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# What is the remainder for (32^32^32)/7 ? Will give the

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What is the remainder for (32^32^32)/7 ? Will give the [#permalink]

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28 May 2007, 04:53
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

What is the remainder for

(32^32^32)/7 ?

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VP
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28 May 2007, 08:12
The remainder is 4! - this post was edited , see followup posts.

Last edited by KillerSquirrel on 29 May 2007, 12:33, edited 1 time in total.

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28 May 2007, 10:17
0.

32^32^32 can be written as
2^5*1024
= 2^5120

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=0
2^6=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

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28 May 2007, 10:24
32*32*32 /7 = 4*4*4 /7 (remainders only)=64/7 = 1 (remainders only).

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28 May 2007, 10:28
vidyasagar wrote:
32*32*32 /7 = 4*4*4 /7 (remainders only)=64/7 = 1 (remainders only).

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

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28 May 2007, 11:23
Dek wrote:
0.

32^32^32 can be written as
2^5*1024
= 2^5120

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=0 Actually 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

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28 May 2007, 11:43
I feel that I need to explain the logic behind my above answer:

As every one knows, if we had multiplication in this problem it would be very easy to solve as vidyasagar pointed out.

(32*32*32)/7 = what is the remainder ?

To find the remainder, all you have to do is to neutralize the sevens and extract the remainders, and then to multiplay it, the result would then be:

4*4*4 = 64

the remainder is 1 = (7*9+1 = 64)

My logic is the same, just in bigger numbers (it's an easy way to find out if you master the basics ! people tend to get uneasy with big numbers)

32^32^32 = 32*32*32...... (n^32^32 times)

extratcing the remainders will give:

4*4*4*4*4....(n^32^32 times)

since 4^16 = 16^8

I can repeat this process to its basic elements and recover the remainder.

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28 May 2007, 11:47
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.

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Re: remainder = 4 ? [#permalink]

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28 May 2007, 12:08
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 ?

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28 May 2007, 13:50
Quote:
What is the remainder for

(32 ^ 32 ^ 32 ) / 7 ?

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, 21:18, edited 3 times in total.

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28 May 2007, 13:50
2^5 does leave a remainder of 4.
But I'm still getting 4 as the remainde.
OA anyone?

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28 May 2007, 14:18
KillerSquirrel..

the progressions here are for the exponents not the values itself..

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

all the above are APs. hence to find out which progression has 5120 as its member we put in the nth of AP formula.

Do you agree ?

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28 May 2007, 21:32
veekayem wrote:
KillerSquirrel..

the progressions here are for the exponents not the values itself..

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

all the above are APs. hence to find out which progression has 5120 as its member we put in the nth of AP formula.

Do you agree ?

I understand ! very nice , solving for AP where

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

so you claim that 2^5120 belongs to group 2 , hence with a remainder of 4,

this is a very clever thinking, but I think it's wrong !!

first of - your whole solution is based on the assumption that:

32^32^32 = 2^5120

I don't think this is correct ! can you please explain that logic ?

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29 May 2007, 09:21
KillerSquirrel wrote:

first of - your whole solution is based on the assumption that:

32^32^32 = 2^5120

I don't think this is correct ! can you please explain that logic ?

This is based on the mathematical law of powers of powers:
i.e. a^b^c = a^(b*c)

Note: With exponents, it works from a bottom-up principle
Therefore a^b^c = (a^b)^c, not a^(b^c)

An example would be:
x^3^4
=(x^3)(x^3)(x^3)(x^3)
=x^12
=x^(3*4)

With the above in mind,
32^32^32
= 32^1024
= (2^5)^1024
= 2^5120

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29 May 2007, 12:29
I took the liberty to summarize the above posts:

the remainder is clearly 4.

(32^32^32)/7 = what is the remainder ?

Method one

32^32^32 = 32^(32*32) = 2^(5*32*32) = 2^5120

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, 13:13, edited 1 time in total.

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29 May 2007, 21:38
Reminder is 1

In a question like above, reminder can be found out by multiplying reminders and dividing that with the original number

32/7= reminder 4
and we get 3 -4's and 4*4*4 =64

Divide 64/7 reminder 1.

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29 May 2007, 22:17
vijay2001 wrote:
Reminder is 1

In a question like above, reminder can be found out by multiplying reminders and dividing that with the original number

32/7= reminder 4
and we get 3 -4's and 4*4*4 =64

Divide 64/7 reminder 1.

Is this true in any case?

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30 May 2007, 05:04
That is a super-duper simplification
But does it work. We seem to have contradictions.
What is the OA

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30 May 2007, 05:37
vijay2001 wrote:
Reminder is 1

In a question like above, reminder can be found out by multiplying reminders and dividing that with the original number

32/7= reminder 4
and we get 3 -4's and 4*4*4 =64

Divide 64/7 reminder 1.

I think that the question ask for the remainder when (32^32^32)/7 not when (32*32*32)/7.

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30 May 2007, 13:07
Oops! I wonder if I ever get over my carelessness..

4! cannot be the reminder since the reminder has to be less than 7 and 4! is 24.

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30 May 2007, 13:07

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