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In dividing a number by 585

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In dividing a number by 585

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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

The OA is E.

I would like to someone Expert help me with this PS question. I don't understand why is E the correct answer.

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GMAT/MBA Expert

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Quote:
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

The OA is E.

I would like to someone Expert help me with this PS question. I don't understand why is E the correct answer.
Hi LUANDATO,
Let's take a look at your question.

The question states that when the number is divided by 5 the remainder is 4.
Let the number be x and q be the quotient, then

$$x=5p+4 --- (i)$$

The successive division by 9 result into the remainder 8.
$$p=9q+8 --- (ii)$$

Similarly, the successive division by 13 result into the remainder 12.
$$q=13r+12 --- (iii)$$

Now substitute value of q in eq(ii)
$$p=9q+8$$
$$p=9(13r+12)+8$$
$$p=117r+108+8$$
$$p=117r+116$$

Now replace this value of p in eq(i)
$$x=5p+4$$
$$x=5(117r+116)+4$$
$$x=585r+580+4$$
$$x=585r+584$$

Which represents that 585 is the divisor, r is the quotient and 584 is the remainder.
Therefore, when the number is divided by 585, the remainder is 584.

Hence, Option E is correct.

I am available, if you'd like any follow up.

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GMAT/MBA Expert

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LUANDATO 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
Here's a different approach....

----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

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