When the positive integer n is divided by 25, the remainder

n = 25k + 13
n = 20m +3

25k + 13 = 20m +3
25k + 10 = 20m
5k + 2 = 4m
m=(5k+2)/4 - m has to be integer.
k has to be even and not divisible by 4

k=2,6,10 ==> 63, 163, 263 - our magic integers

I missing you at this part:

m=(5k+2)/4 - m has to be integer.
k has to be even and not divisible by 4

Why does k have to be even, and not divisible by 4?

1. m has to be an integer.
2. (5k+2) has to be even and divisible by 4
3. (5k+2) is even when 5k is even. Therefore k is even.
4. if 5k is divisible by 4, (5k+2) will not divisible by 4: 5*4i+2=4*(5i)+2. Therefore, (5k+2) has not to be divisible by 4 => 5k has not to be divisible by 4
5. k is even and indivisible by 4.
6: 2,6,10....

When the positive integer n is divided by 25, the remainder is 13. What is the value of n?

Given that \(n=25q+13\), so n could be 13, 38, 63, 88, 113, 138, 163, ...

(1) n < 100. n could be 13, 38, 63, or 88. Not sufficient.

(2) When n is divided by 20, the remainder is 3 --> \(n=20p+3\). From this n can be 3, 23, 43, 63, 83, 103, 123, 143, 163, ... Hence, n, among other values, can be 63 or 163. Not sufficient.

(1)+(2) The only value of n which fits both statements is 63. Sufficient.

Answer: C.

What I meant was we are given n = 25q + 13 => possible values 13,38,63,88,113

statement 2

n=20p + 3 => 3,23,43,63,83,103

For the second statement all the values in red don't satisfy the first equation only 63 does... How do you find another value quickly to claim insufficiency?

don't we have to match the equations here?

Well, there cannot be only one value that satisfies both \(n=25q+13\) (13, 38, 63, 88, 113, ...) and \(n=20p+3\) (3, 23, 43, 63, 83, 103, ...).

Next, there is a way to derive general formula for \(n\) (of a type \(n=mx+r\), where \(x\) is a divisor and \(r\) is a remainder) based on above two statements:

Divisor \(x\) would be the least common multiple of above two divisors 25 and 20, hence \(x=100\).

Remainder \(r\) would be the first common integer in above two patterns, hence \(r=63\).

Therefore general formula based on both statements is \(n=100m+63\). Hence n can be 63, 163, 263, ...

For more about this concept check: manhattan-remainder-problem-93752.html#p721341, when-positive-integer-n-is-divided-by-5-the-remainder-is-90442.html#p722552, when-the-positive-integer-a-is-divided-by-5-and-125591.html#p1028654

Hope this helps.

Really neat trick... I'm gonna save this absolutely brilliant... Thanks Bunuel!

\(\left\{ \matrix{\\
n \ge 1\,\,{\mathop{\rm int}} \hfill \cr \\
n = 25Q + 13,\,\,Q \ge 0\,\,{\mathop{\rm int}} \hfill \cr} \right.\)

\(? = n\)

\(\left( 1 \right)\,\,n < 100\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,{\rm{Q = 0}}\,\,\,\, \Rightarrow \,\,\,\,\,n = 13 \hfill \cr \\
\,{\rm{Take}}\,\,{\rm{Q = 1}}\,\,\,\, \Rightarrow \,\,\,\,\,n = 38 \hfill \cr} \right.\)

\(\left( 2 \right)\,\,\left\{ \matrix{\\
n = 20K + 3,\,\,K \ge 0\,\,{\mathop{\rm int}} \hfill \cr \\
n = 25Q + 13,\,\,Q \ge 0\,\,{\mathop{\rm int}} \hfill \cr} \right.\,\,\,\, \Rightarrow \,\,\,25Q + 10 = n - 3\,\, = \,\,20K\)

\(\Rightarrow \,\,\,25Q + 10\,\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,20\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,Q = 2\,\,\, \Rightarrow \,\,? = n = 63 \hfill \cr \\
\,{\rm{Take}}\,\,Q = 6\,\,\, \Rightarrow \,\,? = n = 163 \hfill \cr} \right.\)

\(\left( {1 + 2} \right)\,\,\,? = n = 63\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Hi - How is 83 possible - as the condition in the question mentions the remainder to be 13 when divided by 25. In case of 83- the remainder would be 8?

n can be 83 from \(n=20p+3\). Edited the solution to make this clearer. Thank you!