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If a is a positive integer and if remainders of 4 and 6 are obtained

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If a is a positive integer and if remainders of 4 and 6 are obtained  [#permalink]

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New post 22 Feb 2018, 20:44
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If a is a positive integer and if remainders of 4 and 6 are obtained  [#permalink]

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New post 22 Feb 2018, 22:13
Bunuel wrote:
If a is a positive integer and if remainders of 4 and 6 are obtained when 89 and 125, respectively, are divided by a, then a =

(A) 7

(B) 9

(C) 15

(D) 17

(E) 19


From the above question stem, we get the following equations
xa + 4 = 89 -> (1)
ya + 6 = 125 -> (2)

(1) -> xa = 85
(2) -> ya = 119

Dividing (2) by (1) we get \(\frac{x}{y} = \frac{5}{7}\)

If x=5 and y=7, then the value of a is 17(Option D)
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Re: If a is a positive integer and if remainders of 4 and 6 are obtained  [#permalink]

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New post 23 Feb 2018, 01:15
Bunuel wrote:
If a is a positive integer and if remainders of 4 and 6 are obtained when 89 and 125, respectively, are divided by a, then a =

(A) 7

(B) 9

(C) 15

(D) 17

(E) 19


We'll go one option at a time and eliminate impossible answers.
This is an Alternative approach.

If 89/a gives a remainder of 4 that means that a divides 89 - 4 = 85.
(A) divides 84, (B) divides 90, (C) divides 90 and (E) divides 76 (and 95) so none of them divide 85.

Our answer must be (D).
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If a is a positive integer and if remainders of 4 and 6 are obtained  [#permalink]

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New post 23 Feb 2018, 02:07
Bunuel wrote:
If a is a positive integer and if remainders of 4 and 6 are obtained when 89 and 125, respectively, are divided by a, then a =

(A) 7

(B) 9

(C) 15

(D) 17

(E) 19


check with options

89 mod 7 != 4
89 mod 9 != 4
89 mod 15 != 4
89 mod 17 = 4
125 mod 17 = 6

(D) imo
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Re: If a is a positive integer and if remainders of 4 and 6 are obtained  [#permalink]

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New post 23 Feb 2018, 05:07
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Bunuel wrote:
If a is a positive integer and if remainders of 4 and 6 are obtained when 89 and 125, respectively, are divided by a, then a =

(A) 7

(B) 9

(C) 15

(D) 17

(E) 19


When 89 is divided by a then the remainder is 4
i.e. 89-4=85 must be divisible by a


When 125 is divided by a then the remainder is 6
i.e. 125-6=119 must be divisible by a

i.e. a must be a factor of 85 as well as of 119

HCF of 85 and 119 = 17

Hence a = 17

Answer: option D
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Re: If a is a positive integer and if remainders of 4 and 6 are obtained  [#permalink]

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New post 15 Mar 2018, 06:47
Bunuel wrote:
If a is a positive integer and if remainders of 4 and 6 are obtained when 89 and 125, respectively, are divided by a, then a =

(A) 7

(B) 9

(C) 15

(D) 17

(E) 19


89 = nq + 4

\(\frac{85}{q}\)= n -----I

125 = np + 6

\(\frac{119}{p}\) = n -----II

From I & II, n =\(\frac{85}{q}\)= \(\frac{119}{p}\)

17 is the number which divides both 85 & 119.

(D)
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Re: If a is a positive integer and if remainders of 4 and 6 are obtained  [#permalink]

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New post 15 Mar 2018, 07:00
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Re: If a is a positive integer and if remainders of 4 and 6 are obtained  [#permalink]

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New post 15 Mar 2018, 08:13
Bunuel wrote:
If a is a positive integer and if remainders of 4 and 6 are obtained when 89 and 125, respectively, are divided by a, then a =

(A) 7

(B) 9

(C) 15

(D) 17

(E) 19


Max possible Divisors are 85 & 119 , leaving remainder of 4 & 6

Find the HCF of 85 & 119 : 1 & 17

Answer must be (D) 17
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If a is a positive integer and if remainders of 4 and 6 are obtained  [#permalink]

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New post 15 Mar 2018, 09:57
Bunuel wrote:
If a is a positive integer and if remainders of 4 and 6 are obtained when 89 and 125, respectively, are divided by a, then a =

(A) 7

(B) 9

(C) 15

(D) 17

(E) 19


let quotients=q and p
a=85/q=119/p
17 is only common factor
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If a is a positive integer and if remainders of 4 and 6 are obtained   [#permalink] 15 Mar 2018, 09:57
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