When an even integer p is divided by 7, the remainder is 6. Which of

Question

When an even integer p is divided by 7, the remainder is 6. Which of the following, when added to p, gives a sum that leaves a remainder of 4 when divided by 14?

(A) 10
(B) 14
(C) 16
(D) 26
(E) 32

e-GMAT Question of the Week #10

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BullRunner · Accepted Answer

If an integer leaves remainder 4 with 14,it will leave the same remainder when divided by 7.
So,we need to find a number that leaves rem 4 when div by 7
A.7p+6+10=>rem. 2(when div by 7)
B.7p+6+14=>rem. 6
C.7p+6+16=>rem. 1
D.7p+6+26=>rem. 4 (Eureka!)
E.7p+6+32=>rem. 3
Ans=D

EgmatQuantExpert · Answer

Solution 
  Given:  
 • p is an even integer ………. (1)
• When p is divided by 7, the remainder is 6 …………. (2)

To find:  
 • Which of the given answer choices, when added to p, gives a sum that leaves a remainder of 4 when divided by 14

Approach and Working:   
 • From statement (2), we can write ‘p’ in the form, N = Q*D + R, as:
 o p = 7a + 6, 
• And, from statement (1), we know that p is n even integer.
 o Implies, 7a + 6 is an even integer
o Since, 6 is an even number, 7a must be even  
o For 7a to be even, 
  ‘a’ must be even, since, 7 is an odd number   
o Thus, let’s assume a = 2b, where b is a positive integer
o Therefore, p = 7 * 2b + 6 = 14b + 6 ………………………………………. (3) 
• Now, let’s assume the number that has to be added to p as ‘n’,
 o Thus, the sum will be = p + n = 14b + 6 + n …………………………… (4)
o This can be written as, 14b + (n+2) + 4 
• We are given that the sum leaves a remainder of 4, when divided by 14,
 o That is, 14b + (n+2) + 4 leaves a remainder of 4, when divided by 14
o Implies, 14b + (n+2) must be divisible by 14
  14b is always divisible by 14
 Thus, n+2 must be divisible by 14
 For this, ‘n’ can take the values of the form, 12 + 14k, where k is an integer

Verifying the options, only option D, which is equal to 26, satisfies this condition.

Hence, the correct answer is option D.

Answer: D

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chetan2u · Answer

so p = 2n, that is an even number
p=2n=7k+6 so k has to be an even integer
the above basically means that when p is divided by 14 or (7k), the remainder is 6..
and when a number is added, remainder 4 is left..

therefore 6+new number -4 = new number +2 should be div by 14
 (a)  10........ 10+2=12... NO
 (b)  14........ 14+2=16... NO
 (c)  16........ 16+2=18... NO
 (d)  26........ 26+2=28.. YES
 (e)  32........ 32+2=34... NO

ans D

BARUAH · Answer

20( even num when divided by 7 gives remainder 6 ) + 26 = 46, divided be 14 leaves remainder 4. 
Ans.D. 26

Sent from my Redmi Note 3 using  GMAT Club Forum mobile app

shelleyjohnsam · Answer

From the question -
P = 7I + 6 ......1)
P + x = 14I + 4 ......2)
Where I is a constant

Therefore, x=7I - 2 ...... From 1 & 2

From the options D) 7x4-2 = 26

Posted from my mobile device

Mona10031990 · Answer

P= 7q+6 ------> 1

Now ,Let S = 14t+4 -----> 2

Equation 2- Equation 1

S-P = 14t+4-7q-6 
 P+New number-P = 7(q-2t)-2
 New number= 7*interger-2
 New number+2= 7*interger

Above equation implies that (new number + 2) is divisible by 7 , hence check which answer option +2 is divisible 
by 7

ANS: D

gracie · Answer

least value of p=6
let x=number added to p
q=quotient
6+x=14q+4→
14q-x=2
we know that x and q are even
if q=2, then x=26
D

Izzyjolly · Answer

Here is my approach.
Let x is number added.
p = 7a+6 , a is non-negative integer
p+x = 14b+4 , b is non-negative integer
--> 7a+6+x = 14b+4 --> x+2 = 14b-7a --> (x+2) is divisible by 7 --> Only D satisfies.
Answer D.

Abhishek009 · Answer

Least possible value of p = 6

NOw, check using options

(A)  \frac{6 + 10}{14}  = Remainder 2
(B)  \frac{6 + 14}{14}  = Remainder 6
(C)  \frac{6 + 16}{14}  = Remainder 8 
 (D)  \frac{6 + 26}{14}  = Remainder 4 
 (E)  \frac{6 + 32}{14}  = Remainder 8

Thus, Answer must be (D)

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When an even integer p is divided by 7, the remainder is 6. Which of

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When an even integer p is divided by 7, the remainder is 6. Which of

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