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Re: Remainder Problem, divisible by 5? [#permalink]
14 Nov 2009, 19:01

Hi,

I'm not sure if this is correct, but my answer is D. Both put together can be used to answer the question.

Explanation: Since 1 says x-y gives a remainder of 1 when divided by 5, this gives an option that x-y has to end in either 3 or 8. 2 says that x+y gives a remainder of 3 when divided by 5. Hence x+y has to end in either 1 or 6. If we try the possible combinations for x and y then it can only be when they end with 2 and 1 respectively. So if we square up numbers ending with 2 and 1 the unit digits will be 4 and 1 and hence sum to 5 which is divisible by 5.

Re: Remainder Problem, divisible by 5? [#permalink]
14 Nov 2009, 19:07

avinarvind wrote:

Hi,

I'm not sure if this is correct, but my answer is D. Both put together can be used to answer the question.

Explanation: Since 1 says x-y gives a remainder of 1 when divided by 5, this gives an option that x-y has to end in either 3 or 8. 2 says that x+y gives a remainder of 3 when divided by 5. Hence x+y has to end in either 1 or 6. If we try the possible combinations for x and y then it can only be when they end with 2 and 1 respectively. So if we square up numbers ending with 2 and 1 the unit digits will be 4 and 1 and hence sum to 5 which is divisible by 5.

Hence x^2+y^2 is divisible by 5.

Please correct me if I am wrong

I'm getting answer E

1. x-y/5 gives remainder 1

therefore x-y=1, 6, 11, 16, 21 insufficient because we don't know x or y

Re: Remainder Problem, divisible by 5? [#permalink]
14 Nov 2009, 21:55

I am not sure if I understood the above explanation.

But from what I see, the question is whether x^2+y^2 is divsible by 5.

So as per the explanation x^2 + y^2 is divisible by 5 since y=1 and x has to end in 7. The unit digit will always be 9+1=0 and hence divisible by 5. So answer is 'D'.

Re: Remainder Problem, divisible by 5? [#permalink]
15 Nov 2009, 04:10

Is x^2+y^2 divisible by 5? 1). When x-y is divided by 5, the remainder is 1 2). When x+y is divided by 5, the remainder is 3

Answer:

Statement 1: This tells us two things- 1) Either one of x and y is a multiple of 5 or 2) Both x and y are non -multiples of 5-( x-y=can result in multiple of 5 or non multiple of 5 ) Both x and y CANNOT be multiples of 5 -Not sufficient

Statement 2: This tells us the same thing as 1 1) Either one of x and y is a multiple of 5 or 2) Both x and y are non -multiples of 5( x+y=can result in multiple of 5 or non multiple of 5 ) Both x and y CANNOT be multiples of 5 -Not sufficient

Combining (1) and (2) Either one of x and y is a multiple of 5 or Both x and y are non -multiples of 5 this means either x2 or y2 is a multiple of 5 or both are non multiples of 5 x2+y2- this can result in a multiple of 5 or a non multiple - so Not sufficient

Re: Remainder Problem, divisible by 5? [#permalink]
15 Nov 2009, 07:38

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ctrlaltdel wrote:

Is x^2+y^2 divisible by 5? 1). When x-y is divided by 5, the remainder is 1 2). When x+y is divided by 5, the remainder is 3

First I must say this is not GMAT question. As every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

But let's deal with question as it's written. x^2+y^2 to be divisible by 5 it should be an integer first and then should have the last digit 0 or 5:

(1) \(x-y=5n+1\) --> \((x-y)^2=25n^2+10n+1\) --> \(x^2+y^2=25n^2+10n+1+2xy\). Now 25n^2+10n is divisible by 5, but what about 1-2xy? Not sufficient.

(2) \(x+y=5m+3\) --> \((x+y)^2=25m^2+30m+9\) --> \(x^2+y^2=25m^2+30m+9-2xy\). The same here. Not sufficient.

(1)+(2) Add the equations: \(2(x^2+y^2)=25n^2+10n+25m^2+30m+10=25(n^2+m^2)+10(n+3m)+10\).

\(x^2+y^2=\frac{25(n^2+m^2)}{2}+5(n+3m)+5\). Well if \(n^2+m^2\) is even, \(x^2+y^2\) is an integer and it's divisible by 5. But what if \(n^2+m^2\) is odd, then \(x^2+y^2\) is not an integer at all, hence not divisible by 5.

Can we somehow conclude that \(n^2+m^2\) is even? If we are not told that x and y are integers then not.

Answer: E.

To check this consider following pairs of x an y: x=4.5, y=3.5 and x=7, y=1, \(x=4.5\), \(y=3.5\) \(x-y=4.5-3.5=1\) first statement holds true, 1 divided by 5 remainder 1. \(x+y=4.5+3.5=8\) second statement holds true, 8 divided by 5 remainder 3. \(x^2+y^2=4.5^2+3.5^2=32.5\) not divisible by 5.

\(x=7\), \(y=1\) \(x-y=7-1=6\) first statement holds true, 6 divided by 5 remainder 1. \(x+y=7+1=8\) second statement holds true, 8 divided by 5 remainder 3. \(x^2+y^2=7^2+1^2=50\) is divisible by 5.

Two different answers. Not sufficient. Answer: E.

Now let's see if we were told that x and y are positive integers, as GMAT does. Still would be quite hard 700+ question.

Last step: \(x^2+y^2=\frac{25(n^2+m^2)}{2}+5(n+3m)+5\). Can we now conclude that that \(n^2+m^2\) is even?

\(x-y=5n+1\) and \(x+y=5m+3\). Add them up: \(2x=5(n+m)+4\) --> \(x=\frac{5(n+m)}{2}+2\). As x is an integer n+m must be divisible by 2, hence either both are even or both are odd, in any case \(n^2+m^2\) is even and divisible by 2. Hence \(x^2+y^2=\frac{25(n^2+m^2)}{2}+5(n+15m)+5\) is an integer and divisible by 5.

Re: Is x^2+y^2 divisible by 5? [#permalink]
10 Nov 2014, 00:27

In the Last step, this is done:

x-y=5n+1 and x+y=5m+3. Add them up: 2x=5(n+m)+4 --> \(x=\frac{5(n+m)}{2}+2\). As x is an integer n+m must be divisible by 2, hence either both are even or both are odd, in any case n^2+m^2 is even and divisible by 2. Hence \(x^2+y^2=\frac{25(n^2+m^2)}{2} + 5(n+3m) + 5\) is an integer and divisible by 5.

Once we assume the question states that x and y are +ve integers, x^2 and y^2 have to be integers. In that case can we not skip all of the above, because the underlined rule applies for (n^2+m^2) in the aforementioned Last step ( i.e \(x^2+y^2=\frac{25(n^2+m^2)}{2} + 5(n+3m) + 5\)) too, right? I mean, As x^2+y^2 is an integer, \((n^2+m^2)\) must be to be divisible by 2, right?

Thank you

gmatclubot

Re: Is x^2+y^2 divisible by 5?
[#permalink]
10 Nov 2014, 00:27

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...