When positive integer n is divided by 13, the remainder is 2. When n

Question

When positive integer n is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

chetan2u · Accepted Answer

the equation that can be formed is 13x+2=8y+5..
13x-3=8y...
as we can see x can take only odd values as the RHS will always be even..
Also x can take values till 13 as 13*14>180..
now we have to substitue x as 1,3,5,7,9,11,13...
once we find 7 fitting in , any other value need not be checked as every 4th value will give us answer so next value will be 15..
ans 1.. B

earnit · Answer

N = 13q+2  (possible values: 2,15,28,41,54,67,80, 93 ,106....
 N = 8z+5  (possible values: 5,13,21,29,37,45,53,61,69,77,85, 93

Therefore, N = (13*8)k+93 
 N = 104k+93  (possible values:93,197)

Clearly only one value is possible that is less than 180.

Answer:B

AmoyV · Answer

n=13Q1+2
n=8Q2+5

13Q1+2=8Q2+5
13Q1=8Q2+3

When we list the multiples of 13, only when n=91 can we have a remainder of 3 when n is divided by 8.
OR
We will get a multiple of 13 only when Q2=11

Answer: B

chetan2u · Answer

hi 
you do not have to check all multiples of 13.. the multiple has to be odd so Q1 has to be odd.. so check for only 1,3,5,7,9.. .. may save some time

Harley1980 · Answer

And we can save even more time:
all numbers that will be multiple to 13 and 8+3 will be in periods 104 (13 * 8 = 104 )
so if our first number is 91 than next will be 91 + 104 = 195 and next 195 + 104 = 299 ans so on.

And we know that we have diapason from 1 to 180 
so 180 - 104 = 76 
if we have one number that will be divisible on 13 and 8 + 3
and that number will be less that 76 than we have 2 numbers 
if we don't have such number before 76 than we have only one number in this diapason.

for example we can solve the same task but change 8 + 3 on 8 + 4 
the first number that will be divisible on 13 and 8 + 4 it's 52
this number before 76 so we have two numbers before 180 
second number is 52 + 104 = 156

This is not very useful on this task but if we will have diapson not 1..180 but 1...800 for example
it'll be very useful shortcut.

Bunuel · Answer

MAGOOSH OFFICIAL SOLUTION:

The LCM of 8 and 13 is 104. Hence there cannot be more than 2 such values less than 180. Options (D) and (E) are out of the window for sure.

The number n should be of the following two forms:
n = 8a + 5
n = 13b + 2

In a given bunch of numbers, there will be many more numbers of the form (8a + 5) and fewer of the form (13b + 2) so let’s start with a number of the form (13b + 2).
If b = 0, n = 2. Is it of the form (8a + 5)? No. n/8 gives a remainder of 2, not 5.
If b = 1, n = 15. Is it of the form (8a + 5)? No. n/8 gives a remainder of 7, not 5.
If b = 2, n = 28. Is it of the form (8a + 5)? No. n/8 gives a remainder of 4, not 5.
If b = 3, n = 41. Is it of the form (8a + 5)? No. n/8 gives a remainder of 1, not 5.
If b = 4, n = 54. Is it of the form (8a + 5)? No. n/8 gives a remainder of 6, not 5.
If b = 5, n = 67. Is it of the form (8a + 5)? No. n/8 gives a remainder of 3, not 5.
If b = 6, n = 80. Is it of the form (8a + 5)? No. n/8 gives a remainder of 0, not 5.
If b = 7, n = 93. Is it of the form (8a + 5)? Yes! n/8 gives a remainder of 5.

The smallest value of n is 93. The next value of n = 93 + 104 = 197 i.e. greater than 180. Hence there is just one value of n less than 180.

Answer: B.

Abhishek009 · Answer

Possible Values of n will be

FOR SET 13 ={ 15 , 28 , 41 , 54 , 67 , 80 , 93 , 106 , 119 , 132, 145, 158 , 171 }

FOR SET 8 = { 13 , 21 , 29 , 37 , 45 , 53, 61 , 69 , 77, 85 ............. Go no further }

For set 8 there is a patter to observe the numbers are all ODD numbers , but for set 13 the alternate numbers are Odd Numbers....

So, now check the odd numbers in set 13 which when divided by 8 leaves remainder 5 { Here start from 93 as you have already checked numbers upto 85 and no number is common }

93/8 = 11*8 + 5 { This is our number }

Check further ( only the odd numbers ) you cant find any ODD number....

Hence we have only the number 93 which is common to both SET 13 and SET 8 , so this is our number !!

Answer will be only one possible number and that is 93

aserghe1 · Answer

Trying over 5 numbers until finding the one that works seems inefficient. Here's what I did...

From the given info, we have these 2 equations:
(1) n = 13x + 2
(2) n = 8y + 5

Note that the above are equivalent to:
(1*) n = 13x - 11
(2*) n = 8y - 3

Thus, we have
- 13x - 11 = 8y - 3
- 13x = 8y + 8
- 13x = 8(y+1)
So,
- (y+1) has to be a multiple of 13, and 
- x has to be a multiple of 8

The smallest multiple of 13 is 13 (obviously), so if we substitute y = 12 into (2*), then we get n = 93, which is the smallest value of n.
Similarly, if we substitute x = 8 into (1*), then we get the same answer, n = 93.

You can calculate the LCM of 8 and 13 (which is 104), and see that the next value of n is 197 (93 + 104).
Or, since we found the smallest value of n by using x = 8 or y = 12, we can find the next value by using x = 16 (the next multiple of 8) or y = 25 (which would make y+1 the next multiple of 13). If we substitute x = 16 into (1*), we get n = 197. Similarly, if we substitute y = 25 into (2*), we get the same n = 197.

JeffTargetTestPrep · Answer

We can create the equation:

n = 13Q + 2

So n can be 2, 15, 28, 41, 54, 67, 80, 93, ...

and

n = 8Q + 5

So n can be 5, 13, 21, 29, 37, 45, 53, 61, 69, 77, 85, 93, …

We see that the first number that satisfies both conditions is 93. To find the other numbers, we can keep adding the LCM of 13 and 8, which is 13 x 8 = 104. Therefore, the next value that satisfies both conditions is 93 + 104 = 197. However, 197 is already greater than 180, so we only have one value, namely 93, that is less than 180 and satisfies both conditions.

Answer: B

fskilnik · Answer

1 \leqslant n \leqslant 179\,\,\,\,\,\left( {n\,\,\,\operatorname{int} } ight)\,\,\,\,\left( * ight) 
 n = 13M + 2\,\,\,\,,\,\,\,M\,\,\operatorname{int} \,\,\,\,\left( {	ext{I}} ight) 
 n = 8J + 5\,\,\,\,,\,\,\,J\,\,\operatorname{int} \,\,\,\,\left( {{	ext{II}}} ight)

?\,\,\,:\,\,\,n\,\,\,{	ext{in}}\,\,\,\left( * ight) \cap \left( {	ext{I}} ight) \cap \left( {{	ext{II}}} ight)

\left( * ight)\,\, \cap \,\,\left( {	ext{I}} ight)\,\,\,:\,\,\,\,\,\,1\,\, \leqslant \,\,13M + 2\,\, \leqslant \,\,179\,\,\,\,\,\mathop \Leftrightarrow \limits^{ - \,2} \,\,\,\,\, - 1 \leqslant 13M \leqslant 177\,\,\left( { = 169 + 8} ight) 
 - 1 \leqslant 13M \leqslant 177\,\,\left( { = 169 + 8} ight)\,\,\,\,\,\mathop \Leftrightarrow \limits^{M\,\,\operatorname{int} } \,\,\,0 \leqslant 13M \leqslant 169\,\,\,\,\,\mathop \Leftrightarrow \limits^{:\,\,13} \,\,\,\,0 \leqslant M \leqslant 13

\left( * ight)\,\, \cap \,\,\left( {{	ext{II}}} ight)\,\,\,:\,\,\,\,\,\,\left\{ \begin{gathered}
 \,n - 5\,\,{	ext{divisible}}\,\,{	ext{by}}\,\,8\,\,\,\,\, \Rightarrow \,\,\,\,\,n - 5\,\,\,\,\,{	ext{even}}\,\,\,\,\, \Rightarrow \,\,\,\,\,n\,\,\,{	ext{odd}}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {	ext{I}} ight)} \,\,\,\,\,M\,\,{	ext{odd}} \hfill \
 \,n - 5\mathop = \limits^{\left( {	ext{I}} ight)} 13M - 3\,\,{	ext{divisible}}\,\,{	ext{by}}\,\,8\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{13M - 3}}{2}\,\,\,\,{	ext{divisible}}\,\,{	ext{by}}\,\,4\,\,\,\left( {***} ight)\,\,\,\,\,\, \hfill \ 
\end{gathered} ight.

\left. \begin{gathered}
 M = 1\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{13 - 3}}{2} = 5\,\,\,{	ext{odd}}\,\,\,\left( {***} ight)\,\,\,{	ext{NO}} \hfill \
 M = 3\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{13 \cdot 3 - 3}}{2} = 18\,\,\,\,\left( {***} ight)\,\,\,{	ext{NO}} \hfill \
 M = 5\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{13 \cdot 5 - 3}}{2} = 31\,\,{	ext{odd}}\,\,\,\,\left( {***} ight)\,\,\,{	ext{NO}}\,\,\,\,\,\, \hfill \
 \boxed{M = 7}\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{13 \cdot 7 - 3}}{2} = 44\,\,\,\,\left( {***} ight)\,\,\,{	ext{YES}} \hfill \
 M = 9\,\,\,\,\, \Rightarrow \,\,\,\,{	ext{odd}}\,\,\,\,\left( {***} ight)\,\,\,{	ext{NO}} \hfill \
 M = 11\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{13 \cdot 11 - 3}}{2} = 70\,\,\,\,\left( {***} ight)\,\,\,{	ext{NO}} \hfill \
 M = 13\,\,\,\,\, \Rightarrow \,\,\,\,{	ext{odd}}\,\,\,\,\left( {***} ight)\,\,\,{	ext{NO}} \hfill \ 
\end{gathered} ight\}\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = 1

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

Regards, 
Fabio.

GMATinsight · Answer

13x+2 = 8y+5

Values of the format 13x+2 are 2, 15, 28, 41, 54, 67, 80,  93 , 106...

Values of the format 8y+5 are 5, 13, 21, 29, 37, 45, 53, 61, 69, 77, 85,  93 ...

The first common Solution as highlighted is  93

every next solution will be at a gap that is multiple of 8 as well as a multiple of 13 i.e. at a gap that is LCM of 13 and 8 which is 104

Hence, Next solution = 93+104 = 197 
Hence, Next solution = 197+104 = 301 etc.

But we have to find solutions less than 180 hence 93 is the only possible solution

Answer: Option B

GMATGuruNY · Answer

A quick lesson on remainders:

Onto the problem at hand:

When n is divided by 13, the remainder is 2. 
n = 13a + 2 = 2, 15, 28, 41, 54, 67, 80,  93 ...
 When n is divided by 8, the remainder is 5. 
n = 8b + 5 = 5, 13, 21, 29, 37, 45, 53, 61, 69, 77, 85,  93 ...

Thus, when n is divided by 104 -- the LCM of 13 and 8 -- the remainder will be 93 (the smallest value common to both lists).
n = 104c + 93 =  93 , 197...

In the resulting list, only the value in green is less than 180.

B

RajatGMAT777 · Answer

aserghe1  and All Experts
Thanks for all your comments and thoughts
My question is - deducing a common form of the solution is alot time taking as calculating 104 as LCM is easy but finding the 1st common number for the above 2 patterns took a very long time and is inefficient 
The method by  aserghe1  takes way less time but I am unsure on how did we deduce the equations 1* and 2* 
Thanks in Advance

Krunaal · Answer

The simply subtracted the divisor from (1) and (2) to get 1* and 2*

n = 13x + 2 - 13 = 13x - 11 and n = 8y + 5 - 8 = 8y - 3

BrushMyQuant · Answer

When positive integer n is divided by 13, the remainder is 2.

Theory: Dividend = Divisor*Quotient + Remainder

n -> Dividend
13 -> Divisor
a -> Quotient (Assume)
2 -> Remainders
=> n = 13*a + 2 = 13a + 2

When n is divided by 8, the remainder is 5.

Let quotient be b
=> n = 8b + 5

Method 1: Algebra (Recommended)

How many such values are less than 180? 
 
n = 13a + 2 = 8b + 5
=> 8b = 13a - 3
=> b =  \frac{13a - 3}{8}

Now, only those values of "b" which will give "a" also as an integer will give us the common values of n which satisfy both n = 13a + 2 = 8b + 5 the conditions

8 is even, so 13a - 3 should also be even if we need to get an integer after the division
=> Only odd values of a are possible

a = 1, 13a - 3 = 10,   NOT divisible   by 8
a = 3, 13a - 3 = 10 + 26 = 36,   NOT divisible   by 8
a = 5, 13a - 3 = 36 + 26 = 62,   NOT divisible   by 8
a = 7, 13a - 3 = 62 + 26 = 88,   Divisible   by 8
a = 9, 13a - 3 = 88 + 26 = 114,   NOT divisible   by 8 
a = 11, 13a - 3 = 114 + 26 = 140,   NOT divisible   by 8
a = 13, 13a - 3 = 140 + 26 = 166,   NOT divisible   by 8
Now, n will cross 180

So, only one value of n is possible and i.e. n = 13*7 + 2 = 93

Method 2: Writing common terms

n = 13a + 2 
=> a = 0, n = 13*0 + 2 = 2
=> a = 1, n = 13*1 + 2 = 15
=> n = 2, 15, 28, 41, 54, 67, 80,  93 , 106, 119, 132, 145, 158, 171,...

n = 8b + 5 
=> n = 5, 13, 21, 29, 37, 45, 53, 61, 69, 77, 85,  93 , 101, 109, 117, 125, 133, 141, 149, 157, 165, 173,...

Only common value is 93

So,  Answer will be B 
Hope it helps!

Playlist on Solved Problems on Remainders  here

Watch the following video to MASTER Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

When positive integer n is divided by 13, the remainder is 2. When n

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When positive integer n is divided by 13, the remainder is 2. When n

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