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....
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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
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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
