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 ) × 1285 | → | 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.
Related Questions on Number System
Three numbers are in ratio 1 : 2 : 3 and HCF is 12. The numbers are:
A. 12, 24, 36
B. 11, 22, 33
C. 12, 24, 32
D. 5, 10, 15
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