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If a is a positive integer and if remainders of 4 and 6 are obtained
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22 Feb 2018, 19:44
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If a is a positive integer and if remainders of 4 and 6 are obtained
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22 Feb 2018, 21:13
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
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23 Feb 2018, 00:15
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
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23 Feb 2018, 01:07
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
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23 Feb 2018, 04:07
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 When 89 is divided by a then the remainder is 4 i.e. 894=85 must be divisible by a When 125 is divided by a then the remainder is 6 i.e. 1256=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
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15 Mar 2018, 05:47
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
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15 Mar 2018, 06:00



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Re: If a is a positive integer and if remainders of 4 and 6 are obtained
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15 Mar 2018, 07:13
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
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15 Mar 2018, 08:57
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 D




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