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

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When positive integer n is divided by 3, the remainder is 1. When n is  [#permalink]

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New post 20 Oct 2015, 11:20
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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  [#permalink]

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New post 24 Oct 2015, 22:37
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3
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  [#permalink]

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New post 20 Oct 2015, 11:54
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=3m-1 and p=7t-5

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  [#permalink]

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New post 20 Oct 2015, 22:14
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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 B


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Re: When positive integer n is divided by 3, the remainder is 1. When n is  [#permalink]

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New post 24 Nov 2015, 04:53
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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=3m-1 and p=7t-5

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  [#permalink]

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New post 25 Nov 2015, 00:47
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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  [#permalink]

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New post 25 Nov 2015, 05:49
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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  [#permalink]

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New post 30 Nov 2015, 08:35
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  [#permalink]

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New post 05 Jun 2016, 04:11
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 = 21-19 = 2.
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When positive integer n is divided by 3, the remainder is 1. When n is  [#permalink]

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New post Updated on: 12 Jul 2018, 17:20
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➡
3q-7p=4
least multiple of 3 that is four greater than multiple of 7 is 18
q=6
p=2
n=19
21-19=2=p
B

Originally posted by gracie on 05 Jun 2016, 16:19.
Last edited by gracie on 12 Jul 2018, 17:20, 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  [#permalink]

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New post 29 Jun 2016, 09:16
gracie wrote:
n=1+3(7-1)=19
21-19=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  [#permalink]

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New post 29 Jun 2016, 13:30
Ilomelin wrote:
gracie wrote:
n=1+3(7-1)=19
21-19=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 3--1,4,7,10,13,16,19--
and for 7--5,12,19--
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(d2-1)
I hope this helps
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Re: When positive integer n is divided by 3, the remainder is 1. When n is  [#permalink]

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New post 12 Jul 2017, 15:58
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  [#permalink]

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New post 14 Aug 2018, 00:21
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 &nbs [#permalink] 14 Aug 2018, 00:21
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