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What is the remainder when a is divided by 4? [#permalink]

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19 Dec 2010, 15:22

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What is the remainder when a is divided by 4?

(1) a is the square of an odd integer. (2) a is a multiple of 3.

According to the book, the answer is A.

But according to me, the answer should be C since if we take the first statement and use the value 1 for a, it gives us 1 for the square of 1, what would be the remainder? When we use 3 for a, the square of a will give us 9 which when divided by 4, gives us remainder of 1.

I'm a little confused regarding the correct answer for the below mentioned question:

What is the remainder when a is divided by 4?

(1) a is the square of an odd integer. (2) a is a multiple of 3.

According to the book, the answer is A.

But according to me, the answer should be C since if we take the first statement and use the value 1 for a, it gives us 1 for the square of 1, what would be the remainder? When we use 3 for a, the square of a will give us 9 which when divided by 4, gives us remainder of 1.

Thanks. Kash.

Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

So according to above, when positive integer \(a\) is less than divisor \(d\) then remainder upon division \(a\) by \(d\) is always equals to \(a\), for example 5 divided by 10 yields reminder of 5. So when 1 is divided by 4 remainder is 1.

Or algebraically: 1 divided by 4 can be expressed as \(1=0*4+1\), so \(r=1\).

Back to the original question:

What is the remainder when a is divided by 4?

(1) a is the square of an odd integer --> \(a=(2k+1)^2=4k^2+4k+1\), first two terms (4k^2 and 4k) are divisible by 4 and the third term (1) when divided by 4 yields the remainder of 1. Sufficient.

Or you can try several numbers for \(a\): \(a=1^1=1\) --> 1 divided by 4 yields remainder of 1; \(a=3^1=9\) --> 9 divided by 4 yields remainder of 1; \(a=5^1=25\) --> 25 divided by 4 yields remainder of 1; ...

(2) a is a multiple of 3 --> clearly insufficient as \(a\) can as well be a multiple of 4, 12 for example, and in this case the remainder will be 0, and it also can not be a multiple of 4, 3 for example, and in this case the remainder will be 3. Not sufficient.

Re: MGMAT's Number Properties Data Sufficiency Question [#permalink]

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19 Dec 2010, 15:50

ksear wrote:

Hi everyone,

I'm a little confused regarding the correct answer for the below mentioned question:

What is the remainder when a is divided by 4?

(1) a is the square of an odd integer. (2) a is a multiple of 3.

According to the book, the answer is A.

But according to me, the answer should be C since if we take the first statement and use the value 1 for a, it gives us 1 for the square of 1, what would be the remainder? When we use 3 for a, the square of a will give us 9 which when divided by 4, gives us remainder of 1.

Thanks. Kash.

To answer your specific question, if you divide 1 by 4, the reminder is 1. So the answer is valid.

Re: What is the remainder when a is divided by 4? [#permalink]

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08 Aug 2013, 00:10

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What is the remainder when a is divided by 4?

(1) a is the square of a odd integer. Method #1: a=1 reminder=1; a=9 reminder=1, a=25 reminder=1; a=49 reminder=1... I see a pattern, I am convinced that this is sufficient. Method #2: \(a=(2k+1)^2\) (where k is an integer) \(a=4k^2+4k+1\), \(a=4(k^2+k)+1\) \(a\) is a multiple of four plus one, hence the reminder will be one. Sufficient

(2)a is a multiple of 3. a=3 reminder=3, a=9 reminder=1. Not sufficient.
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Re: What is the remainder when a is divided by 4? [#permalink]

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10 Aug 2013, 04:32

ksear wrote:

What is the remainder when a is divided by 4?

(1) a is the square of an odd integer. (2) a is a multiple of 3.

According to the book, the answer is A.

But according to me, the answer should be C since if we take the first statement and use the value 1 for a, it gives us 1 for the square of 1, what would be the remainder? When we use 3 for a, the square of a will give us 9 which when divided by 4, gives us remainder of 1.

Thanks. Kash.

1^2/4 = (4-3)^2/4 = (4^2 - 2.4.3 + 9)/4 = 4 + 6 + 9/4 = 10 + 9/4 so the remainder from 9/4 we have is 1 . That's how we can realize why 1 appears as the remainder when 1^2 is divided by 4.

Re: What is the remainder when a is divided by 4? [#permalink]

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08 Apr 2016, 12:23

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Re: What is the remainder when a is divided by 4? [#permalink]

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25 May 2017, 07:17

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What is the remainder when p is divided by 4 [#permalink]

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16 Jun 2017, 00:31

shashankism wrote:

What is the remainder when p is divided by 4 ? (1) p is the square of an odd integer. (2) p is a multiple of 3.

(1) p is the square of an odd integer.

Lets try \(3^2, 7^2\)

\(3^2 = 9.\) 9 divided by 4 gives remainder 1.

\(7^2 = 49\) 49 divided by 4 gives remainder 1.

Any square of odd integer divided by 4 will give remainder 1. -------- (I is Sufficient)

(2) p is a multiple of 3.

12 is multiple of 4 hence will give remainder 0. 3 divided by 4 will give remainder 3. 6 divided by 4 will give remainder 2. etc... Therefore multiple values. -------- (II is Not Sufficient)

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Re: What is the remainder when a is divided by 4? [#permalink]

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18 Jun 2017, 23:18

ksear wrote:

What is the remainder when a is divided by 4?

(1) a is the square of an odd integer. (2) a is a multiple of 3.

According to the book, the answer is A.

But according to me, the answer should be C since if we take the first statement and use the value 1 for a, it gives us 1 for the square of 1, what would be the remainder? When we use 3 for a, the square of a will give us 9 which when divided by 4, gives us remainder of 1.

Thanks. Kash.

(1) Let a = (2k+1)^2 as it is square of an odd integer = 4k^2 + 4k +1 . So when it is divided by 4, remainder =1. Sufficient (2) a is a multiple of 3, i.e. 3,6,9,12,15,........ So when it is divided by 4, remainder = 3 , 2, 1, 0, So Not sufficient

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