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If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder

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jade3
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If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder [#permalink] New post   05 Nov 2009, 11:59
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If \(22^3+23^3+24^3+....+87^3+88^3\) is divided by 110 then the remainder will be

A. 55
B. 1
C. 33
D. 0
E. 44

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Re: If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder [#permalink] New post   05 Nov 2009, 21:21
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jade3 wrote:
If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder will be

A. 55
B.1
C.33
D.0
E.44


(a^n + b^n ) is divisible by a+b , given that n is odd.

we need to form pairs of numbers whose sum is 110.

(22^3 + 88^3 ) + (23^3 + 87^3 ) ...... + 55^3

so all pairs are divisibly by 110 , which leave 55^3/110 , and the remainder for this is 55.

so IMO A.
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Re: If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder [#permalink] New post   05 Nov 2009, 22:18
jax91 wrote:
jade3 wrote:
If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder will be

A. 55
B.1
C.33
D.0
E.44


(a^n + b^n ) is divisible by a+b , given that n is odd.

we need to form pairs of numbers whose sum is 110.

(22^3 + 88^3 ) + (23^3 + 87^3 ) ...... + 55^3

so all pairs are divisibly by 110 , which leave 55^3/110 , and the remainder for this is 55.

so IMO A.


Correct Jax91. the answer is 55
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Re: If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder [#permalink] New post   06 Nov 2009, 00:26
I wud differ with the final answer. if jade3's solution method is agreed upon, we get 55^3/110. which is (55*55*55)/(55*2) or (55*55)/2. Thus the odd number divided by 2 will result into remainder 1.

In that case the OA should B.
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Re: If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder [#permalink] New post   06 Nov 2009, 03:01
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kalpeshchopada7 wrote:
I wud differ with the final answer. if jade3's solution method is agreed upon, we get 55^3/110. which is (55*55*55)/(55*2) or (55*55)/2. Thus the odd number divided by 2 will result into remainder 1.

In that case the OA should B.


The colored step cannot be used because, while finding the remainder you cannot reduce the numbers(Ex:when you divide 32/64 the remainder is 32, But if you reduce 32/64 to 1/2 you get the remainder 1)

According to remainder theory

If P be the product of N1, N2, N3…

And let D divide P, then the remainder of P/D would be same as remainder of (Mod[N1/D]*Mod[N2/D]*Mod[N3/D]*…..)/D where Mod denotes the remainder operation

In our case Remainder of (55*55*55/110)= Remainder of (3025*55/110)
Remainder of(3025/110) = 55
Remainder of(55/110)=55
Now according to the remainder theorem Remainder of (55*55*55/110)= Remainder of (55*55/110)=55
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Re: If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder [#permalink] New post   06 Nov 2009, 05:09
Thanks Jade3. I acknowledge my mistake. +1 from me!!
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Re: If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder [#permalink] New post   06 Nov 2009, 21:01
Can somebody prove this statement for me please:
(a^n + b^n ) is divisible by a+b , given that n is odd.

Thanks
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Re: If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder [#permalink] New post   06 Nov 2009, 21:56
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a^3+b^3 = (a+b)(a^2-ab+b^2).
a^5+b^5 = (a+b){a^4 - (a^3)b + (a^2)(b^2) - a(b^3) + b^4)}
....
Like wise,
a^n+b^n [If n is odd] = (a+b){a^(n-1) - [a^(n-2)]*b + [a^(n-3)]*b^2 +.....+ b^n-1}

Ezinis, Guess this should be clear. You can try solving for yourself.
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Re: If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder [#permalink] New post   06 Nov 2009, 22:09
ezinis wrote:
Can somebody prove this statement for me please:

(a^n + b^n ) is divisible by a+b , given that n is odd.

Thanks


Given that n is odd:

(a^n + b^n ) = a^(n-1) - a^(n-2)b + a^(n-3)b^2 - a^(n-4) b^3 + ........... - ab^(n-1) + b^n

Suppose n = 5:
(a^5 + b^5 ) = (a + b) (a^4 - a^3b + a^2b^2 - ab^3 + b^4)

Suppose n = 7:
(a^7 + b^7 ) = (a + b) (a^6 - a^5b + a^4b^2 - a^3b^3 + a^2b^4 - ab^5 + b^6)
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Re: If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder [#permalink] New post   Updated on: 07 Nov 2009, 11:03
ezinis wrote:
Can somebody prove this statement for me please:
(a^n + b^n ) is divisible by a+b , given that n is odd.

Thanks


Let us try backwards
Let n be odd then

(x+y)(x^(n-1) - x^(n-2) y + ... --x y^(n-2) + y^(n-1)) = (x^n-x^(n-1)y+……..-x y^(n-1)+y^n)=x^n+y^n
Because, you will notice that most of the terms would cancel in pairs; and the result would be x^n+y^n

You can try for n=3

(x+y)(x^2-x y+y^2)=x^3-x^2 y+ x y^2+ y x^2-y^2 x+y^3

You notice that the second term and fourth term cancel each other
Similarly You notice that the third term and fifth term cancel each other

You get the result as x^3+y^3

Now can try for x=5 yourself.

Originally posted by jade3 on 06 Nov 2009, 22:29.
Last edited by jade3 on 07 Nov 2009, 11:03, edited 2 times in total.
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Re: If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder [#permalink] New post   07 Nov 2009, 10:38
thanks guy +1. good one to remember
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Re: If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder [#permalink] New post   09 May 2011, 04:28
interesting one indeed.
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Re: If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder [#permalink] New post   15 Jan 2019, 11:18
I don't get how you all guys get the remainder of 55^3/110 without any complicated calculations. Can someone explain?

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Re: If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder [#permalink]
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