If a is a positive integer and 81 divided by a results in a

If a is a positive integer and 81 divided by a results in a remainder of 1, what is the value of a?

Given: \(81=qa+1\) --> \(qa=80\) --> so, \(a\) is a factor of 80: 2, 4, 5, 8, 10, 16, 20, 40, or 80.

(1) The remainder when a is divided by 40 is 0 --> this basically tells us that \(a\) is a multiple of 40, so \(a\) could be either 40 or 80 (from the above). Not sufficient.

(2) The remainder when 40 is divided by a is 40 --> this basically means that \(a\) is greater than 40 (for example, the remainder upon division 40 by 50 is 40). Now, the only factot of 80 which is greater than 40 is 80 itself. So, \(a=80\). Sufficient.

Answer: B.

Hope it's clear.
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Dear Bunuel,
i am not able to understand the st. 2 analysis. (ofcourse i understood what you have written.)
one doubt remains like: if it is said x/y leaves a remainder of 10, can i consider values of x/y as 10/20 and like that?

Yes, 10 divided by 20 yields the remainder of 10.

THEORY:
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).

For example positive integer n is divided by 25 yields the remainder of 13 can be expressed as: \(n=25q+13\). Now, the lowest value of \(q\) can be zero and in this case \(n=13\) --> 13 divided by 25 yields the remainder of 13. Generally when divisor (25 in our case) is more than dividend (13 in our case) then the reminder equals to the dividend. For example:
3 divided by 24 yields a reminder of 3 --> \(3=0*24+3\);
or:
5 divided by 6 yields a reminder of 5 --> \(5=0*6+5\),

Hope it's clear.
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Thanks for the solution

I have one doubt ...
as per the 2nd statement when 40 is divided by a remainder is 40 i.e. 40 = aq +40 or aq =0 ?

can you please explain this also ..

Yes, "the remainder when 40 is divided by a is 40" can be written as \(40=ap+40\) --> \(ap=0\). What is the point of confusion here?
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Given a > 0 & 81 = ak + 1

a = ?

Statement 1:

The remainder when a is divided by 40 is 0

a = 40p = 40 or 80
to satisfy 81 = ak + 1

Statement 1 is Not Sufficient.

Statement 2:

The remainder when 40 is divided by a is 40

40 = aq + 40, hence a > 40 & has to satisfy 81 = ak + 1

Hence a = 80

Statement 2 is Sufficient.

Answer B.


Thanks,
GyM
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Target question: What is the value of a?

Given: 81 divided by a results in a remainder of 1
In other words, 81 is 1 greater than some multiple of a
This means that 80 is some multiple of a
Another way to say this is: a is a divisor of 80
The divisors of 80 are: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80
So, the possible values of a are: 2, 4, 5, 8, 10, 16, 20, 40, 80
ASIDE: I omitted 1 from the list, since it does not satisfy the condition of getting a remainder of 1 when 81 is divided by 1.

Statement 1: The remainder when a is divided by 40 is 0.
The possible values of a are: 2, 4, 5, 8, 10, 16, 20, 40, 80
As you can see, there are two possible values of a satisfy statement 1.
Case a: a = 40. This works, because 40 divided by 40 leaves remainder 0. In this case, the answer to the target question is a = 40
Case b:a = 80. This works, because 80 divided by 40 leaves remainder 0. In this case, the answer to the target question is a = 80
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: The remainder when 40 is divided by a is 40.
The possible values of a are: 2, 4, 5, 8, 10, 16, 20, 40, 80
As you can see, ONLY ONE possible value of a satisfies statement 2.
When a = 80, we get a remainder of 40 when 40 is divided by 80
So, the answer to the target question is a = 80
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent

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Very nice conceptual problem!

\(a \ge 1\,\,{\mathop{\rm int}} \,\,\,\,\,\left( * \right)\)

\(81 = M \cdot a + 1\,\,,\,\,M\mathop \ge \limits^{\left( * \right)} 1\,\,{\mathop{\rm int}} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,M \cdot a = 80\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,a \le 80\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{positive}}\,\,{\rm{divisor}}\,\,{\rm{of}}\,\,80\,\,\,\,\left( {**} \right)\)

\(? = a\)


\(\left( 1 \right)\,\,a = 40J\,\,,\,\,\,J\mathop \ge \limits^{\left( * \right)} 1\,\,\,{\mathop{\rm int}} \left\{ \matrix{\\
\,{\rm{Take}}\,\,J = 1\,\,\,\,\left( {M = 2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 40\,\,\,\,{\rm{viable}} \hfill \cr \\
\,{\rm{Take}}\,\,J = 2\,\,\,\,\left( {M = 1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 80\,\,\,\,{\rm{viable}} \hfill \cr} \right.\)


\(\left( 2 \right)\,\,40 = N \cdot a + \underline {40} \,\,,\,\,N\,\,{\mathop{\rm int}} \,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\left\{ \matrix{\\
\,a\,\, > \,\,\underline {40} \hfill \cr \\
\,N = 0 \hfill \cr} \right.\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\,a = 80\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{SUFF}}.\)


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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