How many trailing Zeroes does 53! + 54! have?

Can someone please explain why we have another trailing 0 in 55

Will it not be 00000 * 55 and this should not add another zero to the calculation??

Thanks alot

We don't have another trailing 0 in 55. Rather, 55 will contribute to another 0. Here's why:

The concept is based on how may pairs of 2 and 5 can be formed. Each pair will contribute one 0. Let's look at few small scale examples:

1. (22)*5 = (2*11)*5 = 110 -> one 0 because there is one 2-5 pair.
2. (220)*5 = (2*2*5*11)*5 = 1100 -> two 0s because there are two 2-5 pairs.
3. (10)*5 = (2*5)*5 = 50 -> one 0 because there is one 2-5 pair.
4. (100)*5 = (2*2*5*5)*5 = 500 -> two 0s because there are two 2-5 pairs.
5. (10)*10 = (2*5)*2*5 = 100 -> two 0s because there are two 2-5 pair.

In the problem, we have 53!*55.
55 = 5*11 -> it has no 2.
53! has a whole bunch of 2s (49 to be exact). One of the 49 2s will pair with one 5 from 55 to add another 0.

Hope this helps.

53! + 54! = 53! ( 1 + 54 )

53! ( 55 )

53! will have 12 trailing zeroes

53/5 = 10
10/5 =2

53! ( 55 ) = 53! ( 11 * 5 )

53! ( 11 * 5 ) will have 13 trailing zeroes ( 53 will have 12 trailing zeroes + 1 trailing zero for one additional 5 in 55 )

Hence correct answer will be (B) 13

PS : Good to see you back Mr Austin...

This is what i did =>
Let p=53!+54!
TO get the number of 10's we need to get the number of fives
Reason=> a 10 is formed via a 2 and a 5.
since number of 2's are abundant in quantity => no. of 5's act as the limiting agent.
p=53![1+54]=53!*55
number of 5's in 53!=> 53/5 + 53/25 +...=> 10+2=> 12
but wait 55 has one more five.
hence total number of fives are => 13
Hence number of trailing zeroes has to be 13

So its B

Tricky question! Though the number of 0s in both 53! and 54! will be the same because 54 doesn't bring in any additional 5, the problem is that we don't know what kind of numbers are getting added. If we are adding the numbers such as:

...200000...
...800000...

we will end up having another trailing 0 in the sum. Hence, we cannot find the number of trailing 0s from the sum.

Therefore, the step of taking 53! common becomes mandatory to ensure that we have all in factors terms.

53! * 55

As found above, we see that 53! gives 12 5s and we get another one from 55, so in all 13 5s.

Answer (B)

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