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50! = 50*49*48*47....3*2*1

We have to find the highest power of 7 present in 50!
since we can see 50! is a multiple of all consecutive integers in descending order from 50 to 1, its not difficult to figure out how many times 7 will appear.

Multiples of 7 present in 50! are: 49, 42, 35, 28, 21, 14, 7.
So 7 multiples out of which 49 has not one, but two 7's.
So total number of 7's = 8

Hence answer is A
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This is a fairly straightforward question which tests you on the ‘Maximum power in a factorial’ concept. If you have solved sufficient questions on this model, this question will take you less than a minute to solve.

We know that, 50! = 50 * 49 * 48 * …… * 7 * 6 * 5 * 4 * 3 * 2 * 1. If we are trying to find out the highest value of 7^n such that it is a factor of 50!, essentially, we are trying to find out how many times the number 7 appears in the expansion of 50!.

As you can see, in the expansion of 50!, 7 appears 7 times, because there are 7 multiples of 7 between 1 and 50. Correct? But, do not forget that the number 49 is a square of 7. It will therefore contribute an additional 7.
So, the highest value of n such that 7^n can completely divide 50!, is 8. The correct answer option is A.

We will also demonstrate the shortcut to find out the ‘Maximum power in a factorial’. But, bear in mind that this can be applied in finding out the maximum power of a Prime number only.

Attachment:
24th May 2019 - Reply 2.JPG
24th May 2019 - Reply 2.JPG [ 25.49 KiB | Viewed 5342 times ]

On an ending note, it is always better to use the shortcut method, especially when the factorial given is that of a large number. The first method that we used, although good, cannot be used practically; it is meant only to understand the concept better.

Hope this helps!
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Factors of 7 leading up to 50 are 7, 14, 21, 28, 35, 42, and 49. Each of these contains one factor of 7, except for 49 which contains 2. Thus we add up all the factors of 7 to get 1 + 1 + 1 + 1 + 1 + 1 + 2 = 8 total factors of 7. Thus, our answer is 8! There are many other ways to solve this problem as well, this is just one example.
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