[shanks4mba]

Originally Posted by Abhinav90

how to check?
Why did you accepted 13/2=6.5 and not 23/3 = 7.666 ?
plz edify !

see actually i m testing for authenticity
A/B here both shud be a less than the base system.
for eg. in base system 10 all digits are frm 0 to 9
hence A/B = 2/3 or 4/5 is possible but 11/2 is not ...same thing

[madnikhil]

How many sets of three or more consecutive odd numbers can be formed such that their sum is 500?

[Abhinav90]

Originally Posted by madnikhil

How many sets of three or more consecutive odd numbers can be formed such that their sum is 500?

n(2a+[n-1]d)/2 = 500
Hence n(2a+[n-1]d) =1000
Hence a= (500/n) - [n-1] [d=2]

Now to let n divide 500,n has to be its factors
Therefore 500=2^2 * 5^3
Hence no. of factors = 3*4 =12

But acc. to question,n cannot be 1 or 2.It cannot be 250 or 500
So we are left wid 4,5,10,20,25,50,100,125
Check by inserting it in the above equation.
For n=4, a=125-3=122 (Not possible as number has to be odd)

Checking all,I am getting only one set for n=10,a=41(Not taking into account negative integers,Else Two)
Wats the OA?

[jessica08]

Hey puys,Please help me find this!

Find the number of consecutive zeroes at the end of the number (10!+50!+90!+100!)^(15!+25!+50!)

[uhhh...]

Originally Posted by jessica08

Hey puys,Please help me find this!

Find the number of consecutive zeroes at the end of the number (10!+50!+90!+100!)(15!+25!+50!)

is same as as the power of 5 in 10! 15! = 5

[jessica08]

Originally Posted by uhhh...

is same as as the power of 5 in 10! 15! = 5

Dude,the question is edited...
Please go thru it once again...

[er.shaan887]

Originally Posted by jessica08

Hey puys,Please help me find this!

Find the number of consecutive zeroes at the end of the number (10!+50!+90!+100!)^(15!+25!+50!)

i think the answer shud b 2(15!+25!+50!).....
if its right then i will explain the approach...

[uhhh...]

Originally Posted by jessica08

dude,the question is edited...
Please go thru it once again...

2(15!+25!+50!)

[bishoo123]

Originally Posted by jessica08

Hey puys,Please help me find this!

Find the number of consecutive zeroes at the end of the number (10!+50!+90!+100!)^(15!+25!+50!)

Is it 2*(15!+25!+50!) ?

[jessica08]

2(15!+25!+50!) is the answer.
Can any of u please explain the approach?