by krishnan.anju1987 Tue Aug 14, 2012 3:21 pm
What are the answer options for this question. I proceeded something like this but I am nearly sure that something was wrong which made the numbers too large. Would be better if I could give it a try after looking at the answer options. However, I am unable to solve this one as of now.
The method I used.
11!*22!*33!...1010!
now 92*11=1012. Also, in this case, the number of 10s can be found by finding the number of 5s which become the limiting factor. Hence, if we see 11!, the number of 5s in it is 2(10,5). Similarly 22 has 4 (20,10,15,5)5s. Now the whole pattern can be divided into ranges* number of times they appear in the pattern.
92(1 to 11)*91(11 to 22)* 90(22 to 33)*89(33 to 44) range and so on. The number of 5s must be added. Hence, It would be 92*2+91*2+90*2+...+1*2/ Removing 2 from this equation. 2(92+91+90+...1). This will be (92*93/2)*2 or simply 92*93.
But if we take
If we take 110!
it looks like 109*108*107*106*105*104*103*102*101*100*99!. In this expression 99! consists of 18 5s. The rest of the expansion has 3 5s. SO for every increase of 100 1 extra 5 has to be added and for 1000 2 extra 5s. Till 1010 11 extra 5s must be added. Hence total value is 8567((92*93)+11). Since I don't have the options I am not sure whether my approach is right.
It would be great if someone could verify this or let me know my mistake in case I am making one.