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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:

A. 0
B. 1
C. 69
D. 35
E. 42

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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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3
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, 02:30
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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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 divided by 70 leaves remai  [#permalink]

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11 Sep 2018, 22:16
2
rahul16singh28 wrote:
If $$x = (16^3 + 17^3 + 18^3 + 19^3)$$, then x divided by 70 leaves remainder of

1. 0
2. 1
3. 69
4. 35
5. 42

Note:- $$a^n+b^n$$ is divisible by (a+b) when n is ODD.

In line with the above, $$a^3+b^3$$ is divisible by (a+b)

a) $$17^3 + 18^3$$ is divisible by 17+18=35
b) $$16^3 + 19^3$$ is divisible by 16+19=35
c) Sum of unit digits of 17^3 + 18^3, 3+2=5, Sum of unit digits of 16^3 + 19^3, 6+9=15. So the given expression is an EVEN number

Hence from (a),(b), and (c), we have
$$x = (16^3 + 17^3 + 18^3 + 19^3)$$=$$17^3 + 18^3+16^3 + 19^3=35k+35p=35*2*y=70y$$
So, the given expression when divided by 70 leaves a remainder 0.

Ans. (A)

Another approach:-
$$a^n+b^n+c^n+d^n$$ is divisible by (a+b+c+d) when n is ODD , a, b, c, and d are in A.P.(Arithmetic progression)
Here 16,17,18, and 19 are in AP. n=3, which is ODD’
Hence, $$x = (16^3 + 17^3 + 18^3 + 19^3)$$ is divisible by (16+17+18+19=70)

Therefore, remainder is zero.
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Re: If x = (16^3 + 17^3 + 18^3 + 19^3), then x divided by 70 leaves remai  [#permalink]

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11 Sep 2018, 22:36
a^3+b^3=(a+b)(a^2+b^2-ab)
16^3+19^3=35k
17^3+18^3=35k
So x=35k +35k=70k
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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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12 Sep 2018, 03:43
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.

Hi....

would the following logic work?

16^3 + 17^3 is iv by 33
18^3 + 19^3 is div by 37

therefore

16^3 + 17^3 + 18^3 + 19^3 is div by 33+37 or 70

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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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12 Sep 2018, 05:23
Mansoor50 wrote:
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.

Hi....

would the following logic work?

16^3 + 17^3 is iv by 33
18^3 + 19^3 is div by 37

therefore

16^3 + 17^3 + 18^3 + 19^3 is div by 33+37 or 70

regards

a is divisible by 3 (say a = 6)
and b is divisible by 5 (say b = 5)

Does this mean (a + b = 11) is divisible by 8? It is certainly not necessary.
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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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12 Sep 2018, 06:04
Mansoor50 wrote:
AnkitK wrote:
If x = (16^3+17^3+18^3+19^3), then x when divided by 70 leaves a remainder of :-

Hi....

would the following logic work?

16^3 + 17^3 is iv by 33
18^3 + 19^3 is div by 37

therefore

16^3 + 17^3 + 18^3 + 19^3 is div by 33+37 or 70

regards

a is divisible by 3 (say a = 6)
and b is divisible by 5 (say b = 5)

Does this mean (a + b = 11) is divisible by 8? It is certainly not necessary.

Thanks!!! i should have tested the assumption myself....!!
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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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12 Sep 2018, 09:42
1
This question can be solved by focusing on only the units digit of x and the answer choices. The units digit of the answer choices are all unique. So, by knowing only the units digit of x we can know the units digit of the right answer.

This would NOT work if the units digit of the answer choices were all the same - in this case say 0, 10, 20, 30, and 40. But, of course, the GMAT is a test of your ability to think and simplify, not of your ability to do a bunch of arithmetic, so this is the type of thing you want to be on the lookout for.

So in this case

The units digit of 16 cubed is 6
The units digit of 17 cubed is 3
The units digit of 18 cubed is 2
The units digit of 19 cubed is 9

6 + 3 + 2 + 9 = 20. So, x has a units digit of 0. So, the remainder of x divided by 70 will also have a units digit of 0. A is the only answer choices that works.

A is correct.

Think before you math
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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] 12 Sep 2018, 09:42
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