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e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3203
Question of the Week- 10 ( When an even integer p is divided by 7,...)  [#permalink]

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10 00:00

Difficulty:   45% (medium)

Question Stats: 66% (02:26) correct 34% (02:36) wrong based on 176 sessions

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Question of the Week #10

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?

Options

(a) 10
(b) 14
(c) 16
(d) 26
(e) 32

To access all the questions: Question of the Week: Consolidated List

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Originally posted by EgmatQuantExpert on 05 Aug 2018, 22:42.
Last edited by EgmatQuantExpert on 13 Aug 2018, 00:14, edited 2 times in total.
Math Expert V
Joined: 02 Aug 2009
Posts: 8340
Re: Question of the Week- 10 ( When an even integer p is divided by 7,...)  [#permalink]

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EgmatQuantExpert wrote:
Question of the Week #10

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?

Options

(a) 10
(b) 14
(c) 16
(d) 26
(e) 32

To access all the QOW questions: Question of the Week: Consolidated List

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
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GMAT 1: 660 Q48 V34 Re: Question of the Week- 10 ( When an even integer p is divided by 7,...)  [#permalink]

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1
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
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Re: Question of the Week- 10 ( When an even integer p is divided by 7,...)  [#permalink]

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EgmatQuantExpert wrote:
Question of the Week #10

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?

Options

(a) 10
(b) 14
(c) 16
(d) 26
(e) 32

To access all the QOW questions: Question of the Week: Consolidated List
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
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Re: Question of the Week- 10 ( When an even integer p is divided by 7,...)  [#permalink]

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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

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Joined: 12 May 2018
Posts: 13
Re: Question of the Week- 10 ( When an even integer p is divided by 7,...)  [#permalink]

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1
[As Dividend = Divisor*Quotient + Remainder ]

P= 7q+6 ------> 1 [q is an integer]

Now ,Let S = 14t+4 -----> 2 [ t is an integer & S= P+ New number]

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 [ 26+2=28]
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Re: Question of the Week- 10 ( When an even integer p is divided by 7,...)  [#permalink]

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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 [even + even = even]
o For 7a to be even,
 ‘a’ must be even, since, 7 is an odd number [odd * even = even]
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.

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Question of the Week- 10 ( When an even integer p is divided by 7,...)  [#permalink]

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EgmatQuantExpert wrote:
Question of the Week #10

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?

Options

(a) 10
(b) 14
(c) 16
(d) 26
(e) 32

least value of p=6
q=quotient
6+x=14q+4→
14q-x=2
we know that x and q are even
if q=2, then x=26
D
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Re: Question of the Week- 10 ( When an even integer p is divided by 7,...)  [#permalink]

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EgmatQuantExpert wrote:
Question of the Week #10

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?

Options

(a) 10
(b) 14
(c) 16
(d) 26
(e) 32

To access all the questions: Question of the Week: Consolidated List

Here is my approach.
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.
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Re: Question of the Week- 10 ( When an even integer p is divided by 7,...)  [#permalink]

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1
EgmatQuantExpert wrote:
Question of the Week #10

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?

Options

(a) 10
(b) 14
(c) 16
(d) 26
(e) 32

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

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