When positive integer x is divided by 3, the remainder is 1; when x is

Question

When positive integer x is divided by 3, the remainder is 1; when x is divided by 5, the remainder is 1. If x+y is divisible by both 3 and 5, which of the following is the least value of positive integer y?

A. 4
B. 6
C. 14
D. 16
E. 30

When integer x is divided by 3, the remainder is 1; when x is divided by 5, the remainder is 1. If x+y is divisible by both 3 and 5, which of the following is the least value of positive integer y?

imeanup · Answer

x=3q+1 
 x=5p+1 
Value of  x  may be  {16,31,46,...}  the  HCF  ,or  GCF  of  {3,5}=15 
The least possible number  (x+y) = (16+14)=30 
Therefore,  y  must be  14 .

IMO  Ans C

TarunKumar1234 · Answer

Given, When positive integer x is divided by 3, the remainder is 1; when x is divided by 5, the remainder is 1. If x+y is divisible by both 3 and 5, which of the following is the least value of positive integer y?

As x is divisible by 3 and 5 with remainder 1. So, x is divisible by 15 and remainder is 1. so, x belongs to 1, 16, 31....

For x+y to be divisible by 15, we need y= 14

Going by option, min value of y= 14 then x =1
So, I think C.

V2richandu · Answer

As per the question when x is divided by 3 and 5 ,in both cases the remainder is 1.
That means x+1 is divisible by 3 and 5. So y must be 1,but it's not there in options, so the next number divisible by both 3 and 5 (or divisible by 15) is x+1+15 i.e. x+16.
So y = 16.

So answer must be (D)

Posted from my mobile device

Sush19947 · Answer

Since x+y is divisible by both 3 and 5, x+y should be a multiple of 15 (i.e. 0,15,30,etc.) Now, x can take the values 1,16,31, etc. Since y cannot be negative, x+y cannot be 0.
Therefore, 
if x=1 then x+y=15 and y=14 
if x=16 then x+y=30 and y=14

so for any possible value of x and x+y the least possible value of y is 14.

Hope this helps.

BrushMyQuant · Answer

When positive integer x is divided by 3, the remainder is 1 and when x is divided by 5, the remainder is 1.

As x leaves same remainder (1) when divided by both 3 and 5
=> x = 1 + Multiple of LCM(3,5)
=> x = 1 + 15k (where k is an integer)
=> x = 15k + 1

Now,  x + y has to be a multiple of both 3 and 5 
=> x + y will be a multiple of LCM(3,5) = 15

x = 15k + 1 and is (15-1=) 14 less than a multiple of 15
=> Least value of y will be 14

So,  Answer will be C 
Hope it helps!

Watch the following video to MASTER Remainders

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

When positive integer x is divided by 3, the remainder is 1; when x is

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

When positive integer x is divided by 3, the remainder is 1; when x is

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards