Solved Example #16
Tuesday, February 19th, 2013
Question : Find the No of zeroes at the end of 25! +26! + 27! + 28! + 30!
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
Number of zeroes is given by the sum of the quotients obtained by successive division of n by 5.
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
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.
Note to self: Start making crazy questions to get more students to join online courses