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When positive integer n is divided by 3, the remainder is 1. When n is
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20 Oct 2015, 12:20
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When positive integer n is divided by 3, the remainder is 1. When n is divided by 7, the remainder is 5. What is the smallest positive integer p, such that (n + p) is a multiple of 21? (A) 1 (B) 2 (C) 5 (D) 19 (E) 20
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Re: When positive integer n is divided by 3, the remainder is 1. When n is
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24 Oct 2015, 23:37
When positive integer n is divided by 3, the remainder is 1 i.e., n=3x+1 values of n can be one of {1, 4, 7, 10, 13, 16, 19, 22..............49, 52, 59..................}
Similarly, When n is divided by 7, the remainder is 5..i.e., n=7y+5 values of n can be one of {5, 12, 19, 26, 32, 38, 45, 52, 59........}
combining both the sets we get n={19, 52, ...........}
What is the smallest positive integer p, such that (n + p) is a multiple of 21 or 21x i.e., {21, 42, 63, 84..........}. in case of n=19 p=2 in case of n=52 p=11 . . . . so on
as we observe the value of p increases as the value of n increases. so for min value of p, we take min value of n. B is the answer.




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Re: When positive integer n is divided by 3, the remainder is 1. When n is
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20 Oct 2015, 12:54
Bunuel wrote: When positive integer n is divided by 3, the remainder is 1. When n is divided by 7, the remainder is 5. What is the smallest positive integer p, such that (n + p) is a multiple of 21?
(A) 1 (B) 2 (C) 5 (D) 19 (E) 20 n+p=3a+1+p = 7b+5+p = 21m =>p+1=3m and p+5=7t =>p=3m1 and p=7t5 P is positive so m>=1 and t>=1 p is the smallest positive when m=1 and t=1 => p=2 Ans: B



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When positive integer n is divided by 3, the remainder is 1. When n is
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20 Oct 2015, 23:14
Bunuel wrote: When positive integer n is divided by 3, the remainder is 1. When n is divided by 7, the remainder is 5. What is the smallest positive integer p, such that (n + p) is a multiple of 21?
(A) 1 (B) 2 (C) 5 (D) 19 (E) 20 given number is of the form 3x + 1 and 7y + 5 \(=> 3x + 1 = 7y + 5\) \(=> x = (7y + 4) / 3\) the lowest integer values which satisfies this are y = 2 and x = 6 and number = 7y + 5 = 19 therefore lowest value to be added to 19 make it divisible by 21 is 2 answer choice Bkudos if you like the explanation



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Re: When positive integer n is divided by 3, the remainder is 1. When n is
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24 Nov 2015, 05:53
Could shed some light on how the second 2) is derived , many thanks 1) n+p=3a+1+p = 7b+5+p = 21m 2) =>p+1=3m and p+5=7t 3) =>p=3m1 and p=7t5
P is positive so m>=1 and t>=1 p is the smallest positive when m=1 and t=1 => p=2



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Re: When positive integer n is divided by 3, the remainder is 1. When n is
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25 Nov 2015, 01:47
Bunuel wrote: When positive integer n is divided by 3, the remainder is 1. n = { 4 , 7 , 10 , 13, 16, 19 ........} Bunuel wrote: When n is divided by 7, the remainder is 5. n = { 12 , 19 ........} Bunuel wrote: What is the smallest positive integer p, such that (n + p) is a multiple of 21? 19 + p = 21 { Smallest multiple of 21 }So, p = 2 Hence answer is (B) 2
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Re: When positive integer n is divided by 3, the remainder is 1. When n is
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25 Nov 2015, 06:49
Many thanks abishek009 really appreciate it



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Re: When positive integer n is divided by 3, the remainder is 1. When n is
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30 Nov 2015, 09:35
n/3 = q +1 n/7= q + 5
n+p : possible values ; 21,42,63 we need smallest value of p
now guess value of n based on values of n+p i.e, close to 21,42,63
try 20; fails first
try 40 first ; ok second condition ok therefore n can be 40 in which case p has to be 2



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Re: When positive integer n is divided by 3, the remainder is 1. When n is
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05 Jun 2016, 05:11
The smallest integer that leaves a remainder 1 when divided by 3 and remainder of 5 when divided by 7 is 19.
Smallest multiple of 21 is 21 itself hence smallest value of P = 2119 = 2.



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When positive integer n is divided by 3, the remainder is 1. When n is
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Updated on: 12 Jul 2018, 18:20
When positive integer n is divided by 3, the remainder is 1. When n is divided by 7, the remainder is 5. What is the smallest positive integer p, such that (n + p) is a multiple of 21?
(A) 1 (B) 2 (C) 5 (D) 19 (E) 20
n=3q+1 n=7p+5 3q+1=7p+5➡ 3q7p=4 least multiple of 3 that is four greater than multiple of 7 is 18 q=6 p=2 n=19 2119=2=p B
Originally posted by gracie on 05 Jun 2016, 17:19.
Last edited by gracie on 12 Jul 2018, 18:20, edited 4 times in total.



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Re: When positive integer n is divided by 3, the remainder is 1. When n is
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29 Jun 2016, 10:16
gracie wrote: n=1+3(71)=19 2119=2 p=2 Hello, could you please explain how you got to that equation? Greetings.
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When positive integer n is divided by 3, the remainder is 1. When n is
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29 Jun 2016, 14:30
Ilomelin wrote: gracie wrote: n=1+3(71)=19 2119=2 p=2 Hello, could you please explain how you got to that equation? Greetings. hi, Ilomelin in looking at the possible values of n for 31,4,7,10,13,16,19 and for 75,12,19 I noted that for 3, 19 is the 7th option for 7, 19 is the 3rd option and derived an equation let r=remainder d=divisor n=r1+d1(d21) I hope this helps gracie



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Re: When positive integer n is divided by 3, the remainder is 1. When n is
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12 Jul 2017, 16:58
Bunuel wrote: When positive integer n is divided by 3, the remainder is 1. When n is divided by 7, the remainder is 5. What is the smallest positive integer p, such that (n + p) is a multiple of 21?
(A) 1 (B) 2 (C) 5 (D) 19 (E) 20 Since when n is divided by 3, the remainder is 1, n could be: 1, 4, 7, 10, 13, 16, 19, 22, … Since when n is divided by 7, the remainder is 5, n could be: 5, 12, 19, 26, … We can see that n = 19 satisfies both division/remainder criteria. And if p = 2, we have n + p = 19 + 2 = 21, which is a multiple of 21. Answer: B
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Re: When positive integer n is divided by 3, the remainder is 1. When n is
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14 Aug 2018, 01:21
1) 3K+1 = 1,4,7,10,13,16,19,22,25........ 2) 7N+5 = 5,12,19...... so the this number should be of form 21M+19 and therefore for n+p to be a multiple of 21 we need to add 2 ANS B



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Re: When positive integer n is divided by 3, the remainder is 1. When n is
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19 Dec 2018, 12:20
Easiest way is to just plug in.
A) if p=1 then n=20 >>> 20/3=6 r.2  incorrect as the question says the remainder has to be 1 B) if p=2 then n=19 >>> 19/3=6 r.1 & 19/7=2 r.5  everything fits
Answer: B




Re: When positive integer n is divided by 3, the remainder is 1. When n is
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