divisibility problem

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divisibility problem

by Xbond » Tue Apr 14, 2009 11:49 am
Hi there,

Someone can help me to understand and resolve this problem. I have some difficulties to understand the concept of divisibility and I completly lost with remainder.


When the positive integer x is divided by 4, is the remainder equal to 3?

(1) When x is divided by 2, the remainder is 1

(2) x is divisible by 3
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by DanaJ » Tue Apr 14, 2009 12:40 pm
Sometimes picking numerical examples can help.

1. Pick the two numbers to test this. The first one will be 5: when 5 is divided by 2, the remainder is 1 and the same goes here for dividing 5 by 4.
But if you pick 7, then the remainder of dividing this number by 2 will again be 1. However, dividing 7 by 4 yields a remainder of 3.
Since you can't eliminate any of the two cases, 1 is insufficient.

2. Again, pick 6, 9, 12, 15. In the first case, the remainder will be 2, in the second the remainder will be 1, in the third case the remainder will be 0, in the fourth case the remainder will be 3. As you can see there are plenty of cases to choose from, so 2 is insufficient as well.

Now, put both cases together and you get that you have a number that is divisible by 3 and that is odd (since the remainder of dividing it by 2 is 1).
Pick some numbers to solve it quickly: 15 and 9 - remainders 3 and 1 respectively. So this isn't enough either, with the answer being E.

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by vittalgmat » Tue Apr 14, 2009 2:32 pm
Excellent explanation Dana,
E for me as well, coz one cannot get a unique remainder when x is divided by 4 with the given restrictions/properties.

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by lunarpower » Tue Apr 21, 2009 12:06 am
remember the following: the PATTERNS in REMAINDER problems will emerge fairly early when you plug in numbers.
there are two kinds of problems that are especially suited to plugging in numbers and looking for patterns:
* digit problems
* remainder problems

pattern recognition is always a good backup strategy, but, on these two problem types, it's pure gold.

therefore, if you don't IMMEDIATELY realize a good theoretical way to do a remainder problem, you should get on the number plugging RIGHT AWAY.

in fact, i got an 800 on this thing, and i would immediately jump to plug-in / pattern recognition on problems such as this one. so don't sweat it if you don't find the theory method.

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statement (1):
just dredge up an exhaustive list of numbers that satisfy this criterion. if division by 2 leaves a remainder of 1, then the number in question must be 1 more than a multiple of 2.
therefore,
1, 3, 5, 7, ...
(you could also figure out this list by pure experimentation)

try these numbers in the prompt question:
x = 1 --> no (the remainder is 1)
x = 3 --> yes

if you don't like dividing numbers that are less than 4 by 4, then just try 5 and 7. 5 gives "no", and 7 gives "yes".

so, insufficient.

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statement (2)
3, 6, 9, 12, ...

3 --> yes
6 --> no
insufficient.

if you don't like dividing numbers that are less than 4 by 4, your first "yes" will occur at x = 15, which is not that far down the list.

so, insufficient.

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together:
take the numbers that appear on both lists. (if you didn't make very long lists - which you certainly didn't have to, since the first 2 numbers of each list are enough to resolve the issue - then you may well have to extend the lists to do this)

those numbers are
3, 9, 15, 21, ...

try them
3 --> yes
9 --> no
still insufficient.

if you don't like dividing numbers that are less than 4 by 4, then just try 9 and 15, which give "no" and "yes" respectively.
still insufficient.

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ans (e)
Ron has been teaching various standardized tests for 20 years.

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by vittalgmat » Tue Apr 21, 2009 12:20 am
Thanks Ron for the useful insight.
Long time no see and welcome back!!

rgds

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by Xbond » Wed Apr 22, 2009 11:44 am
Many thks guys

as usual Ron, you are the best

Xbond