arora007 wrote:x,q and r are positive integers and x/17 =q remainder r.
What is the remainder when x is divided by 34?
1) q is odd
2) r=5

x/17 =q remainder r

1) q is odd
Assume x = 17
q=1,r=0
Reminder of 17/34 = 17
Assume x= 18
q=1,r=1
Reminder of 18/34 = 18
Hence INSUFF

2) r=5
Assume x = 22
q=1, r= 5
Reminder of 17/34 = 17
Assume x = 39
q=2, r= 5
Reminder of 39/34 = 5
hence INSUFF

Combining:
Assume x = 22
q=1, r= 5
Reminder of 22/34 = 22
Assume x = 56
q=3, r= 5
Reminder of = 22
Hence INSUFF

pick C.

Made a Typo. Edited back...

The answer is C

If remain = 5 and quotient is odd number as given, according to periodicity of 17 and 34, this information is enough to tell us that the remainder will always be 22.

Regards
Archie

archiekins2007 wrote:The answer is C

If remain = 5 and quotient is odd number as given, according to periodicity of 17 and 34, this information is enough to tell us that the remainder will always be 22.

Regards
Archie

Hey archie your solution is correct. I tried by putting dummy values in the expressions to solve.
Can u please explain what is the periodicity of 17 and 34. I didnt understand what you meant exactly.

Ishaan

Hey Ishaan,
Its difficult to put in words but I can try:)
Multiples of 34 will repeat itslef periodically for ever even multiple of 17. So if you draw on a sheet, you will see that if we know that a number divided by 17 has and eve or an odd quientient along with a given remainder, it will always generate the same remainder for 34 as well.
Its kind of like the same approach you used when you put in nunbers, just by drawing it on a number-line.

Regards
Archie

OA is C

archiekins2007 wrote:Hey Ishaan,
Its difficult to put in words but I can try:)
Multiples of 34 will repeat itslef periodically for ever even multiple of 17. So if you draw on a sheet, you will see that if we know that a number divided by 17 has and eve or an odd quientient along with a given remainder, it will always generate the same remainder for 34 as well.
Its kind of like the same approach you used when you put in nunbers, just by drawing it on a number-line.

Regards
Archie

Hey archie,
thanks buddy, i got the idea. since 34 is even multiple of 17 (i.e. 2) the remainder will be the same as that of 17 for all even multiples of 34. And when we divide 34 by odd multiples of 34 we will get the same remainder (22).

I think i will note down this point in my flash cards. Its a very important deduction which can save much time in exam.
Thanks again.
Ishaan

using periodicity of a number to determine remainder (r):
please can someone give simpler examples...

Am confused at the explanations given above: e.g
if x/17 = q + r given: x/17 = 1 + r(5)
then for x/34 ==> r = (17 +r )?, where 17 + r = 22, right?

How could be do it algebaricaly ?

That is to set a formula as X = 17q + R and X + 34q1 + r1 ..

still don't know how the equation you cited above clarifies my question.
Well, one thing I know is this:
x = 34q + r.....................where x, q, and r are positive integers.
the remainder (r) as above will yield the same value for 17 if
x = 17Q + r only if Q = 2q (even multiple). Also, 17Q <= 34q

likewise for:
x = 68 + R
we expect that for "x = 17(4) + r "
R = r because 68 is an even multiple of 17