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Senior Manager  V
Joined: 22 Feb 2018
Posts: 420
If a and b are positive integers and a > b, what is the remainder when  [#permalink]

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5 00:00

Difficulty:   75% (hard)

Question Stats: 51% (02:04) correct 49% (01:45) wrong based on 75 sessions

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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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Director  G
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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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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.

Math Expert V
Joined: 02 Aug 2009
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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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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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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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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

Pls Hit kudos if you like the solution

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

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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 $$2$$.

(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.
$$remainder:\frac{(a-b)^2}{9}…\frac{(9n+2)^2}{9}…\frac{9n+4+36n}{9}…\frac{4}{9}$$
$$remainder=4$$ 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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