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Question of the Week 10 ( When an even integer p is divided by 7,...)
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Updated on: 13 Aug 2018, 00:14
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eGMAT Question of the Week #10When 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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Re: Question of the Week 10 ( When an even integer p is divided by 7,...)
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05 Aug 2018, 23:07
EgmatQuantExpert wrote: eGMAT Question of the Week #10When 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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Re: Question of the Week 10 ( When an even integer p is divided by 7,...)
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06 Aug 2018, 00:15
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,...)
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06 Aug 2018, 01:45
EgmatQuantExpert wrote: eGMAT Question of the Week #10When 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,...)
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06 Aug 2018, 02:17
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) 7x42 = 26
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Re: Question of the Week 10 ( When an even integer p is divided by 7,...)
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06 Aug 2018, 08:29
[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 SP = 14t+47q6 P+New numberP = 7(q2t)2 New number= 7*interger2 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,...)
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16 Aug 2018, 00:34
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: • 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. Answer: D
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Question of the Week 10 ( When an even integer p is divided by 7,...)
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18 Aug 2018, 09:13
EgmatQuantExpert wrote: eGMAT Question of the Week #10When 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 let x=number added to p q=quotient 6+x=14q+4→ 14qx=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,...)
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18 Aug 2018, 10:21
EgmatQuantExpert wrote: eGMAT Question of the Week #10When 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. Let x is number added. p = 7a+6 , a is nonnegative integer p+x = 14b+4 , b is nonnegative integer > 7a+6+x = 14b+4 > x+2 = 14b7a > (x+2) is divisible by 7 > Only D satisfies. Answer D.
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Re: Question of the Week 10 ( When an even integer p is divided by 7,...)
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18 Aug 2018, 11:29
EgmatQuantExpert wrote: eGMAT Question of the Week #10When 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 8Thus, Answer must be (D)
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