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What will be the remainder when 13^7 + 14^7 + 15^7 + 16^7 is

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What will be the remainder when 13^7 + 14^7 + 15^7 + 16^7 is [#permalink]

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New post 05 Aug 2008, 00:51
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

What will be the remainder when 13^7 + 14^7 + 15^7 + 16^7 is divided by 58?
57
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Re: funny remainder... [#permalink]

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New post 05 Aug 2008, 01:28
D 0.

2.5 minutes :roll:

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Re: funny remainder... [#permalink]

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New post 05 Aug 2008, 01:42
Is this really a GMAT Q?

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Re: funny remainder... [#permalink]

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New post 05 Aug 2008, 01:47
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13 = 58 - 45
14 = 58 - 44
15 = 58 - 43
16 = 58 - 42

13^7 + 14^7 + 15^7 + 16^7
= (58-45)^7 + (58-44)^7 + (58-43)^7 + (58-42)^7

if we open this ... in all (58-x)^7. there be 7 terms with powers of 58 and one term with only x
all powers of 58 will be divisible by 58
we just need to check 45+44+43+42 = 174 also divisible by 58

remainder 0

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Re: funny remainder... [#permalink]

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New post 05 Aug 2008, 01:55
durgesh79 wrote:
13 = 58 - 45
14 = 58 - 44
15 = 58 - 43
16 = 58 - 42

13^7 + 14^7 + 15^7 + 16^7
= (58-45)^7 + (58-44)^7 + (58-43)^7 + (58-42)^7

if we open this ... in all (58-x)^7. there be 7 terms with powers of 58 and one term with only x
all powers of 58 will be divisible by 58
we just need to check 45+44+43+42 = 174 also divisible by 58

remainder 0


Thanks ... makes sense!

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Re: funny remainder... [#permalink]

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New post 05 Aug 2008, 03:57
durgesh79 wrote:
13 = 58 - 45
14 = 58 - 44
15 = 58 - 43
16 = 58 - 42

13^7 + 14^7 + 15^7 + 16^7
= (58-45)^7 + (58-44)^7 + (58-43)^7 + (58-42)^7

if we open this ... in all (58-x)^7. there be 7 terms with powers of 58 and one term with only x
all powers of 58 will be divisible by 58
we just need to check 45+44+43+42 = 174 also divisible by 58

remainder 0


I am not clear on this. The x terms all have power 7. How do you deal with that ?
If 45+44+43+42 = 174 is divisible by 58, is 45^7 + 44^7 + 43^7...... also divisible by 58 ?

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Re: funny remainder... [#permalink]

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New post 05 Aug 2008, 04:11
bhushangiri wrote:
durgesh79 wrote:
13 = 58 - 45
14 = 58 - 44
15 = 58 - 43
16 = 58 - 42

13^7 + 14^7 + 15^7 + 16^7
= (58-45)^7 + (58-44)^7 + (58-43)^7 + (58-42)^7

if we open this ... in all (58-x)^7. there be 7 terms with powers of 58 and one term with only x
all powers of 58 will be divisible by 58
we just need to check 45+44+43+42 = 174 also divisible by 58

remainder 0


I am not clear on this. The x terms all have power 7. How do you deal with that ?
If 45+44+43+42 = 174 is divisible by 58, is 45^7 + 44^7 + 43^7...... also divisible by 58 ?


If you expand any ploynomial such that (x - y ) ^n, you will always find x in all terms except last. So in this Q all you have to ensure is that sum of last terms is divisble by 58 or not. If yes then sum is divisible by 58

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Re: funny remainder... [#permalink]

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New post 05 Aug 2008, 04:17
nmohindru wrote:

If you expand any ploynomial such that (x - y ) ^n, you will always find x in all terms except last. So in this Q all you have to ensure is that sum of last terms is divisble by 58 or not. If yes then sum is divisible by 58


I know that rule - binomial expansion. x will always be present in all terms except the last. But the last term will have power n.
In context of current question, the last terms are not 45, 44, 43, and 42. But 45^7, 44^7, 43^7, and 42^7

I also know 45+44+43+42 = 174 is divisible by 58.
But does that mean 45^7 + 44^7 + 43^7 + 42^7 is also divisible by 58 ?

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Re: funny remainder... [#permalink]

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New post 05 Aug 2008, 04:40
bhushangiri wrote:
nmohindru wrote:

If you expand any ploynomial such that (x - y ) ^n, you will always find x in all terms except last. So in this Q all you have to ensure is that sum of last terms is divisble by 58 or not. If yes then sum is divisible by 58


I know that rule - binomial expansion. x will always be present in all terms except the last. But the last term will have power n.
In context of current question, the last terms are not 45, 44, 43, and 42. But 45^7, 44^7, 43^7, and 42^7

I also know 45+44+43+42 = 174 is divisible by 58.
But does that mean 45^7 + 44^7 + 43^7 + 42^7 is also divisible by 58 ?


Hmmm .... you are right

I was trying another thing according to which

\(a^n + b^n + c^n + d^n\) is divisible by a + b + c + d if n is odd.

\(a^3 + b^3 = (a + b) (a^2 -ab + b^2)\)

This will give us the answer

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Re: funny remainder... [#permalink]

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New post 05 Aug 2008, 04:46
nmohindru wrote:

Hmmm .... you are right

I was trying another thing according to which

\(a^n + b^n + c^n + d^n\) is divisible by a + b + c + d if n is odd.

\(a^3 + b^3 = (a + b) (a^2 -ab + b^2)\)

This will give us the answer



Cud u plz elaborate on the rule ? What is the rule, and under what conditions does it hold ?

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Re: funny remainder... [#permalink]

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New post 05 Aug 2008, 04:56
bhushangiri wrote:
nmohindru wrote:

Hmmm .... you are right

I was trying another thing according to which

\(a^n + b^n + c^n + d^n\) is divisible by a + b + c + d if n is odd.

\(a^3 + b^3 = (a + b) (a^2 -ab + b^2)\)

This will give us the answer



Cud u plz elaborate on the rule ? What is the rule, and under what conditions does it hold ?


The rule is

\(a^n + b^n + c^n + d^n\) is divisible by a + b + c + d if n is odd.

because \(a^n + b^n + c^n + d^n\) when n is odd will become (a + b + c + d) (a^n - a^n-1*b ....... )

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Re: funny remainder... [#permalink]

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New post 05 Aug 2008, 05:10
nmohindru wrote:

The rule is

\(a^n + b^n + c^n + d^n\) is divisible by a + b + c + d if n is odd.

because \(a^n + b^n + c^n + d^n\) when n is odd will become (a + b + c + d) (a^n - a^n-1*b ....... )


I think there are additional constraints, i can think of atleast one.. that a,b,c,d need to be consecutive numbers. Coz it doesnt work with a=1, b=2, c=4. but works with 1,2,3 or 2,3,4 or 1,2,3,4.

But thanks buddy. Thats todays gain.. one new rule.

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Re: funny remainder... [#permalink]

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New post 05 Aug 2008, 05:15
bhushangiri wrote:
nmohindru wrote:

The rule is

\(a^n + b^n + c^n + d^n\) is divisible by a + b + c + d if n is odd.

because \(a^n + b^n + c^n + d^n\) when n is odd will become (a + b + c + d) (a^n - a^n-1*b ....... )


I think there are additional constraints, i can think of atleast one.. that a,b,c,d need to be consecutive numbers. Coz it doesnt work with a=1, b=2, c=4. but works with 1,2,3 or 2,3,4 or 1,2,3,4.

But thanks buddy. Thats todays gain.. one new rule.


Yes you are right .... works only for consecutive numbers

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Re: funny remainder... [#permalink]

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New post 05 Aug 2008, 05:19
arjtryarjtry wrote:
What will be the remainder when 13^7 + 14^7 + 15^7 + 16^7 is divided by 58?
57
1
30
0
28


If you know modular arithmetic:

13^7 + 14^7 + 15^7 + 16^7 is clearly even --> divisible by 2

If it's also divisible by 29, it's divisible by 58.

Notice
16 ~ -13 mod 29
15 ~ -14 mod 29

(~ means 'congruent to')

13^7 + 14^7 + 15^7 + 16^7 ~ 13^7 + 14^7 + (-14)^7 + (-13)^7 mod 29
~ 0 mod 29

So the expression is divisible by 2 and 29, and thus by 58.
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Re: funny remainder... [#permalink]

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New post 05 Aug 2008, 06:24
good discussion..

another method came to my mind while driving home today ...

13^7 + 14^7 + 15^7 + 16^7
= (14.5-1.5)^7 + (14.5-0.5)^7 + (14.5+0.5)^7 + (14.5+1.5)^7

if you open it all terms with odd powers of 1.5 and 0.5 will cancel each other.....

for every Postive term with 7C0, 7C2, 7C4 and 7C6, we can take out 4*14.5 as a common factor .... (i did that on paper) and guess what is 4*14.5... its 58 ... :wink:

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Re: funny remainder... [#permalink]

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New post 05 Aug 2008, 07:49
durgesh79 wrote:
good discussion..

another method came to my mind while driving home today ...

13^7 + 14^7 + 15^7 + 16^7
= (14.5-1.5)^7 + (14.5-0.5)^7 + (14.5+0.5)^7 + (14.5+1.5)^7

if you open it all terms with odd powers of 1.5 and 0.5 will cancel each other.....

for every Postive term with 7C0, 7C2, 7C4 and 7C6, we can take out 4*14.5 as a common factor .... (i did that on paper) and guess what is 4*14.5... its 58 ... :wink:


too good professor.. :-D

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Re: funny remainder... [#permalink]

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New post 05 Aug 2008, 08:01
bhushangiri wrote:
too good professor.. :-D


not that good... :oops:
i think i just got lucky with this question .... there are faults in both of my approaches ... somehow got the right answer :P
moral : try solving maths question while driving, your approach may be wrong, but you'll get the right answer :lol:

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Re: funny remainder... [#permalink]

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New post 05 Aug 2008, 18:48
Just wondering. How do you rate the difficulty on this question if it was on that gmat?

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Re: funny remainder... [#permalink]

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New post 12 Aug 2008, 07:05
a^n + b^n = (a+b)(a^n-a^(n-1)*b+...+b^n)

13^7+16^7 = 29*(13^6 - 13^5*16+....+16^6) ->29*an odd number (as 13^6 is odd while the others are all even)
Similarly 14^7+15^7= 29*an odd number

13^7+14^7+15^7+16^7=29*an even number=29*2k=58k
Thus, the sum is divisible by 58

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Re: funny remainder... [#permalink]

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New post 12 Aug 2008, 07:32
IanStewart wrote:
arjtryarjtry wrote:
What will be the remainder when 13^7 + 14^7 + 15^7 + 16^7 is divided by 58?
57
1
30
0
28


If you know modular arithmetic:

13^7 + 14^7 + 15^7 + 16^7 is clearly even --> divisible by 2

If it's also divisible by 29, it's divisible by 58.

Notice
16 ~ -13 mod 29
15 ~ -14 mod 29

(~ means 'congruent to')
13^7 + 14^7 + 15^7 + 16^7 ~ 13^7 + 14^7 + (-14)^7 + (-13)^7 mod 29
~ 0 mod 29

So the expression is divisible by 2 and 29, and thus by 58.


I couldn't understand the highlighted part.

Could you elaborate
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Re: funny remainder...   [#permalink] 12 Aug 2008, 07:32

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