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If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai

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If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai [#permalink]

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14 Jul 2011, 01:45
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If $$x = (16^3+17^3+18^3+19^3)$$, then x when divided by 70 leaves a remainder of:

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Re: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai [#permalink]

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14 Jul 2011, 02:21
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AnkitK wrote:
If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remainder of :-

OA. 0

We know:
$$a^3 + b^3 = (a + b)*(a^2 - ab + b^2)$$

$$16^3 + 19^3 = (35)(16^2 - 16*19 + 19^2)$$
$$17^3 + 18^3 = (35)(17^2 - 17*18 + 18^2)$$

Therefore, $$(16^3+17^3+18^3+19^3) = 35(16^2 - 16*19 + 19^2 + 17^2 - 17*18 + 18^2)$$

$$(16^2 - 16*19 + 19^2 + 17^2 - 17*18 + 18^2)$$ is a multiple of 2 since we have (Even - Even + Odd + Odd - Even + Even). Hence the entire expression is Even.
Therefore, x is divisible by 70.

Why did I think of algebraic identities? - because exponents were 3 for all the terms and there seemed to be no +1, -1 symmetry. The numbers weren't close to a multiple of 70 and were too big. Most people don't know the squares/cubes of 16, 17 etc.
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Re: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai [#permalink]

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14 Jul 2011, 02:26
Good work karishma!!cheers
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Re: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai [#permalink]

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14 Jul 2011, 02:30
AnkitK wrote:
If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remainder of :-

OA. 0

Hi Ankit, I dont think GMAT will ask this , but to answer your question , the best way to do this is
a^+b^3 = (a+b)(a^2 -ab +b^2)
( 16^3+19^3) + (17^3+18^3)
(16+17)( 16^2 -16*19+19^2) + (17+18)(17^2-17*18 +18^2)

35*( 16^2 -16*19+19^2) (17^2-17*18 +18^2)

.
so we can re write the equation as
35^(even -even+odd)(even-even+odd)
odd*odd = even = atleast one 2

35*2 (...........) (...........)
when this equation is divided by 70 , we will have 0 as the remainder.

hope this helps.
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Re: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai [#permalink]

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14 Jul 2011, 13:51
Thanks for solution
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Re: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai [#permalink]

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15 Jul 2011, 23:33
AnkitK wrote:
If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remainder of :-

OA. 0

A good to know question IMO but still I would take it this way-

At first glance I read a symmetry here as I break the $$A^3 + B^3$$formula- pair of 16, 19 and of 17,18 give me a total of 35 (something useful considering I am looking for divisibility by 70. Now my job is half done but I have still to check whether the remaining is even - to check for divisibility by 2).

after opening the formula $$A^3 + B^3$$ nd rearranging terms one would get-

Lets recall - $$(A^3 + B^3) = (A+B)(A^2 + B^2 - A*B)$$..
so pair 1 gives- $$(16^2 + 19^2 - 16*19 ) (16+19) = 35 * (16^2 + 19^2 - 16*19 )$$
pair 2 gives- $$(17^2 + 18^2 - 17*18) (17+18) = 35 * (17^2 + 18^2 - 17*18)$$
now lets add them, so we have got now-
$$(35) * (16^2 + 17^2 + 18^2 + 19^2 - 16*19 - 17*18) -$$
to get down finally we need to determine whether $$(16^2 + 17^2 + 18^2 + 19^2 - 16*19 - 17*18)$$ is divisible by 2 - (Even + Odd + Even + Odd - Even - Even).
You see we are done here (obviously remainder is big ZERO
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Re: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai [#permalink]

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16 Jul 2011, 02:17
Thanks Karishma! good technique to remember for similar questions.
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Re: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai [#permalink]

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16 Jul 2011, 02:22
Karishma.. Thanks a lot for the solution and technique used..
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Re: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai [#permalink]

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Re: If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remai   [#permalink] 03 Oct 2017, 06:18
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