$N = 204 \times 221 \times 238 \times 255 \times . . . \times 850.$ How many consecutive zeroes will be there at the end of this number N?

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Solution

The correct option is B 10
$N = 204 \times 221 \times 238 \times 255 \times . . . \times 850 = (17 \times 12) \times (17 \times 13) \times (17 \times 14) \times (17 \times 15) \times . . . \times (17 \times 50)$
We are required to count the number of 5s in N = number of zeroes in N.
To count the number of 5s, we can count it from
$[(17 \times 1) \times (17 \times 2) . . . (17 \times 12) \times (17 \times 13) \times (17 \times 14) \times (17 \times 15) \times . . . (17 \times 50)] a n d t h e n s u b s t r a c t t h e n u m b e r o f 5 s i n [(17 \times 1) \times (17 \times 2) . . (17 \times 11)] [(17 \times 1) \times (17 \times 2) . . . (17 \times 12) \times (17 \times 13) \times . . . (17 \times 50)] = 12$
Number of 5s in $[(17 \times 1) \times (17 \times 2) . . (17 \times 11)] = 2$
Hence, number of 5s in $(17 \times 12) \times (17 \times 13) \times (17 \times 14) \times (17 \times 15) \times . . . (17 \times 50) = 12 - 2 = 10$

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