Tmoni26 wrote:
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.