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Re: Quick solutions to this

by Ian Stewart » Thu Jul 31, 2008 2:37 am
[email protected] wrote:Hi

Does anyone know a quick way of solving this problem

If j and k are positive integers, j – 2 is divisible by 4 and k – 5 is divisible by 4, all of the following could be the value of j – k EXCEPT:

A) 43
B) 33
C)21
D)13
E) 5

OA is A.

Thanks in advance
Remember that if you add or subtract two multiples of 4, you'll always get a multiple of 4 (and there's nothing special about '4' here; that's true when you add or subtract any two multiples of the same number). We know that j-2 and k-5 are multiples of 4, so:

(j-2) - (k-5) = j - k + 3 must be a multiple of 4.

Since 43+3 = 46 is not a multiple of 4, A is correct.

You might also notice that if j-k+3 is a multiple of 4, so is j-k+3-4 = j-k-1, so the remainder will be 1 when j - k is divided by 4.
Last edited by Ian Stewart on Thu Jul 31, 2008 4:06 am, edited 1 time in total.
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by Mixaela » Thu Jul 31, 2008 2:53 am
Karen,

The base assumption that we work with is that if x divisible by 4 and y is divisible by 4 then x-y is divisible by 4.

In our case we have:

x = j-2 is divisible by 4
y = k - 5 is divisible by 4

x - y = j -2 -(k-5) = j - 2 - k + 5 = (j-k) + 3 is divisible by 4.

Now we just need to check each case:

A) j - k = 43 => 43 + 3 = 46 is NOT divisible by 4
b) j - k = 33 => 33 + 3 = 36 is divisible by 4
c) j - k = 21 => 21 + 3 = 24 is divisible by 4
d) j - k = 13 => 13 + 3 = 16 is divisible by 4
e) j - k = 5 => 5 + 3 = 8 is divisible by 4.

The obvious answer is A

Regards,
Mihaela
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by VP_RedSoxFan » Thu Jul 31, 2008 7:55 am
I really like the approach by Mixaela. You'll just have to prove to yourself that subtracting two multiples of 4 leaves a difference that is still a multiple of 4.

You can either know this to be true or you can prove it to yourself with a quick example on test day, i.e. 24-16=8, etc.
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