The number 1, 3, 5, 7. . . . . 99 and 128 are multiplied together. The number of zeros at the end of the product must be -
A. 19
B. 22
C. 7
D. Nil
Answer: Option C
Solution (By Examveda Team)
→( 1, 3, 5, 7. . . . . 99 ) × 128| 5 | → | 51 |
| 15 | → | 51 |
| 25 | → | 52 |
| 35 | → | 51 |
| 45 | → | 51 |
| 55 | → | 51 |
| 65 | → | 51 |
| 75 | → | 52 |
| 85 | → | 51 |
| 95 | → | 51 |
| × | ||
| 128 | → | 27 |
512 + 27 will make zero but since 2 comes 7 times, so only 7 zero will come.
This Question Belongs to Arithmetic Ability >> Number System
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