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a, b are positive integers. The remainder of a to be divided by 8 is 4
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09 Nov 2014, 07:21
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a, b are positive integers. The remainder of a to be divided by 8 is 4 and the remainder of b to be divided by 6 is 2. Which is possible to be the remainder of a*b to be divided by 48 a) 2 b) 6 c) 8 d) 12 e) 20
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Re: a, b are positive integers. The remainder of a to be divided by 8 is 4
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10 Nov 2014, 01:20
h31337u wrote: a, b are positive integers. The remainder of a to be divided by 8 is 4 and the remainder of b to be divided by 6 is 2. Which is possible to be the remainder of a*b to be divided by 48
a) 2 b) 6 c) 8 d) 12 e) 20 The remainder of a to be divided by 8 is 4: a = 8q + 4. a could be 4, 12, 20, 24, ... The remainder of b to be divided by 6 is 2: b = 6p + 2. b could be 2, 8, 14, 20, ... If a =4 and b = 2, then ab = 8 and 8 divided by 48 yields the remainder of 8. Answer: C.
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a, b are positive integers. The remainder of a to be divided by 8 is 4
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13 Nov 2014, 01:29
\(\frac{a}{8}\) >>> Remainder = 4 \(\frac{b}{6}\) >>> Remainder = 2 \(\frac{a*b}{8*6}\) >> As numbers & divisors get multiplied, remainder will also get multipliedRemainder = 4*2 = 8 Answer = V
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Re: a, b are positive integers. The remainder of a to be divided by 8 is 4
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09 Nov 2014, 09:42
h31337u wrote: a, b are positive integers. The remainder of a to be divided by 8 is 4 and the remainder of b to be divided by 6 is 2. Which is possible to be the remainder of a*b to be divided by 48
a) 2 b) 6 c) 8 d) 12 e) 20 Hi Bunuel Could you please explain the solution? I got up to this a=8p+4 b=6q+2 a*b=48pq+16p+24q+8 when a*b is divided by 48,then 48pq is divisible by 48, 16p could be divisible by 48 (16*3) , 24 q could be divisible by 48 (24*2) so 8 divided by 48 > Possible reminder 8 Could you please confirm if this is a correct thinking? Thanks !



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Re: a, b are positive integers. The remainder of a to be divided by 8 is 4
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12 Nov 2014, 07:28
Bunuel wrote: h31337u wrote: a, b are positive integers. The remainder of a to be divided by 8 is 4 and the remainder of b to be divided by 6 is 2. Which is possible to be the remainder of a*b to be divided by 48
a) 2 b) 6 c) 8 d) 12 e) 20 The remainder of a to be divided by 8 is 4: a = 8q + 4. a could be 4, 12, 20, 24, ... The remainder of b to be divided by 6 is 2: b = 6p + 2. b could be 2, 8, 14, 20, ... If a =4 and b = 2, then ab = 8 and 8 divided by 48 yields the remainder of 8. Answer: C. I understand the case when a=4 b=2, but what about a=12 b=2, then ab=24. When this 24 is divided by 48, then the remainder is 24.



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Re: a, b are positive integers. The remainder of a to be divided by 8 is 4
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12 Nov 2014, 07:46
h31337u wrote: Bunuel wrote: h31337u wrote: a, b are positive integers. The remainder of a to be divided by 8 is 4 and the remainder of b to be divided by 6 is 2. Which is possible to be the remainder of a*b to be divided by 48
a) 2 b) 6 c) 8 d) 12 e) 20 The remainder of a to be divided by 8 is 4: a = 8q + 4. a could be 4, 12, 20, 24, ... The remainder of b to be divided by 6 is 2: b = 6p + 2. b could be 2, 8, 14, 20, ... If a =4 and b = 2, then ab = 8 and 8 divided by 48 yields the remainder of 8. Answer: C. I understand the case when a=4 b=2, but what about a=12 b=2, then ab=24. When this 24 is divided by 48, then the remainder is 24. Yes but 24 is not among the options...
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Re: a, b are positive integers. The remainder of a to be divided by 8 is 4
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12 Sep 2016, 19:19
Bunuel wrote: h31337u wrote: a, b are positive integers. The remainder of a to be divided by 8 is 4 and the remainder of b to be divided by 6 is 2. Which is possible to be the remainder of a*b to be divided by 48
a) 2 b) 6 c) 8 d) 12 e) 20 The remainder of a to be divided by 8 is 4: a = 8q + 4. a could be 4, 12, 20, 24, ... The remainder of b to be divided by 6 is 2: b = 6p + 2. b could be 2, 8, 14, 20, ... If a =4 and b = 2, then ab = 8 and 8 divided by 48 yields the remainder of 8. Answer: C. HI the series are 4,12,20....44 2,8,14....44 the first number happens to be 44 and the second one is 68, if we divide 68 by 48 we get the reminder 20 which is option E, i ant denying option A cant be the ans but option E also could be the ans



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Re: a, b are positive integers. The remainder of a to be divided by 8 is 4
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12 Sep 2016, 20:16
why will you divide 68 by 48? Sent from my iPhone using GMAT Club Forum mobile app



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Re: a, b are positive integers. The remainder of a to be divided by 8 is 4
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12 Sep 2016, 20:50
i misunderstood the stem, thanks for prompt response



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Re: a, b are positive integers. The remainder of a to be divided by 8 is 4
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12 Sep 2016, 22:38
h31337u wrote: a, b are positive integers. The remainder of a to be divided by 8 is 4 and the remainder of b to be divided by 6 is 2. Which is possible to be the remainder of a*b to be divided by 48
a) 2 b) 6 c) 8 d) 12 e) 20 Hi Two ways to do it... a=8x+4.. b=6y+2.. 1) convenient way.. Take x and y as 0, and you will get a*b as 4*2=8 C. 2) a*b=(8x+4)(6y+2)=4(2x+1)*2(3y+1)=8(2x+1)(6y+2)..... So a*b is a multiple of 8.. Also the divider is a multiple of 8.. So remainder has to be a multiple of 8.. While doing this , I just thought of this method. It should work for all cases, I believe. Only C is a multiple of 8.. C
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Re: a, b are positive integers. The remainder of a to be divided by 8 is 4
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06 Sep 2017, 15:38
PareshGmat wrote: \(\frac{a}{8}\) >>> Remainder = 4
\(\frac{b}{6}\) >>> Remainder = 2
\(\frac{a*b}{8*6}\) >> As numbers & divisors get multiplied, remainder will also get multiplied
Remainder = 4*2 = 8
Answer = V sorry, you are wrong. you need to erase this post so that other people will not get wrong. 5/3 has remainder 2. 8/7 has remainder 1 but 40 / 21 does not have remainder 2



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a, b are positive integers. The remainder of a to be divided by 8 is 4
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02 Nov 2018, 01:47
We can also apply the rule: remainder of a product is the product of remainders
thus in this case:
Rem(ab)> rem(a) x rem(b) > 4 x 2 > 8
where a>8p+4 and with p as 0, a is 4. and rem(4/48) is 4 same for b
the pitfall for me was that i got stuck with the algebra..which was not needed




a, b are positive integers. The remainder of a to be divided by 8 is 4
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02 Nov 2018, 01:47






