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Math Expert V
Joined: 02 Sep 2009
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When x is divided by 3, the remainder is 2, and when y is divided by 7  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 35% (03:10) correct 65% (02:42) wrong based on 135 sessions

### HideShow timer Statistics When x is divided by 3, the remainder is 2, and when y is divided by 7, the remainder is 4. What is the remainder when x + y is divided by 21?

(1) x^2 divided by 7 leaves a remainder of 4.
(2) y – 4 is divisible by 3.

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Re: When x is divided by 3, the remainder is 2, and when y is divided by 7  [#permalink]

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1
Bunuel wrote:
When x is divided by 3, the remainder is 2, and when y is divided by 7, the remainder is 4. What is the remainder when x + y is divided by 21?

(1) x^2 divided by 7 leaves a remainder of 4.
(2) y – 4 is divisible by 3.

X can be 2,5 and y can be 18 so reminder is different in both..SO E
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Re: When x is divided by 3, the remainder is 2, and when y is divided by 7  [#permalink]

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Bunuel wrote:
When x is divided by 3, the remainder is 2, and when y is divided by 7, the remainder is 4. What is the remainder when x + y is divided by 21?

(1) x^2 divided by 7 leaves a remainder of 4.
(2) y – 4 is divisible by 3.

can someone give a solution with equations ?
Math Expert V
Joined: 02 Aug 2009
Posts: 7763
When x is divided by 3, the remainder is 2, and when y is divided by 7  [#permalink]

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Takdir wrote:
Bunuel wrote:
When x is divided by 3, the remainder is 2, and when y is divided by 7, the remainder is 4. What is the remainder when x + y is divided by 21?

(1) x^2 divided by 7 leaves a remainder of 4.
(2) y – 4 is divisible by 3.

can someone give a solution with equations ?

Hi,

if you are looking for solution with equation..

the main statement tells us..
A) x=3a+2.... x could be 2,5,8,11,14,17....
B) y=7b+4.... y could be 4,11,18,25...

statements..
I) $$x^2=7c+4$$..... $$x^2$$ could be 4,11,18,25.....
so x can be $$\sqrt{4},\sqrt{11},\sqrt{18},\sqrt{25}$$... or $$2,\sqrt{11},\sqrt{18},5$$...
not required to go beyond this as we see ATLEAST two values of x possible 2,5...
insuff

II. it will give you some value of y..

insuff

combined..

whatever be the value of y..
y+2 and y+5 will surely give different REMAINDER when div by 21..

hence insuff

E
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Re: When x is divided by 3 the remainder is 2, and when y is divided by 7  [#permalink]

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Hi,

x when divided by 3, Remainder is 2.
x = 3a + 2

y when divided by 7, remainder is 4.
y = 7b + 4

Remainder when x + y divided by 21?
x + y = 3a + 2 + 7b+ 4
x + y = 3a + 7b+ 6.
If a is divisible by 7 and b is divisible by 3 then Remainder = 6.

A. x2 / 7 , Remainder = 4.
x = (3a + 2). x2 = 9a2 +6a + 4.
a is divisible by 7, so that remainder is 4.

B. y - 4 is divisible by 3
y = 7b + 4 ; y-4 = 7b
7b is divisible by 3, so b is divisible by 3.

C. a = divisible by 7, b = divisible by 3.
x + y = 3a + 7b + 6 when divided by 21 should give remainder 6.

The OA is E, Please can someone tell why C is incorrect?
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Location: United States
Re: When x is divided by 3 the remainder is 2, and when y is divided by 7  [#permalink]

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Akshay_Naik wrote:
Hi,

x when divided by 3, Remainder is 2.
x = 3a + 2

y when divided by 7, remainder is 4.
y = 7b + 4

Remainder when x + y divided by 21?
x + y = 3a + 2 + 7b+ 4
x + y = 3a + 7b+ 6.
If a is divisible by 7 and b is divisible by 3 then Remainder = 6.

A. x2 / 7 , Remainder = 4.
x = (3a + 2). x2 = 9a2 +6a + 4.
a is divisible by 7, so that remainder is 4.

B. y - 4 is divisible by 3
y = 7b + 4 ; y-4 = 7b
7b is divisible by 3, so b is divisible by 3.

C. a = divisible by 7, b = divisible by 3.
x + y = 3a + 7b + 6 when divided by 21 should give remainder 6.

The OA is E, Please can someone tell why C is incorrect?

Try the numbers x=5 and y=4 it gives a remainder of 9 when divided by 21
But when you take x=23 and y=4, it gives a remainder of 6 when divided by 21
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Re: When x is divided by 3, the remainder is 2, and when y is divided by 7  [#permalink]

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Bunuel chetan2u
So what I don't understand is that my algebraic solution would indicate C as the correct answer. Please let me know where I went wrong:

Stem: $$x=3q+2$$ and $$y=7p+4$$ $$–>$$ $$x+y=3q+7p+6$$
Hence, our question becomes: is $$3q+7p=21k$$. This is the case whenever $$3q$$ is divisible by $$7$$ and $$7p$$ is divisible by $$3$$.

Statement I: $$x^2=9q^2+12q+4=7L+4 –> q=7L$$
Still, Not sufficient as no information is given about y

Statement II: $$y-4=3m$$. Now from the stem, we know that $$y=7p+4$$ hence, this statement tells us that $$7p=3m$$
Still, Not sufficient as no information is given about x

Statement I&II combined:
From I) we know that $$3q$$ is divisible by $$7$$, hence, $$3q$$ must be a multiple of $$21$$
From II) we know that $$7p$$ is divisible by $$3$$, hence, $$7p$$ must be a multiple of $$21$$

Thus, $$x+y=(3q+7p)+6= 21b+6$$

Please let me know where I went wrong!
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Re: When x is divided by 3, the remainder is 2, and when y is divided by 7  [#permalink]

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no one has any idea why above solution is incorrect?
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Joined: 02 Aug 2009
Posts: 7763
Re: When x is divided by 3, the remainder is 2, and when y is divided by 7  [#permalink]

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2
ghnlrug wrote:
Bunuel chetan2u
So what I don't understand is that my algebraic solution would indicate C as the correct answer. Please let me know where I went wrong:

Stem: $$x=3q+2$$ and $$y=7p+4$$ $$–>$$ $$x+y=3q+7p+6$$
Hence, our question becomes: is $$3q+7p=21k$$. This is the case whenever $$3q$$ is divisible by $$7$$ and $$7p$$ is divisible by $$3$$.

Statement I: $$x^2=9q^2+12q+4=7L+4 –> q=7L$$
Still, Not sufficient as no information is given about y

Statement II: $$y-4=3m$$. Now from the stem, we know that $$y=7p+4$$ hence, this statement tells us that $$7p=3m$$
Still, Not sufficient as no information is given about x

Statement I&II combined:
From I) we know that $$3q$$ is divisible by $$7$$, hence, $$3q$$ must be a multiple of $$21$$
From II) we know that $$7p$$ is divisible by $$3$$, hence, $$7p$$ must be a multiple of $$21$$

Thus, $$x+y=(3q+7p)+6= 21b+6$$

Please let me know where I went wrong!

You are wrong here..

Statement I: $$x^2=9q^2+12q+4=7L+4 –> q=7L$$
Still, Not sufficient as no information is given about y

$$x^2=9q^2+12q+4=7L+4 –>9q^2+12q=7L......3q(3q+4)=7L$$..
When q is multiple of 7, 3q(3q+4)=21a(21a+4)=7L.... possible
Here 3q is divisible by 7
When 3q+4 is multiple of 7, say q is 1....3*1(3*1+4)=3*(7).... possible
Here 3q=3 and it is not divisible by 7
_________________ Re: When x is divided by 3, the remainder is 2, and when y is divided by 7   [#permalink] 04 Jan 2019, 05:03
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