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# In dividing a number by 585, a student employed the method of short di

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Manager
Joined: 09 Jan 2016
Posts: 107
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In dividing a number by 585, a student employed the method of short di  [#permalink]

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01 Jul 2017, 07:08
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Difficulty:

75% (hard)

Question Stats:

56% (02:20) correct 44% (02:26) wrong based on 147 sessions

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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

A. 24
B. 144
C. 288
D. 292
E. 584
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Concentration: Technology, Strategy
Schools: ISB '19, IIMA , IIMB, XLRI
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Re: In dividing a number by 585, a student employed the method of short di  [#permalink]

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01 Jul 2017, 07:20
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1
Chemerical71 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

Let the number be $$x$$.

$$x$$ is successively divided by $$5$$, $$9$$ and $$13$$ (factors $$585$$) and got the remainders $$4$$, $$8$$, $$12$$ respectively.

Difference of all the divisors and remainders respectively is $$1$$.

$$Divisor - Remainder = 1$$

$$5-4 = 1$$

$$9-8 = 1$$

$$13-12 = 1$$

Therefore; $$585 - Remainder = 1$$

$$Remainder = 585 - 1 = 584$$

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Joined: 11 Sep 2015
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Re: In dividing a number by 585, a student employed the method of short di  [#permalink]

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25 Oct 2017, 10:10
Top Contributor
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Chemerical71 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

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Joined: 09 Nov 2015
Posts: 33
Re: In dividing a number by 585, a student employed the method of short di  [#permalink]

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20 Nov 2017, 00:46
1
The two solutions posted above are based on the fact that there is a consistent relationship between each of the divisors and the corresponding remainder (divisor is 1 more than remainder in each case). But what if this is not the case? To illustrate this, I have composed an identical problem but with different values for divisors and remainders:

"A certain number is successively divided by 2, 3 & 4 (the factors of 24) and the remainders are 1, 2 & 0 respectively. What will the remainder be if the number is divided outright by 24?"

As you can see, there is no discernible pattern between the divisors and remainders. My solution (which can be applied to all problems of this nature) is:

Let N be the number and Q1, Q2 & Q3 be the quotients for the first, second and third divisions respectively.
2Q1 + 1=N ........ (i)
3Q2 + 2=Q1 ...... (ii)
4Q3 + 0=Q2 .......(iii)
We also know that 24Q + R=N where Q is the quotient and R is the remainder when the number N is divided outright by 24. Now, it is a fact that when a number 'n' is successively divided by the factors of another number 'd', the last quotient is the same as the quotient when 'n' is divided outright by 'd'. This can easily be proved algebraically or verified by some simple examples (for instance, n=18, d=6, d1=2 and d2=3).
Therefore, 24Q3 + R=N
From (ii) and (iii), we can deduce that Q3=(Q1-2)/12
Thus, 24(Q1-2)/12 +R=N ......(iv)
So, from (iv) and (i), we have: 2(Q1-2) + R=2Q1 + 1; R=5. (N=53, Q1=26, Q2=8 and Q3=2)

This approach will also yield the correct result (remainder=584) when applied to the original problem. Hope I have been able to provide a clear explanation.
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Re: In dividing a number by 585, a student employed the method of short di  [#permalink]

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17 Jan 2019, 23:08
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Re: In dividing a number by 585, a student employed the method of short di   [#permalink] 17 Jan 2019, 23:08
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