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.