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When positive integer y is divided by 51, the remainder is x. What is the remainder when y is divided by 17?

(1) When y is divided by 34 the remainder is zero.
(2) x = 17

Given - y is divided by 51, the remainder is x.

y = 51Q+x (Q is an integer) or y = 17 (3Q) + X. ; have to find out the remainder when y is divided by 17.

Statement 1:

y is a multiple of 34. Values that suit this are: 0, 34, 102 etc. All of these are divisible by 17 (as 34 = 17 x 2).

Hence sufficient as remainder for each of the values is 0.

Statement 2:

x = 17.

From the question stem we have y = 17 (3Q) + x. i.e. y = 17 (3Q) + 17.

This means that y = 17 (3Q+1). Always a multiple of 17.

Sufficient.

Answer D.
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Bunuel PKN gmatbusters chetan2u VeritasKarishma

I have a query regarding solution provided by CV7585

Quote:
Given - y is divided by 51, the remainder is x.

y = 51Q+x (Q is an integer) or y = 17 (3Q) + X. ; have to find out the remainder when y is divided by 17.

Statement 1:

y is a multiple of 34. Values that suit this are:0, 34, 102 etc. All of these are divisible by 17 (as 34 = 17 x 2).

Hence sufficient as remainder for each of the values is 0.

Is not the smallest multiple of a number (ignoring negative integer values as appropriate to this Q) the number itself and not zero?
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adkikani
Bunuel PKN gmatbusters chetan2u VeritasKarishma

I have a query regarding solution provided by CV7585

Quote:
Given - y is divided by 51, the remainder is x.

y = 51Q+x (Q is an integer) or y = 17 (3Q) + X. ; have to find out the remainder when y is divided by 17.

Statement 1:

y is a multiple of 34. Values that suit this are:0, 34, 102 etc. All of these are divisible by 17 (as 34 = 17 x 2).

Hence sufficient as remainder for each of the values is 0.

Is not the smallest multiple of a number (ignoring negative integer values as appropriate to this Q) the number itself and not zero?

Hi adkikani,

MULTIPLE=ANY NUMBER*AN INTEGER(+/-/0)

Here, y is a multiple of 34 IMPLIES THAT
y=any number (34)* AN INTEGER(+/-/0)

Hope it helps.

You may raise further queries (if any)
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Bunuel
When positive integer y is divided by 51, the remainder is x. What is the remainder when y is divided by 17?

(1) When y is divided by 34 the remainder is zero.
(2) x = 17

y/51 = q + x/51 rearranged y = 51q + x

1)
y/34 = b + 0/34 rearranged y = 34b + 0 if we let b = 0 then y = 0 and 0/34 remainder 0
If we let b= 1 then y = 34 and 34/34 = 1.0 remainder zero.

Now since 0,34,68 etc.. are multiples of 17 so the remainder will be zero.
Sufficient

2) x = 17

from question stem if we let q = 0 we get y= 17

17/17 = 1.0 remainder zero

If q = 1 then y = 68 and 68/17 = 4.0 remainder zero

Sufficient

Answer choice D.

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adkikani
Bunuel PKN gmatbusters chetan2u VeritasKarishma

I have a query regarding solution provided by CV7585

Quote:
Given - y is divided by 51, the remainder is x.

y = 51Q+x (Q is an integer) or y = 17 (3Q) + X. ; have to find out the remainder when y is divided by 17.

Statement 1:

y is a multiple of 34. Values that suit this are:0, 34, 102 etc. All of these are divisible by 17 (as 34 = 17 x 2).

Hence sufficient as remainder for each of the values is 0.

Is not the smallest multiple of a number (ignoring negative integer values as appropriate to this Q) the number itself and not zero?


If you try 2*0 = 0 also 0/2 = 0
2*1 = 2

Etc..

Zero is a multiple of every number except itself.

If you also take a look at this link it would help.

https://gmatclub.com/forum/is-0-zero-to ... 04179.html

Posted from my mobile device
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When positive integer y is divided by 51, the remainder is x. What is the remainder when y is divided by 17?

(1) When y is divided by 34 the remainder is zero.
(2) x = 17[/quote]

The question states, y = (51a + x), y=17b + ?
From (1), (51a+x) divided by 34 gives remainder 0. That means, 51a & x, both are multiple of 34.
34 = 17*2. Thus (51a+x) divided by 17 will give remainder 0
From (2), y= (51a+ 17) = (17*3a + 17) divided by 17 will give remainder 0.
Thus answer is D. Both should be sufficient.
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