Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 24 Jul 2009
Posts: 73
Location: United States

Is x^2+y^2 divisible by 5? [#permalink]
Show Tags
14 Nov 2009, 18:56
1
This post received KUDOS
7
This post was BOOKMARKED
Question Stats:
44% (02:05) correct
56% (01:23) wrong based on 174 sessions
HideShow timer Statistics
Is x^2+y^2 divisible by 5?
(1) When xy is divided by 5, the remainder is 1 (2) When x+y is divided by 5, the remainder is 3



Intern
Joined: 12 Nov 2009
Posts: 7

Re: Remainder Problem, divisible by 5? [#permalink]
Show Tags
14 Nov 2009, 20:01
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 xy gives a remainder of 1 when divided by 5, this gives an option that xy 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



VP
Joined: 05 Mar 2008
Posts: 1469

Re: Remainder Problem, divisible by 5? [#permalink]
Show Tags
14 Nov 2009, 20:07
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 xy gives a remainder of 1 when divided by 5, this gives an option that xy 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. xy/5 gives remainder 1 therefore xy=1, 6, 11, 16, 21 insufficient because we don't know x or y 2. x+y/5 gives remainder 3 therefore x+y = 3, 8, 13, 18, 23 insufficient the formula is y = qx + r equation 1: xy = 5q + 1 equation 2: x+y = 5q + 3 combine equations and solve x+y = x  y  1 + 3 2y = 2 y = 1 if y = 1 x can be 7 7+1 = 8 and divided by 5 gives remainder 3 71 = 6 and divided by 5 gives remainder 1 however 17 + 1 = 18 and divided by 5 gives remainder 3 171 = 16 and divided by 5 gives remainder 1 so answer E



Intern
Joined: 12 Nov 2009
Posts: 7

Re: Remainder Problem, divisible by 5? [#permalink]
Show Tags
14 Nov 2009, 22:55
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'.



Intern
Joined: 26 Sep 2009
Posts: 10

Re: Remainder Problem, divisible by 5? [#permalink]
Show Tags
15 Nov 2009, 05:10
Is x^2+y^2 divisible by 5? 1). When xy 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( xy=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
Answer (E)



VP
Joined: 05 Mar 2008
Posts: 1469

Re: Remainder Problem, divisible by 5? [#permalink]
Show Tags
15 Nov 2009, 08:12
just realized..i didn't even answer the question...I apologize..I think the answer should be d not e



Math Expert
Joined: 02 Sep 2009
Posts: 39723

Re: Remainder Problem, divisible by 5? [#permalink]
Show Tags
15 Nov 2009, 08:38
7
This post received KUDOS
Expert's post
5
This post was BOOKMARKED
ctrlaltdel wrote: Is x^2+y^2 divisible by 5? 1). When xy 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) \(xy=5n+1\) > \((xy)^2=25n^2+10n+1\) > \(x^2+y^2=25n^2+10n+1+2xy\). Now 25n^2+10n is divisible by 5, but what about 12xy? Not sufficient. (2) \(x+y=5m+3\) > \((x+y)^2=25m^2+30m+9\) > \(x^2+y^2=25m^2+30m+92xy\). 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, \(x=4.5\), \(y=3.5\) \(xy=4.53.5=1\) 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. \(x=7\), \(y=1\) \(xy=71=6\) 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? \(xy=5n+1\) 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. In this case answer C.Hope it's clear.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 24 Jul 2009
Posts: 73
Location: United States

Re: Remainder Problem, divisible by 5? [#permalink]
Show Tags
15 Nov 2009, 08:45
Bunuel, you rock...



Intern
Joined: 18 Nov 2009
Posts: 2

Re: Remainder Problem, divisible by 5? [#permalink]
Show Tags
18 Nov 2009, 20:53
That was tough, you are scaring me. I just joined yesterday with a hope that i will get good DS questions. This is too tough



Senior Manager
Joined: 21 Mar 2010
Posts: 310

Re: Remainder Problem, divisible by 5? [#permalink]
Show Tags
26 Feb 2011, 09:26
Hmm im thinking the key to any remainder question is the equation! will try to implement this as a thumb rule!



Intern
Joined: 10 Jan 2012
Posts: 2

Re: no.prop [#permalink]
Show Tags
11 Jan 2012, 02:34
1 case (xy)/5 > 1 (Remainder) (xy)^2/5> 1(Remainder) nothing can be deduced for remainder of x^2+y^2 divided by 5 insufficient
2 case
(x+y)/5>3 (remainder) (x+y)^2/5>4(remainder)
insufficient as nothing can be deduced for remainder of x^2+y^2 divided by 5
however adding two cases 2(x^2+y^2)/5> remainder 0 (1+4=5 divisible by 5) hence divisible by 5
hence both the cases are required
alternatively take x=12, y=6 and cross check
hence answer is c
what is OA



Manager
Joined: 03 Oct 2009
Posts: 62

Re: Is x^2+y^2 divisible by 5? 1). When xy is divided by 5, the [#permalink]
Show Tags
17 Jan 2012, 22:59
Is x^2+y^2 divisible by 5? 1). When xy is divided by 5, the remainder is 1
Insufficient
2). When x+y is divided by 5, the remainder is 3
Insufficient
1 + 2
When (xy)^2 is divided by 5, remainder will be 1
(xy)^2 = (x^2+y^2  2xy) = 5p + 1 > A

When (x+y)^2 is divided by 5, remainder will be 4
(x+y)^2 = (x^2+y^2 + 2xy) = 5k + 4 > B
A + B
2(x^2+y^2) = 5(p+k)+5 2(x^2+y^2) = 5(p+k+1) (x^2+y^2) = (5/2)(p+k+1) > for (x^2+y^2) to be integer , (p+k+1) has to be divisible by 2.
Hence, (x^2+y^2) is divisible by 5.



Intern
Joined: 05 Jun 2005
Posts: 2
Concentration: Strategy, Finance
GMAT Date: 04052014
WE: Engineering (Manufacturing)

Re: Is x^2+y^2 divisible by 5? [#permalink]
Show Tags
27 Feb 2014, 10:26
Bunuel, you are awesome....damn.,..that is such a great explanation!!!!



Intern
Joined: 23 Aug 2014
Posts: 42
GMAT Date: 11292014

Re: Is x^2+y^2 divisible by 5? [#permalink]
Show Tags
10 Nov 2014, 01:27
In the Last step, this is done:
xy=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




Re: Is x^2+y^2 divisible by 5?
[#permalink]
10 Nov 2014, 01:27







