ananyboss
January 15, 2013, 1:37pm
13883
@pirateiim478
@pirateiim478 said: For case you need to divide by 2!. Rest is correct!!sum of consecutive numbers
In how many ways 200 can be represented by sum of two consecutive - Even - Numbers
- 0 Ways ( not possible ) possibility -- (98+102, 96+104, etc )
ananyboss
January 15, 2013, 1:45pm
13884
@karan20 said:
For case you need to divide by 2!. Rest is correct!!14 is correctIn how many ways 200 can be represented as1. sum of consecutive numbers ----- 2 Ways 2. Sum of consecutive even numbers --- 4 ways 3.Sum of consecutive odd numbers ----- 4 Way
@karan20
Can you tell the approach !
budokai001
January 15, 2013, 1:52pm
13887
@pirateiim478 said: @karan20 Can you explain your approach for even and odd consecutive numbers? Last 2 digits of 32 ^222 => 7^222%25 = 7^2 %25 =
24
rkshtsurana
January 15, 2013, 1:53pm
13888
@pirateiim478 said: @karan20 Can you explain your approach for even and odd consecutive numbers?
What are last 2 digits of 32^222 ?
76 * 32 *32 which is eventually 76 *2^10 which is last two digit of 2^10,, so 24
pirateiim478
January 15, 2013, 2:09pm
13889
A century can end with .......
karan20
January 15, 2013, 2:18pm
13891
@karan20 said:
No. of ways N can be written as sum of 2 or more consecutive nos. = ( no. of odd factors of N ) -
200 = 5^2 x 2^3
No. of odd factors = 3
There fore NO. of ways 200 can be written as sum of consecutive nos. = 2
Consecutive odd nos. = 3 ( 47-53, 99-101, 1-21 )
Consecutive even nos. = 1 ( 18-32 ) { 4 nos. on either of 25 }
wats d OA?
@ananyboss PS: made error while calculating earlier
bs0409
January 15, 2013, 2:21pm
13892
@pirateiim478 said: For case you need to divide by 2!. Rest is correct!!
14 is correct
200=2*4*25=2^3*5^2
1)No. of odd factors=3
So ans=2
2)100=2^2*5^2
represent 100 as a sum of +ve integers and multiply by 2 both sides to represent 200 as a sum of even natural numbers.
3)There shld ve even no. of odd numbers
Let them be 2k-m + 2k-(m-2) + ........2k-1 + 2k+1 + 2k+3 +.........2k+m=2k(m+1)=200
or, k(m+1)=100=2^2*5^2
k=25,m=3
k=50,m=1
k=10,m=9
3 solutions
karangarcia
January 15, 2013, 2:33pm
13894
Find the numbers of Zeroes in 1!*2!*3!............100!
pirateiim478
January 15, 2013, 2:50pm
13896
@KaranGarcia said: Find the numbers of Zeroes in 1!*2!*3!............100!
5!-9! =1 zero each,10!-14!=>2 each...20!-24! 4 each and 25!-29! 6 each There is jump of 1 zero because extra 5 is added. Similarly calc upto 100! with care when multiples of 25 occur like 50,75,100.. Total =1124 zeroes
ravi.theja
January 15, 2013, 2:52pm
13898
@pavimai said: N! is completely divisible by 13^52.what is sum of the digits of the smallest such number N
first number where u get a 13 is 13 ! let x be the number which gets us 52 13's..
enigma001
January 15, 2013, 2:54pm
13899
@pirateiim478 said: @karan20 Can you explain your approach for even and odd consecutive numbers?
What are last 2 digits of 32^222 ?
(2^5)^222 = (2^10)^111= (2^10)^(odd number) = 24
pirateiim478
January 15, 2013, 2:55pm
13900
Similar question: how many zeros are there at the end of
Hint: write n as n+1-1 :)
Post your approach. It will be useful for puys who dont know approach