When positive integer n is divided by 5, the remainder is 2; when n is

Question

When positive integer n is divided by 5, the remainder is 2; when n is divided by 6, the remainder is 3. If 0<n<100, how many possible values of n are there?

A. 1
B. 2
C. 3
D. 4
E. 5

BrushMyQuant · Accepted Answer

Theory: Dividend = Divisor*Quotient + Remainder

When positive integer n is divided by 5, the remainder is 2
=> n = 5x + 2 [ x being the quotient] ...(1)

When positive integer n is divided by 6, the remainder is 3
=> n = 6y + 3 [ y being the quotient] ...(2)

There are multiple ways to solve the problem. Two of them are listed below

Method 1: Compare the two equations for n (Faster Method)
n = 5x + 2 = 6y + 3
=> x = \(\frac{6y+1}{5}\)
So, only those values of y which will give x as integer will give common values for n in (1) and (2)
We need to get units digit of 6y as 4 or 9 to make 6y+1 a multiple of 5
So, y can be 4 giving x as \(\frac{6*4+1}{5}\) = 5 => n = 6*4 + 3 = 27
y can also be 9 giving 6y as 54 and x as \(\frac{6*9+1}{5}\) = 11. => n = 6*9 + 3 = 57
y can be 14 giving x as an integer. n = 6*14 + 3 = 87
y = 19 will become too big for n < 100
So, 3 values of n such that 0 < n < 100 are possible.

Method 2: Write all values of n in both the cases and compare them
n = 5x + 2
Put x=0,1,2,3,4... and find the corresponding values of n
x=0, n = 5*0 + 2 = 2 and so on..
So values n will be
2, 7, 12, 17, 22,27.... So numbers ending with 2 and 7

n = 6y + 3
Put y=0,1,2,3,4... and find the corresponding values of n
x=0, n = 6*0 + 3 = 3 and so on..
So values n will be
3, 9, 15, 21, 27,....
If we write all the numbers then we can find the common numbers as 27,57,87

So, Answer will be C
Hope it helps!

Watch the following video to MASTER Remainders

yashikaaggarwal · Answer

When positive integer n is divided by 5, the remainder is 2; 
Here Value of n can be = 2,7,12,17,22, 27 ....................................

when n is divided by 6, the remainder is 3. 
Here Value of n can be = 3,9,15,21, 27 ,33....................................

The first no. common to both constraint is 27
Therefore, N is a multiple of 27
multiple of 27 = 27,54,81,108..........
If 0<n<100,
only 3 values lie in the given constraint (0<n<100)

Answer is C

Archit3110 · Answer

given when n is divided by 5 we get remainder 2 so possible values of n 
2,7,12,17,22, 27 ,32,37,42,47,52, 57 ,62,67,72,77,82, 87 ,92,97
also given that when n is divided by 6 remainder is 3 possible values
3,9,15,21, 27 ,33,39,45,51, 57 ,63,69,75,81, 87 ,93,99
common values of n ; 27 , 57, 87 , 
option c

Saasingh · Answer

C.

We can see that the difference between divisor and remainder = 3 in both cases. Thus n is of the form LCM(5,6) - 3 = 30k-3.

Thus, n can be = 27,57,87

hellothere88 · Answer

hi do you mind explaining how do you get -3?

Saasingh · Answer

Hi.

Sure. 
The -3 is the difference between the divisor and remainder. As you can see the difference is 3 for both the pairs of divisors-remainder. 
5-2 = 3 and 6-3= 3. Commond difference needs to be subtracted from lcm.

Besides, you can also solve this by creating 2 separate equations using divident = divisor * k + remainder 
And then solving both eqns to get the smallest such number. 
Once you have the smallest number, keep adding the LCM of two divisors until you reach the limit.

Posted from my mobile device

ramlala · Answer

When n is devided by 6 remainder is 3.

Possible values.. 3,9,15,21, 27 
Hey we got the 1st value as 27....(deivded by 5, remainder is 2)

Now add LSM of 6 & 5...i.e. 30 to 27

27+30=57, 57+30=87, 87+30=117>100

Total values are 27, 57 & 87

Ans:  C

ScottTargetTestPrep · Answer

Solution:
 
Since both 2 and 3 are 3 less than 5 and 6, respectively, the smallest positive integer that satisfies the required conditions is 3 less than LCM(5, 6), i.e, 30 - 3 = 27. The subsequent integers will be 30 more than the previous integer. In other words, the possible values of n are 27, 57, 87, 117, and so on. However, since n is less than 100, there are only 3 possible values of n.

Alternate Solution:
 
Since the remainder when n is divided by 5 is 2, we can write n = 5k + 2 for some positive integer k. Similarly, since the remainder when n is divided by 6 is 3, we can write n = 6s + 3 for some positive integer s. Now, notice that n + 3 = 5k + 5 is divisible by 5 and also n + 3 = 6s + 6 is divisible by 6. Thus, n + 3 is divisible by LCM(5, 6) = 30.

If n + 3 = 30, then n = 27. 
If n + 3 = 60, then n = 57. 
If n + 3 = 90, then n = 87.

The next possible value for n + 3 (which is 120) yields a value for n which is greater than 100; hence, there are 3 possible values for n.

Answer: C

bv8562 · Answer

Hi  ScottTargetTestPrep , can I apply the above highlighted method in an equation such as this:

x = 4q+3 (4-3 = 1)
x = 9p+4 (9-4 = 5)

LCM od 4 and 9 = 36

Thus, x = 36*k + (36-5) = 36k + 31

Is this the right application? Also, could you please explain the rationale behind it?

shivanks · Answer

yaa ans is C but not in that manner multiple of 27 is 81 and dividing 81 by 5 then reminder is not 2, or same with 54 when divided by 5 then reminder is not 2...
one common pattern is 30k-3 = n (k= 1,2,3..)

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When positive integer n is divided by 5, the remainder is 2; when n is

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When positive integer n is divided by 5, the remainder is 2; when n is

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