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Am not sure you read this correctly. ^ signifies raised to the power of. You have used multiplication(*) here or is there a super-duper simplification I just cant see

Now we find a pattern for powers of 2 divided by 7 2^0 = 0 2^1=0 2^2=0 2^3=1 2^4=2 2^5=0Actually 32/7 leaves 4 as reminder 2^6=1 2^7 = 2 2^8 = 4 2^9=1 Here the repetition starts. Every 3 nos. So 5120/3 gives a rmainder of 2. Since every 3rd power is 1. this should be 0

Ans. 0

32^32^32 = 2^1600

We substract 2 from 1600 and get 1598 which we divide on 3 and get 2 as reminder => means that we have to take #2 from the pattern you discovered => reminder is 2

2 power 1 - remainder 2
2 power 2 - remainder 4
2 power 3 - remainder 1
2 power 4 - remainder 2
2 power 5 - remainder 4
2 power 6 - remainder 1
and so on...

Basically we need to find out the remainder of 2 power 5120.

We can separate the series above like this:

Remainder 2 series - 2^1,2^4,2^7....
Remainder 4 series - 2^2,2^5,2^8....
Remainder 1 series - 2^3,2^6,2^9....

So if we find out into which series 5120 falls we know the remainder.
This can be done by using the progressions formula. nth term of a AP is = a + (n-1)d.

Putting values for each we get that 5120 belongs to the remainder 4 series.

2 power 1 - remainder 2 2 power 2 - remainder 4 2 power 3 - remainder 1 2 power 4 - remainder 2 2 power 5 - remainder 4 2 power 6 - remainder 1 and so on...

Basically we need to find out the remainder of 2 power 5120.

We can separate the series above like this:

Remainder 2 series - 2^1,2^4,2^7.... Remainder 4 series - 2^2,2^5,2^8.... Remainder 1 series - 2^3,2^6,2^9....

So if we find out into which series 5120 falls we know the remainder. This can be done by using the progressions formula. nth term of a AP is = a + (n-1)d. Putting values for each we get that 5120 belongs to the remainder 4 series.

Hence remainder is 4.

Aren't you suppose to use GP formula with exponents ?

32^1 = 32 - units digit = 2
31^2 = 32*32 - units digit = 4
32^3 = 32*32*32 - units digit = 8
32^4 = 32*32*32*32 - units digit = 6
32^5 = 32*32*32*32*32 - units digit = 2

the pattern is: 2,4,8,6,2,4,8,6, etc. at the 32nd term the cycle ends on a 6.
now, the cycle restarts for 32 more iterations. but this time the cycle starts at 6: 6, 2, 4, 8, 6, 2, 4, 8 etc. the 32nd term ends on a 8. my answer, 8/7 = 1

Please post the OA and the OR

Last edited by ggarr on 28 May 2007, 20:18, edited 3 times in total.

2 power 1 - remainder 2
2 power 2 - remainder 4
2 power 3 - remainder 1
2 power 4 - remainder 2
2 power 5 - remainder 4
2 power 6 - remainder 1
and so on...

The progressions would be:

1,4,7...... is 5120 in here ? If so remainder 2
2,5,8...... is 5120 in here ? If so remainder 4
3,6,9...... is 5120 in here ? If so remainder 1

using Arithmetic Progression formula:

AP = a + (n-1)d

a = first value = 1,2,3

d = difference = 3

will give you:

Group 1 (numbers with remainder 2) 5120 = 1 + (n-1)*3 ,n = 1,707.3333
Group 2 (numbers with remainder 4) 5120 = 2 + (n-1)*3, n = 1,707
Group 3 (numbers with remainder 1) 5120 = 3 + (n-1)*3, n = 1,706.666

2^5120 is parst of group 2, hence the remainder is 4.

Method two

32^32^32

32^1024 where (32 is 7*4+4) and we need only the remainder:

4^1024 = 16^512 where (16 is 7*2+2) and we need only the remainder:

2^512 = 4^256 = 16^128 where (16 is 7*2+2) and we need only the remainder:

2^128 = 4^64 = 16^32 where (16 is 7*2+2) and we need only the remainder:

2^32 = 4^16 = 16^8 where (16 is 7*2+2) and we need only the remainder:

2^8 = 4^4 = 16^2 where (16 is 7*2+2) and we need only the remainder:

2^2 = 4

the remainder is 4.

thanks all (AdrianG, veekayem, Dek, vidyasagar, ggarr, waldeck55) for the fruitful discussion !!

we are still waiting for the OA !

Last edited by KillerSquirrel on 30 May 2007, 12:13, edited 1 time in total.