Solved Example #16

To understand this, let us understand the basic idea first

What will be the number of 0s at the end of a + b + c would depend upon the least number of 0s that any one of a or b or c has.

For eg: 300 + 120000 + 17272730 will end in 1 zero

But, if they have the same number of zeroes, we will also have to consider the last non-zero digit.

For eg: 12000 + 161237000 + 1212331000 will not end in 3 zeroes but in 4 zeroes because the last non-zero digits 2, 7 and 1 will add up to generate an extra zero.

If you understood the above part, read on.

We first need to figure out how many zeroes do the factorials individually have

Among the ones mentioned,

25!, 26!, 27! and 28! have 6 zeroes each.

30! has 7 zeroes.

We also need to consider the right most digits of 25!, 26!, 27! and 28!

R(n!) = Last Digit of [ 2^a x R(a!) x R(b!) ]
where n = 5a + b

Using this we get, right most non zero digits of 25! as 4

=> 26! will end in 4*6 or 4

=> 27! will end in 4*7 or 8

=> 28! will end in 8*8 or 4

=> 25! + 26! + 27! + 28! will not end in 6 zeroes but in 7 zeroes

We know that 30! ends in 7 zeroes.

So, the overall number 25! + 26! + 27! + 28! + 30! would end in 7 zeroes

You can read more about these ideas in an old PG post of mine:http://www.pagalguy.com/news/ive-got-power-working-with-factorials-cat-2011-quant-a-17445

Footnote: Let me add that this is way too complicated to be asked in CAT. Even in the unlikely scenario that this does get asked, you should avoid it. This is probably something that a teacher would use to impress students or a coaching institute would include in its test-series / question-a-day / material to scare students into buying their products.