Problems....
The sum of 20 distinct numbers is 801. What is their minimum LCM possible?
Choose one answer.
a. 360
b. 480
c. 42
d. 840
What is the remainder when (17)^36 + (19)^36 is divided by 111?
hi there...
thx a lot for ur solution but i think it's going wrong at the last step...don't u think we hv to apply chinese theorem at the last??
regards,
shaheen
Yes we would have required to find out the values of such numbers for which the reminder is 1 if the reminder was not 1 for any of the numbers
But as here all the reminders are 1 there is no need to do anything else.
any how what else you want to prove using CRT
no! i posted a wrong solution but i could not erase it . the editor wont let me. hence the 1 . sorry for the confusion.
anyways i have a serious doubt which i dont have time to verify !!
i thought xn + yn is divisible by (x+y) but it doesnot seem that way. can some one tell me what is the flaw.
Anyways ,THIS time i am a hell lot busy and i am gonna return only 5 days before CAT . what a mess.
xn + an is exactly divisible by (x + a) if n is odd, but not if n is even
For quick reference you can check other similar conditions @
111 = 3 x 37
Euler Function of 37 is 36 [ 37 x ( 1- 1/37) ]
So both the numbers gives reminder 1 when divided by 1
1736 = 1 mod (36)
1936 = 1 mod (36)
Agaian reminder of 1736 divided by 3
=reminder of ( 18 - 1) 36 divided by 3
=reminder of ( - 1) 36 divided by 3
=1
Similarly for 1936
reminder of 1936 divided by 3
= reminder of (18+1)36 divided by 3
= 1
Thus the answer = 1. 1 + 1. 1 = 2