 5 must be $\geq 1$ (i.e., 1 to 3)

Step 3: Count choices.
\[ \text{Power of 5: } 3 \text{ choices } (1,2,3) \] \[ \text{Power of 3: } 9 \text{ choices } (0 \text{ to } 8) \] \[ \text{Power of 7: } 3 \text{ choices } (0 \text{ to } 2) \] \[ \text{Each of } 11,13,17,19: 2 \text{ choices each} \]

Step 4: Multiply all choices.
\[ = 3 \times 9 \times 3 \times 2^4 \] \[ = 3 \times 9 \times 3 \times 16 = 1296 \]