# find remainder

• Sep 4th 2010, 02:34 AM
nikhil
find remainder
q 1)what is the remainder when 2^2006 is divided by 7
ans 1) 2^3 con 1 mod 7
or 2^2004 con 1 mod 7.......................1
also 2^2 con -3 mod 7.........................2
by 1 and 2
or 2^2006 con -3 mod 7
hence remainder is 4.
note: con stands for congruence

can anyone use the same procedure to find the remainder when 32^[(32)^32] is divided by 7.
my attempt
32 con 4 mod 7
or 32^32 con 4^32 mod 7
or 32^32 =7a+4^32
therefor
32^[(32)^32]=32^[(7a)+4^32] and here I am gone
in the solved example, problem was easy because of 1(2^3 con 1 mod 7) but in question 2 it is not the case.
• Sep 4th 2010, 03:59 AM
undefined
Quote:

Originally Posted by nikhil
q 1)what is the remainder when 2^2006 is divided by 7
ans 1) 2^3 con 1 mod 7
or 2^2004 con 1 mod 7.......................1
also 2^2 con -3 mod 7.........................2
by 1 and 2
or 2^2006 con -3 mod 7
hence remainder is 4.
note: con stands for congruence

can anyone use the same procedure to find the remainder when 32^[(32)^32] is divided by 7.
my attempt
32 con 4 mod 7
or 32^32 con 4^32 mod 7
or 32^32 =7a+4^32
therefor
32^[(32)^32]=32^[(7a)+4^32] and here I am gone
in the solved example, problem was easy because of 1(2^3 con 1 mod 7) but in question 2 it is not the case.

$32^{32^{32}}\equiv 4^{32^{32}}\pmod{7}$

Reduce $32^{32}$ mod 3

$32^{32}\equiv 2^{32}\equiv2^{0}\equiv1\pmod{3}$

$32^{32^{32}}\equiv4^1\equiv4\pmod{7}$
• Sep 6th 2010, 05:02 AM
nikhil
why we reduced 32^32 mod 3 and how did it got implemented.
what I understand (from last line)
32^[(32)^32] con 4 mod 7= (4+7a)^(1+3b) mod 7
how can we deduce that when (4+7a)(4+7a)^(3b) is divided by 7, remainder is 4
• Sep 6th 2010, 05:34 AM
undefined
Quote:

Originally Posted by nikhil
why we reduced 32^32 mod 3 and how did it got implemented.
what I understand (from last line)
32^[(32)^32] con 4 mod 7= (4+7a)^(1+3b) mod 7
how can we deduce that when (4+7a)(4+7a)^(3b) is divided by 7, remainder is 4

$4^3\equiv1\pmod{7}$