Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

Re: remainder = 4 ? [#permalink]
28 May 2007, 11:08

veekayem wrote:

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.