Question
The number 1, 3, 5, 7. . . . . 99 and 128 are multiplied together. The number of zeros at the end of the product must be -
Answer: Option C →( 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.
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