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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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 ..
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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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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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