Count me in this time
@[562370:chetso]: Thanks for reverting back.
Captcha daal daal ke jaan nikal gai. :/
@Predilected said: remainder 3^268/22=????
let x y z be 13p 13 q 13r resp.
p+q+r = 9
but each of p q r has to be +ve
put p' = p+1 ... similarly for q and r
p'+q'+r' = 6
so 8c2 = 28 ways
but when all are equal i.e at p=q=r = 3 x y z wont be co prime
so total 28-1 = 27 triplets
The HCF of three natural numbers x, y and z is 13. If the sum of x, y and z is 117, then how many ordered triplets (x, y, z) exist?
let the numbers be 13a,13b,13c
13a+13b+13c = 117
a+b+c = 9
8c2 = 28
- case when all three are equal
B.Com (H) from DU, 3 yrs work ex in advertising. Am lookin at the top 15 in the US. Might apply to R3 of one or two but mainly targeting R1 in Sept for class of 2012-14