What is the remainder when the positive integer x is divided by 6?

Question

(1) When x is divided by 7, the remainder is 5.
(2) When x is divided by 9, the remainder is 3.

vivekkumar987 · Answer

IMO : E
12, when divided by 7 leaves a remainder 5 & when divided by 9 leaves a remainder 3
So 12/6 remainder 0

75 , when divided by 7 leaves a remainder 5 & when divided by 9 leaves a remainder 3
So 75/6 remainder 3

So the answer is E

arvind910619 · Answer

Imo E 
From the fist condition we can not determine any value of X as there are two variable X=7a+5
a can any value 
From statement 2 
We have X=9b+3
From this also we can deduce whether X is divisible by 6 
If taken together the equation becomes 
63Q+12 
Here Q is any positive integer 
We see that both the statements taken together are not sufficient.

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DUMDUM21 · Answer

If you are solving by taking samples numbers like 12 or 75 then it might take longer time to come up with the number that doesn't meet both constraints. Don't we have a formula for these type of questions?

Bunuel · Answer

Given:

x = 7m + 5 
 x = 9n + 3

We can derive a general formula for the total (of a type  x=dq+r , where  d  is the divisor and  r  is the remainder) based on the two formulas given:  x = 7m + 5  (5,  12 , 19, and so on) and  x= 9n + 3  (2,  12 , 21, and so on).

The divisor  d  would be the least common multiple of the two divisors 7 and 9, hence  d=63  and the remainder  r  would be the first common integer in the two patterns, hence  r=12 :

x = 	ext{(the LCM of m and n)} + 	ext{(the first common value for x)}

So, the general formula based on both pieces of information is  x= 63q+12 . Thus, x could be 12, 75, 138, and so on.

Hope this helps.

HarshavardhanR · Answer

Hi  DUMDUM21 ,

In such cases, one merely has to find the first number that ->

(1) when divided by 7 gives a remainder of 5  AND 
(2) when divided by 9 gives a remainder of 3.

The rest can be found easily. Here is how.

Numbers that give a remainder of 5 when divided by 7 -> 7a + 5 -> 5,  12 , 19, etc.
Numbers that give a remainder of 3 when divided by 9 -> 9b + 3 -> 3,  12 , 21, etc.

12 is the first number that satisfies both the above conditions.

The whole sequence of such numbers (satisfying both conditions) can be arrived at using ->  (lcm of 7 and 9)  (quotient) + 12.

i.e.  63q + 12.

Thus the numbers satisfying both conditions are -> 12, 75, 138, etc.

In essence, find the first number satisfying both, then add lcm to arrive at more such numbers.

Here, 
S1 -> Not sufficient
S2 -> Not sufficient
S1+S2 -> Satisfying both -> 63q + 12 -> 12, 75, etc.

- 12 when divided by 6 gives a remainder of 0.
- But 75 when divided by 6 gives a remainder of 3.
In other words, even with S1 and S2, we do not have enough to arrive at a concrete answer.

Hope this helps. 
___
Harsha
Enthu about all things GMAT | Exploring the GMAT space | My website: gmatanchor.com

What is the remainder when the positive integer x is divided by 6?

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