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If a and b are positive integers and a > b, what is the remainder when

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If a and b are positive integers and a > b, what is the remainder when  [#permalink]

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New post 14 Oct 2018, 08:49
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If \(a\) and \(b\) are positive integers and \(a > b\), what is the remainder when \(a^2 - 2ab + b^2\) is divided by \(9\)?

(1) The remainder when \(a-b\) is divided by \(3\) is \(2\).
(2) The remainder when \(a-b\) is divided by \(9\) is \(2\).

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Re: If a and b are positive integers and a > b, what is the remainder when  [#permalink]

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New post 14 Oct 2018, 09:11
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Princ wrote:
If \(a\) and \(b\) are positive integers and \(a > b\), what is the remainder when \(a^2 - 2ab + b^2\) is divided by \(9\)?

(1) The remainder when \(a-b\) is divided by \(3\) is \(2\).
(2) The remainder when \(a-b\) is divided by \(9\) is \(2\).


From question stem it is asking for \(\frac{(a-b)^2}{9}\)

Statement 1) \(\frac{(a-b)}{3} = q + \frac{2}{3}\) this can be converted to \(a-b = 3q + 2\)

if q = 0 then a-b = 2 and \((a-b)^2\) = 4 as a result it is 4/9

if q = 1 then a-b = 5 and \((a-b)^2\) = 25 as a result it is \(\frac{25}{9} = 2 \frac{7}{9}\)

Insufficient.

Statement 2) \(\frac{(a-b)}{9} = q + \frac{2}{9}\) this can be converted to \(a-b = 9q + 2\)

if q = 0 then a-b = 2 and \((a-b)^2\) = 4 as a result it is 4/9

if q = 1 then a-b = 11 and \((a-b)^2\) = 121 as a result it is \(\frac{121}{9} = 13 \frac{4}{9}\)

Sufficient.

answer choice B
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Re: If a and b are positive integers and a > b, what is the remainder when  [#permalink]

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New post 14 Oct 2018, 09:23
If \(a\) and \(b\) are positive integers and \(a > b\), what is the remainder when \(a^2 - 2ab + b^2\) is divided by \(9\)?
Now \(a^2 - 2ab + b^2=(a-b)^2\)
(1) The remainder when \(a-b\) is divided by \(3\) is \(2\).
cases when reaminder is 2 is when
a-b=2, remainder of 4 when 2^2 is divided by 9
a-b=5, remainder of 7 when 5^2 is divided by 9
a-b=8, remainder of 1 when 8^2 is divided by 9
insuff

(2) The remainder when \(a-b\) is divided by \(9\) is \(2\).
so (a-b)62 will leave a remainder of 2^2=4
suff

B
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Re: If a and b are positive integers and a > b, what is the remainder when  [#permalink]

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New post 14 Jun 2019, 19:50
Dear Salsanousi , chetan2u

I read both your explanations but did not really understand why statement 2 is sufficient, could you please explain in more simplistic manner?

(May be I cannot visualize the statement 2 - remainder is 2 always)
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Re: If a and b are positive integers and a > b, what is the remainder when  [#permalink]

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New post 14 Jun 2019, 20:11
1
Princ wrote:
If \(a\) and \(b\) are positive integers and \(a > b\), what is the remainder when \(a^2 - 2ab + b^2\) is divided by \(9\)?

(1) The remainder when \(a-b\) is divided by \(3\) is \(2\).
(2) The remainder when \(a-b\) is divided by \(9\) is \(2\).


a^2 - 2ab + b^2 = (a - b)^2

(1) a - b = 3k + 2, for any integer k
—> (a - b)^2 = (3k + 2)^2 = 9k^2 + 12k + 4
When divided by 9
—> (9k^2 + 12k + 4)/9 = k^2 + (12k + 4)/9
Many remainders are possible.
—> No unique value - Insufficient

(2) a - b = 9m + 2, for any integer m
—> (a - b)^2 = (9m + 2)^2 = 81m^2 + 36m + 4
When divided by 9
—> (81m^2 + 36m + 4)/9
—> 9m^2 + 4m + 4/9
—> A unique remainder = 4
Sufficient

IMO Option B

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If a and b are positive integers and a > b, what is the remainder when  [#permalink]

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New post 08 Oct 2019, 04:35
Princ wrote:
If \(a\) and \(b\) are positive integers and \(a > b\), what is the remainder when \(a^2 - 2ab + b^2\) is divided by \(9\)?

(1) The remainder when \(a-b\) is divided by \(3\) is \(2\).
(2) The remainder when \(a-b\) is divided by \(9\) is \(2\).


(1) The remainder when \(a-b\) is divided by \(3\) is \(2\): insufic.
\(remainder:\frac{(a-b)^2}{9}…\frac{(3m+2)^2}{9}…\frac{9m+4+12m}{9}…\frac{4+12m}{9}\)
\(m=0:\frac{4}{9}…remainder=4\)
\(m=1:\frac{4+12}{9}=\frac{16}{9}…remainder=7\)

(2) The remainder when \(a-b\) is divided by \(9\) is \(2\): sufic.
\(remainder:\frac{(a-b)^2}{9}…\frac{(9n+2)^2}{9}…\frac{9n+4+36n}{9}…\frac{4}{9}\)
\(remainder=4\)

Answer (B)
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If a and b are positive integers and a > b, what is the remainder when   [#permalink] 08 Oct 2019, 04:35
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