bblast wrote:1> how many zeroes a the and of 100!+200!

ans-24

2> how many zeroes a the and of 100!* 200!

ans-73

1) 100!(1+101*102*.....200)

so number of zeros in the expression is equivalent to number of zeros in 100!;

100/5=20
20/5=4
20+4=24

2) 100!*200!
200/5=40;
40/5=8;
8/5=1;
no. of zeros in 200!=40+8+1=49

hence total no. of zeros = (10^24)*10^49=10^24+49=10^73;

The key is to identify the number of 5's. Manp's method is the best one to adopt in this case.

PS: wondering if GMAT is also going IIM-CAT way!!

hey pankaj, u r right these q's r from number system from a cat book(arun sharma), but its a good idea to solve these, (note-not all cat topics are useful in GMAT-in fact most r not relevant)

To add on there is a 700 level question in MGMAT test series, "how many number of zeroes in 60!? [spoiler](ans-14)[/spoiler]",

so these questions interested me.

bblast wrote:1> how many zeroes a the and of 100!+200!

ans-24

2> how many zeroes a the and of 100!* 200!

ans-73

1. 100 ! will have 24 zeros

100/5=20

20/5=4
_______
20+4=24

200 ! Will definitely have say more Number of zeros but we will not consider those .

The reason being :

Say while adding 2 numbers having 5000 and 50

We get the result 5050.

Upon careful observation we find that the number having less number of zero's in the end is taken into consideration.

Hence 100 ! + 200 ! will have 24 zero's in the end.


2. 100 ! * 200 !

100 ! has 24 zero's in the end ( We just calculated it )

Now Similarly 200 ! will have :


200 / 5 = 40

40 /5 = 8

8 /5 = 1
______________
40 + 8 +1 = 49


Now while multiplying 2 numbers having zero's in the end say 100 ( 2 Zeros ) and 20 ( 1 Zero ) we get result 2000 (3 Zeros )( We observe that the number of zeros in the end is added )

Similarly While multiplying 100 ! * 200 ! We get 24 Zeros to be multiplied with 49 zeros , hence total number of zeros in the end of the number is 24 + 49 = 73