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When integer x is divided by 5 the remainder is 3 and when x is divide

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Bunuel
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When integer x is divided by 5 the remainder is 3 and when x is divide [#permalink] New post   05 Dec 2016, 02:23
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When integer x is divided by 5 the remainder is 3 and when x is divided by 4 the remainder is 3. If x is between 32 and 101, which of the following could be the value of 3x?

A. 249
B. 169
C. 123
D. 83
E. 63
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Re: When integer x is divided by 5 the remainder is 3 and when x is divide [#permalink] New post   05 Dec 2016, 10:42
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Bunuel wrote:
When integer x is divided by 5 the remainder is 3 and when x is divided by 4 the remainder is 3. If x is between 32 and 101, which of the following could be the value of 3x?

A. 249
B. 169
C. 123
D. 83
E. 63


Check the options -

(A) 3x = 249 , so x = 83 ( 3 more than a multiple of 5 & a multiple of 4 )

(B) 3x = 169 , so x = Not a multiple of 3

(C) 3x = 123 , so x = 41 ( 1 more than a multiple of 5 & a multiple of 4 )

(D) 3x = 83 , so x = Not a multiple of 3

(E) 3x = 63 , so x = 21 ( 1 more than a multiple of 5 & a multiple of 4 )

Hence, Correct answer must be (A) 249
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Re: When integer x is divided by 5 the remainder is 3 and when x is divide [#permalink] New post   05 Dec 2016, 02:35
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\(X=5a+3\)
\(X=4b+3\)\(\)

Hence \(X = LCM (4,5)*n + 3\)

\(X = 3, 23, 43, 63, 83 ...\)

\(32 < X < 101\) -----> \(X\) can be 43, 63, and 83

\(3X\) can be 129, 189 and 249.

Answer A.
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Re: When integer x is divided by 5 the remainder is 3 and when x is divide [#permalink] New post   05 Dec 2016, 06:05
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Given x > 32 therefore Answer D and E eliminated. Ans B is not divisible by 3. Hence answer should be either A or C. In A x=83 in B x=41. as A satisfy the given condition of divisibility. Ans A.

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Re: When integer x is divided by 5 the remainder is 3 and when x is divide [#permalink] New post   18 Apr 2018, 08:34
A shorter version, we have:

x = 11y + 3 (1)
x = 19k + 3, k int (2)
y = 19p + r, p int (3)
r = ?

from (1) & (2) 11y + 3 = 19k + 3
-> 11y = 19k

11 and 19 are prime numbers, so choose y = 19, k = 11 (you can choose any values as long as the left hand side = right hand side, i.e. y = 38, k = 22)

when y = 19: from (3) -> 19p + r = 19 -> p = 1, r = 0 -> no remainder
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When integer x is divided by 5 the remainder is 3 and when x is divide [#permalink] New post   Updated on: 19 May 2018, 12:30
Bunuel wrote:
When integer x is divided by 5 the remainder is 3 and when x is divided by 4 the remainder is 3. If x is between 32 and 101, which of the following could be the value of 3x?

A. 249
B. 169
C. 123
D. 83
E. 63


least value of x=3
values of x>32and<101=3+n(5*4)=43,63,83
3 * each value respectively=129,189,249
249
A

Originally posted by gracie on 18 Apr 2018, 10:59.
Last edited by gracie on 19 May 2018, 12:30, edited 2 times in total.
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Re: When integer x is divided by 5 the remainder is 3 and when x is divide [#permalink] New post   19 May 2018, 02:07
A very simple way. It's given
x = 5a + 3
x = 4b + 3

We need to find 3x, so:
3x = 15a + 9
3x = 12b + 9

From the given answers we should pick the one which will be multiple of 15 and 12 after the deduction of 9:
A. 249 -> 249 - 9 = 240, it is multiple of 12 and 15, correct
B. 169
C. 123
D. and E. do no consider since X >=32, therefore 3x >= 96
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Re: When integer x is divided by 5 the remainder is 3 and when x is divide [#permalink] New post   20 Jul 2021, 06:35
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x = 5p + 3 {for p = 0, 1, 2, 3...x = 3, 8, 13, 18... }

x = 4p + 3 {for p = 0, 1, 2, 3...x = 3, 7, 11, 15... }

First overlap number : 3

Next overlap number: 3 + LCM (5, 4 = 20) = 23, 43, 63, 83

Possible values of x = 3 , 23, 43, 63, 83

3x = 3 * 83 = 249

Answer A
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Re: When integer x is divided by 5 the remainder is 3 and when x is divide [#permalink] New post   17 Feb 2024, 05:33
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Re: When integer x is divided by 5 the remainder is 3 and when x is divide [#permalink]
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