A positive integer leaves a remainder 3 on division by 14, and leaves

Question

A positive integer leaves a remainder 3 on division by 14, and leaves a remainder k on division by 35. How many possible values can k take?

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

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GMATinsight · Accepted Answer

14a + 3 = 35b+k

Numbers of the type 14a + 3 will be {3, 17, 31, 45, 59, 73, 87 ...}

When {3, 17, 31, 45, 59, 73, 87 ...} are divided by 35 then remainders are { 3, 17, 31, 10, 24,  3, 17...} respectively

i.e. 5 distinct remainders

Answer: Option E

SaquibHGMATWhiz · Answer

Solution:

let the positive integer be N 
 The positive integer N leaves a remainder 3 on division by 14. So, we can write  N=14p+3  where k is a non-negative integer 
 The positive integer N leaves a remainder k on division by 35. So, we can write  N=35q+k  where q is a non-negative integer and k is the remainder 
 We have to keep in mind that value of  k<35

We can write  14p+3=35q+k 
 ⇒k=14p+3-35q 
 ⇒k=14p-35k+3 
 ⇒k=7(2p-5k)+3 
 ⇒k=7r+3  
 From the above obtained equation  k=7r+3  and the value of k is less than 35 
 So, the possible values of k is when  r=0, 1, 2, 3, 4  and the value of  k=3, 10, 17, 24, 31

Hence the right answer is    Option E

akwenok · Answer

SaquibHGMATWhiz

I don't really get why you say that 35q = 35k and then (2p-5k) = r

Could you please elaborate?

I first missed it by getting 3, 17, 31 only... but then realized that also x: {k, k+35, K+70 ...} so k = x ; k = x - 35 k k = x - 70...

Hence as x: {3, 17, 31, 45, 59, 73, 87} then we have k: { 3, 17, 31, 10, 24 ,  3, 17, 31 , ...} so the repeating starts...

kanikaa9 · Answer

SaquibHGMATWhiz

I have some questions. Note:
1. How did you re-write 35q as 35k?
2. Why is 2p - 5k taken as r?

Please reply!  Bunuel  tagging you if you have a shortcut for this!! Many thanks!

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A positive integer leaves a remainder 3 on division by 14, and leaves

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A positive integer leaves a remainder 3 on division by 14, and leaves

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