IIT JEE , AIEEE Forum - Mathematics , Algebra

nick

give me a easy(objective) method for this..

32^32^32 that is 32 raised to 32 raised to 32 when divided by 7

gives remainder_____________

nano0101

let me try this one

Asmita

d ans. is either 2 or 4........

Asmita

32=4(mod7)

32^32=4^32(mod7)

=2^64(mod7)

now v can easily c tht 2^p(mod7) follows d pattern 2,4,1,2,4,1..........

so 32^32=2(mod7)

32^32^32=2^32(mod7)

=4(mod7)

so v can easiliy c tht d remainder will b either 2 or 4 depending on d no. of tyms d term is raised 2 d power of 32.................

m i correct????

nick

here i got two methods

(1) 32^32^32 can be written as

2^5^1024=2^5120

now 2 leaves remainder 2 when divided by 7
2^2 4

2^3 1

and after this it follows this sequence
therefore
2^5118 will leave remainder 1

2^5119 as 2

and 5^5120 as 4 which is the ans..

plsss rate PEOPLE

nitin62225

ans=4

soln:
it can b written as 4*8¹⁷⁰⁶=4(7+1)¹⁷⁰⁶ so the remainder is 4 when it's divided by 7.

Asmita

ya d ans. is 4 bcoz here d no. of tyms is odd.................

actually i gave my ans. 4 d general case...............

nick

another one

let us consider
32^32
this can be expanded as (28+4)^32
if using binom. theorem we expand it we can take 7 common
but the last term which will be 4^32 will be left

so we can write 32^32 as (7k+4^32)

now (7k+4^32)^32..........1
can be expanded
and 7 again can taken common leaving 4^32^32 only

so 1 can be written as 7m+4^1024

4 when divided by 7 leaves R as 4
4^2 as 2
4^3 as 1
after this it follows this sequence...

so 4^1024 leaves remainder 4 when diveded by 7

hence 1 can be written as 7n+4

plssss rate my efforts

Asmita

a=b(mod c) means tht wen a is divided by c thn b is d remainder.........

tarun_bits

please tell me more about mod and explani more plz

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