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# What is the remainder when y^2 is divided by 6? (1) y is a prime numb

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What is the remainder when y^2 is divided by 6? (1) y is a prime numb  [#permalink]

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02 Apr 2018, 04:54
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What is the remainder when y^2 is divided by 6?

(1) y is a prime number greater than 10.

(2) y is a positive odd number which is NOT divisible by 3.
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What is the remainder when y^2 is divided by 6? (1) y is a prime numb  [#permalink]

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Updated on: 02 Apr 2018, 17:53
Statement 1 : y is a prime number greater than 10.
Any prime no greater than 3 can be represented by 6n+1 or 6n-1
Now $$y ^2 = (6n+1)^2 or (6n-1)^2$$
$$y^2 = (36n^2+12n+1) or (36n^2-12n+1)$$
Hence remainder when y ^2 divisible by 6 = 1
Sufficient.

Statement 2: y is a positive odd number which is NOT divisible by 3.
y can be 7 : y ^2 = 49. remainder when divided by 6 =1
y can be 11: y^2 = 121, remainder when divided by 6 = 1
It will always be 1.
sufficient.

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Originally posted by gmatbusters on 02 Apr 2018, 06:28.
Last edited by gmatbusters on 02 Apr 2018, 17:53, edited 1 time in total.
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What is the remainder when y^2 is divided by 6? (1) y is a prime numb  [#permalink]

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Updated on: 02 Apr 2018, 20:06
amanvermagmat wrote:
What is the remainder when y^2 is divided by 6?

(1) y is a prime number greater than 10.

(2) y is a positive odd number which is NOT divisible by 3.

Statement 1 : y is a prime number greater than 10.
Any prime no greater than 3 can be represented by 6n+1 or 6n-1

By remainder theorem the remainder of y ^2 will product of individual remainders.
for 6n+1 =rem(1) when divided by 6 so ,actual remainder for y ^2 =1*1=1
6n-1 = rem(-1) when divided by 6 so ,actual remainder for y ^2=-1*-1=1
Hence remainder when y ^2 divisible by 6 = 1
Sufficient.

Statement 2: y is a positive odd number which is NOT divisible by 3.
y can be 7 : remainder when divided by 6 =1
y can be 11: when divided by 6 = 1
sufficient.

Originally posted by kunalcvrce on 02 Apr 2018, 08:44.
Last edited by kunalcvrce on 02 Apr 2018, 20:06, edited 1 time in total.
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Re: What is the remainder when y^2 is divided by 6? (1) y is a prime numb  [#permalink]

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02 Apr 2018, 15:56
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Statement 2 is also sufficient

Given: y is a positive odd number which is NOT divisible by 3.
so y=3k+1 where k is even, or y=3k+2 where k is odd.

Case 1 (y=3k+1 where k is even)
$$y^2=(3k+1)^2=9k^2+6k+1$$
since k is even, we can write k=2a where a is an integer, and substitute
$$y^2=9k^2+6k+1=9(2a)^2+6(2a)+1=36a^2+12a+1$$
So remainder = 1 when divided by 6.

Case 2 (y=3k+2 where k is odd)
$$y^2=(3k+2)^2=9k^2+12k+4$$
since k is odd, we can write k=2b+1 where b is an integer, and substitute
$$y^2=9k^2+12k+4=9(2b+1)^2+12(2b+1)+4=9(4b^2+4b+1)+24b+12+4=36b^2+60b+25$$
So remainder = 1 when divided by 6.

Sufficient

Or you could try some numbers that meet the condition, e.g.
1^2=1; remainder=1 when div by 6
5^2=25; remainder=1 when div by 6
7^2=49; remainder=1 when div by 6

I dont know if this method is reliable though.

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Re: What is the remainder when y^2 is divided by 6? (1) y is a prime numb  [#permalink]

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02 Apr 2018, 17:50
Hii
You are right, answer should be D.

I forgot that question is asking for remainser when y^2 is divided by 6, not when y is divided by 6.
Thanks for pointng out.

Kudos to you.

aserghe1 wrote:
Statement 2 is also sufficient

Given: y is a positive odd number which is NOT divisible by 3.
so y=3k+1 where k is even, or y=3k+2 where k is odd.

Case 1 (y=3k+1 where k is even)
$$y^2=(3k+1)^2=9k^2+6k+1$$
since k is even, we can write k=2a where a is an integer, and substitute
$$y^2=9k^2+6k+1=9(2a)^2+6(2a)+1=36a^2+12a+1$$
So remainder = 1 when divided by 6.

Case 2 (y=3k+2 where k is odd)
$$y^2=(3k+2)^2=9k^2+12k+4$$
since k is odd, we can write k=2b+1 where b is an integer, and substitute
$$y^2=9k^2+12k+4=9(2b+1)^2+12(2b+1)+4=9(4b^2+4b+1)+24b+12+4=36b^2+60b+25$$
So remainder = 1 when divided by 6.

Sufficient

Or you could try some numbers that meet the condition, e.g.
1^2=1; remainder=1 when div by 6
5^2=25; remainder=1 when div by 6
7^2=49; remainder=1 when div by 6

I dont know if this method is reliable though.

Posted from my mobile device
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Re: What is the remainder when y^2 is divided by 6? (1) y is a prime numb  [#permalink]

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02 Apr 2018, 23:20
Hi,

Question: Remainder when y^2 divided by 6 ?

Any integer when divided by 6 leaves remainders as 0,1,2,3,4 and 5.

Statement I is sufficient:

Prime number greater than 10.

Yes there is a rule which says, when any prime number greater than 3 can be expressed as 6n+1 or 6n+5.

So according to the rule,

y=6n+1 or y = 6n+5

y^2 = 36n + 12n + 1

or

y^2 = 36n + 60n + 25.

So, both expression when divided by 6, leaves the remainder 1.

Even if you don’t remember this rule(this rule has been discussed in Official GMAT guide), its okay, you just need to try out some values and check out the pattern of y^2. Different pattern then
its not sufficient, if it gives you same result every time then it is sufficient.

Let’s say,

y = 11 , 13, 17, 19

y^2 = 121, 169, 289, 361.

All leaves remainder 1 when divided by 6(Divisibility rule for 6 is it should be divisible by both 3 and 2). So sufficient.

Statement II is sufficient:

y = 1, 5, 7, 11, 13, 17,..

y^2 = 1, 25, 49, 121, 169, 289..

Again, all leaves remainder 1 when divided by 6.

So, each alone are sufficient. So answer is D.

Trying out numbers would be the ideal approach for these types of questions.

Hope this helps.
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What is the remainder when y^2 is divided by 6? (1) y is a prime numb  [#permalink]

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03 Apr 2018, 02:00
amanvermagmat wrote:
What is the remainder when y^2 is divided by 6?

(1) y is a prime number greater than 10.

(2) y is a positive odd number which is NOT divisible by 3.

A point for readers of this thread and specially for the ones who face challenge in QUANT Section

While it's alright to use that Prime number greater than 3 are (6k+1) or (6k-1) form but ONLY IF YOU ALREADY KNOW.

THIS PROPERTY OF PRIME NUMBER IS UN-NECESSARY FOR ANY GMAT RELATED QUESTION

Question : remainder when y^2 is divided by 6?

Statement 1: y is a prime number greater than 10
i.e. y maybe 11, 13, 17. 19, 23 ... etc

Remainder (y^2/6) = R(121/6) or R(169/6) or R(289/6) or ... = 1 always hence . [Check for 3 consecutive cases like shown here]

SUFFICIENT

Statement 2: y is a positive odd number which is NOT divisible by 3

i.e. y may be 5, 7, 11, 13, .. etc
y^2 may be 25, 49, 121, 169 etc
remainder are 1 hence

SUFFICIENT

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What is the remainder when y^2 is divided by 6? (1) y is a prime numb   [#permalink] 03 Apr 2018, 02:00
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