n(n+1)/2 =
n =
now only odd numbers can be expressed as difference of squares of 2 natural numbers
so 44/2 = 22, but we need to take out single number 1
hence 22-1 = 21 total such numbers.....
@Estallar12 yes it should be 21 only imo
(58! - 38!) Highest power of 3 available in expression?
@vbhvgupta said:(58! - 38!) Highest power of 3 available in expression?
38!(58*57*...39 - 1) = x*3^17(58*57*...39 - 1)
the term in the bracket is not divisible by 3.....hence 17 is the highest power of 3 in this expression
@vbhvgupta said:Q
It's 21.
Basically (n+1)^2 - n^2 = 2n+1 = an Odd number. So all odd numbers can be expresed as the sum of two consecutive integers.
Now triangular number is of the form x(x+1)/2. With a little observation u see that for x=44 u will get 1000.
So the values of x satisfying are 2,5,6,9,10,......till 41,42 So total 21 values.
EDIT: Ignore x=1 as 0 is not counted
Basically (n+1)^2 - n^2 = 2n+1 = an Odd number. So all odd numbers can be expresed as the sum of two consecutive integers.
Now triangular number is of the form x(x+1)/2. With a little observation u see that for x=44 u will get 1000.
So the values of x satisfying are 2,5,6,9,10,......till 41,42 So total 21 values.
EDIT: Ignore x=1 as 0 is not counted
(201* 202 * 203* 204* 246 * 247 * 248 * 249)^2 Last 2 digits????
@vbhvgupta said:(58! - 38!) Highest power of 3 available in expression?
Powers of 3 in 58 = 58/3, 19/3, 6/3 = 27.
Powers of 3 in 38 = 38/3, 12/3, 4/3 = 17.
Now, 3^27*a - 3^17*b = 3^17 ( 3^10*a - b) mod 3.
As, 3^10*a - b is not divisible by 3.
Thus, 17 powers of 3 will be there.
@Ibanez said:It's 22.Basically (n+1)^2 - n^2 = 2n+1 = an Odd number. So all odd numbers can be expresed as the sum of two consecutive integers.Now triangular number is of the form x(x+1)/2. With a little observation u see that for x=44 u will get 1000.So the values of x satisfying are 1,2,5,6,9,10,......till 41,42 So total 22 values.
Ans is 21
@vbhvgupta
Should be 21..
Given, n(n+1)/2 = (m+1)^2 - m^2...And, n(n+1)/2
=>n(n+1) = 4m + 2
=> n^2 + n - 4m = 2..
Thus, Possible Solns' for n and m are (2,1), (5,7), (6,10), (9,22), (10,27)....(42, Something)
Thus, a total of 21 pairs..
@Ibanez said:It's 22.y then 3,4 don't then 5,6 do then 7,8 don't. Basically for every four values of x only two will satisfy. With a little observation u see that for x=44 u will get 1000.
Need to ignore zero I guess. ? :splat:
@vbhvgupta said:(201* 202 * 203* 204* 246 * 247 * 248 * 249)^2 Last 2 digits????
Is it 76??
just take remainder by 25 of each term
(01*02*03*04*(-04)*(-03)*(-02)*(01))^2 = 24^4 = 24^even = xx76
@Estallar12 said:Need to ignore zero I guess. ?
Yes indeed. Edited that 😃 Question specifies "natural nos" :P
@Logrhythm said:Is it 76?? just take remainder by 25 of each term(01*02*03*04*(-04)*(-03)*(-02)*(01))^2 = 24^4 = 24^even = xx76
Yar what is the concept behind it....like for some other no's also do we have to divide those no by 25 to get the last 2 digit no????
@vbhvgupta said:(58! - 38!) Highest power of 3 available in expression?
26 for 58! and 17 for 38! ==> 58! - 38! = 3^26 *a - 3^17 * b = 3^17 ( 3 ^9 a - b )
so highest power of 3 is 17 :)
so highest power of 3 is 17 :)