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

Question

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

(1) n < 100
(2) When n is divided by 20, the remainder is 3.

it is c, and i got 63 as n using both statements, but wasn't certain there didn't exist another one.

any quick way to do this.

thx

KillerSquirrel · Answer

When the positive integer n is divided by 25, the remainder is 13.

n = 25*x+13

statement 1

n < 100

n = 13,38,63,88

insufficient

statement 2

when n is divided by 20, the remainder is 3.

n = 20*y + 3

20*y + 3 = 25*x+13

insufficient

both statements

n = 13,38,63,88

n = 20*y + 3 ---> 3,23,43,63,83

n = 63

sufficient

the answer is (C)

walker · Answer

C

1. 38,63,88. insuff.
2. 63,163. insuff.

1&2. 63. suff.

walker · Answer

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

jimmyjamesdonkey · Answer

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?

walker · Answer

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....

Bunuel · Answer

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.

fozzzy · Answer

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?

Bunuel · Answer

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.

fozzzy · Answer

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

fskilnik · Answer

\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.

Uditakaushal1992 · Answer

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?

Bunuel · Answer

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

GmatCracker09 · Answer

KarishmaB  I didn't understand how both will work and not just B, can you help me?

KarishmaB · Answer

n = 25a + 13 When we divide n balls into groups of 25 balls each, 13 are leftover. So n could be 13 or 38 or 63 etc. What is n? (1) n < 100 n can take various values such as 13 or 38 or 63 or 88. Not sufficient alone. (2) When n is divided by 20, the remainder is 3. n = 20b + 3 No common remainder. So we find the first such number by hit and trial. n = 3, 23, 43, 63.... We see that 63 is common to both. So general form of n becomes n = (LCM of 25 and 20)k + 63 = 100k + 63 Now n can be 63 or 163 or 263 etc. Still no unique value GmatCracker09 Not sufficient alone. Using both statements together , now we can see that 63 is the only possible value. Answer (C) Discussion on Division and Remainders: https://youtu.be/A5abKfUBFSc

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

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