How many factors of 19! Are there, whose units digits is 5?
- 1298
- 1296
- 1295
- 1297
- 1291
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Answer: B
Approach: A factor ends in 5 ⇔ it is odd and a multiple of 5. Count factors with power of 2 = 0 and power of 5 ≥ 1.
Solution:
19! = 2¹⁶ × 3⁸ × 5³ × 7² × 11 × 13 × 17 × 19 (powers: 3 → ⌊19/3⌋ + ⌊19/9⌋ = 8; 5 → 3; 7 → 2).
Choices: 2 → 1 (only 0); 3 → 9; 5 → 3 (1 to 3); 7 → 3; 11, 13, 17, 19 → 2 each.
Count = 1 × 9 × 3 × 3 × 2 × 2 × 2 × 2 = 1296.
Why the other options are wrong:
A, C, D, E: off by small amounts — usually from including 5⁰ (gives 1728) or miscounting the power of 3.
Cetking shortcut: Odd part of 19! has (8+1)(3+1)(2+1)·2⁴ = 1728 factors; drop the 5⁰ ones (1/4) ⇒ 1728 × 3/4 = 1296.
Answer: B