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In dividing a number by 585, a student employed the method of short di
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13 Jul 2021, 04:41
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In dividing a number by 585, a student employed the method of short division. He divided the number successively by 5, 9 and 13 (factors 585) and got the remainders 4, 8, 12 respectively. If he had divided the number by 585, the remainder would have been
Re: In dividing a number by 585, a student employed the method of short di
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13 Jul 2021, 04:59
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Carcass wrote:
In dividing a number by 585, a student employed the method of short division. He divided the number successively by 5, 9 and 13 (factors 585) and got the remainders 4, 8, 12 respectively. If he had divided the number by 585, the remainder would have been
A. 24 B. 144 C. 288 D. 292 E. 584
----ASIDE---------------------------- There's a nice rule that say, "If N divided by D equals Q with remainder R, then N = DQ + R" For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2 Likewise, since 53 divided by 10 equals 5 with remainder 3, then we can write 53 = (10)(5) + 3 ----------------------------------------
Let N = the number in question
First notice that each remainder is 1 less than the divisor. When we divide N by 5, the remainder is 4 (one less than 5). When we divide N by 9, the remainder is 8 (one less than 9). When we divide N by 13, the remainder is 12 (one less than 13). This tells us that N+1 is divisible by 5, 9 and 13. Here's why...
When we divide by 5, the remainder is 4. Applying the above rule, we can write: N = 5k + 4 (for some integer k) So: N+1 = 5k + 4 + 1 Simplify to get: N+1 = 5k + 5 Factor to get: N+1 = 5(k + 1) In other words, N+1 is divisible by 5
We can apply the same steps to show that: N+1 is divisible by 9 N+1 is divisible by 13
If N+1 is divisible by 5, 9 and 13, we know that N+1 is divisible by 585 (the product of 5, 9 and 13) So, one possible value of N+1 is 585 If N + 1 = 585, then N = 584
If he had divided the number by 585, the remainder would have been? 584 divided by 585 equals zero with remainder 584 Answer: E
Re: In dividing a number by 585, a student employed the method of short di
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02 Oct 2022, 10:27
Given that a student divided a number by 585 using the method of short division. And he successively divided the number by 5, 9 and 13 (factors 585) and got the remainders 4, 8, 12 respectively. And we need to find the remainder when the he divided the number by 585
He divided 585 by 5, 9 and 13 successively to get the remainders of 4, 8 and 12. Let's try to find the number by going in the reverse direction
Divided a number by 13 to get get 12 remainder
Theory: Dividend = Divisor*Quotient + Remainder
Number -> Dividend 13 -> Divisor a -> Quotient (Assume) 12 -> Remainders => Number which was divided by 13 = 13*a + 12 = 13a + 12
Divided something by 9 to get 13a + 12 as quotient and 8 as remainder => Number which was divided by 9 = 9*(13a + 12) + 8 = 117a + 108 + 8 = 117a + 116
Divided something by 5 to get 117a + 116 as quotient and 4 as remainder => Actual Number = 5*(117a + 116) + 4 = 585a + 580 + 4 = 585a + 584
Remainder when Actual Number is divided by 585
585a + 584 when divided by 585 will give 584 as remainder
So, Answer will be E Hope it helps!
Watch the following video to learn the Basics of Remainders
gmatclubot
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