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Number property/remainders?

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Number property/remainders?

by g3lo » Wed Jan 27, 2016 7:11 pm
If k is an integer such that 56 < k < 66, what is the value of k?
(1) If k were divided by 2, the remainder would be 1.
(2) If k + 1 were divided by 3, the remainder would be 0.

In this question,

How would someone spot the remainders quickly;

For instance; 57,58,59,60,61,62,63,64,65 (how could one spot which numbers have remainder of 1) Or, which numbers have a remainder of 0 when divided by 3 or any other number?

Is there a number property rule I am missing?

Really could use some help,

Thank you,
g3lo
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by manik11 » Wed Jan 27, 2016 7:46 pm
Hi g3lo,

From the prompt you know that K could be any of these values 57,58,59,60,61,62,63,64,65.

Lets check the statements to see if we can find a definite value of K.

Statement 1 : If k were divided by 2, the remainder would be 1

This is also a number properties rule. When you divide an odd no. by 2 you always get a remainder of 1. (For eg. 3/2,7/2,9/2...)

You can always test some numbers from the range of K available.

57/2 would give you quotient of 28 with remainder 1. Same will be true for all the ODD numbers in the range.

So, K could be 57,59,61,63,65 --> No definite value of K --> INSUFFICIENT. Eliminate A & D

Statement 2: If k + 1 were divided by 3, the remainder would be 0

This just means that K+1 is perfectly divisible by 3. The best approach here I guess is to just test out all the numbers in the range.
You just have to add 1 to the number and that sum should be divisible by 3

You can narrow the values of K down to
59 (59+1 = 60) 60 is perfectly divisible by 3
62 (62+1 = 63) 63 is perfectly divisible by 3
65 (65+1 = 66) 66 is perfectly divisible by 3 Be careful here while testing out the numbers. Under pressure when the clock is ticking one might wrongly eliminate the last no. (66) thinking its not in the range. But 66 is the value of K+1 for which K is 65, which is in the range.

Now K could be 59,62,65 --> No definite value of K --> INSUFFICIENT. Eliminate B

Combined from both the statements you still have K = 59 or 65. INSUFFICIENT

Answer : E

Hope this helps!

Let me know the OA. :D
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by g3lo » Wed Jan 27, 2016 7:57 pm
Hi thank you manik11,

That is correct. I understand the approach/process; however, I am having difficulty in finding out to spot the remainder with ease.

The number property (anything divided by 2 that is ODD will always have a remainder of 1 - is of great help). Is there something like this for anything divided by 3 for remainder of 0?

Thank you,
g3lo
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by manik11 » Wed Jan 27, 2016 8:02 pm
g3lo wrote:Hi thank you manik11,

That is correct. I understand the approach/process; however, I am having difficulty in finding out to spot the remainder with ease.

The number property (anything divided by 2 that is ODD will always have a remainder of 1 - is of great help). Is there something like this for anything divided by 3 for remainder of 0?

Thank you,
g3lo
Unfortunately,I don't think there is a rule for remainder when a no. is divided by 3. That said, I personally think that testing out couple of test cases is always the best approach and will serve you better in the long run than remembering a bunch of number property rules.
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by MartyMurray » Wed Jan 27, 2016 11:05 pm
Hi g3lo.

One way to work with multiples and remainders is to use a concept like the following.

Every number can be expressed as x = QN + r.

So, if something is a multiple of 3, then x = 3N + r, and r is 0.

If x is not an exact multiple of 3, then x = 3N + r, and r could be 1 or 2.

Once you start thinking about numbers that way, then you can more easily work with remainders.

For instance, let's look at the question you posted.

If k is an integer such that 56 < k < 66, what is the value of k?

(1) If k were divided by 2, the remainder would be 1.

(2) If k + 1 were divided by 3, the remainder would be 0.

Statement 1 tells us that k = 2N + 1. In other words, k is odd.

There are many odd integers between 56 and 66. So Statement 1 is insufficient.

Statement 2 tells sue that K + 1 = 3N + 0, or k = 3N - 1. You could convert that into k = 3N + 2. Either way will work.

So then you just need to find all the multiples of 3 that are close to 56 and 66 and add 2 to each of them.

If you are not sure which are multiples of three, you could use the rule that says that the sum of the digits of a multiple of 3 will be divisible by 3.

Let's check some numbers.

55: 5 + 5 = 10 Not divisible by 3. So 55 is not divisible by 3 and 55 + 2 = 57 does not work.

56: 5 + 6 = 11 Not divisible by 3. So 56 is not divisible by 3 and 56 + 2 = 58 does not work.

57: 5 + 7 = 12 Divisible by 3. So 57 is divisible by 3 and 57 + 2 = 59 works.

Since multiples of 3 are always 3N + 0, we can go from 57 to the other multiples of 3 by adding 3's.

60 is one. So 60 + 2 = 62 works.

63 must be one also. So 63 + 2 = 65 works.

We get three different possible values of k, 59, 62 and 65. So Statement 2 is insufficient.

Combining the statements, we see that 59 and 65 work for both. So combined the statements are insufficient.

So the correct answer is E.

Do you really need x = QN + r to answer this question? Maybe not, but in any case, seeing numbers that way will help you to see remainders more clearly, and there are GMAT questions the answers to which are most easily found by plugging the appropriate numbers into that remainder formula.
Last edited by MartyMurray on Thu Jan 28, 2016 5:18 am, edited 1 time in total.
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by GMATGuruNY » Thu Jan 28, 2016 5:11 am
g3lo wrote:If k is an integer such that 56 < k < 66, what is the value of k?

(1) If k were divided by 2, the remainder would be 1.
(2) If k + 1 were divided by 3, the remainder would be 0.
Since 56 < k < 66, evaluate the statements by testing the following options for k:
57, 58, 59, 60, 61, 62, 63, 64, 65.

Statement 1:
In other words, k is 1 more than a multiple of 2:
k = 2a + 1, where a is a nonnegative integer.

Thus:
2a + 1 = 57, 58, 59, 60, 61, 62, 63, 64, 65
2a = 56, 57, 58, 59, 60, 61, 62, 63, 64
a = 28, 57/2, 29, 59/2, 30, 61/2, 31, 63/2, 32.
Since a must be an integer, we get:
a = 28, 29, 30, 31, 32.

If a=28, then k = 2*28 + 1 = 57.
If a=29, then k = 2*29 + 1 = 59.
If a=30, then k = 2*30 + 1 = 61.
The resulting values for k -- 57, 59, 61 -- imply that k can be ANY ODD INTEGER between 56 and 66.
Since k can be different values, INSUFFICIENT.

Statement 2:
In other words, k+1 is a multiple of 3:
k + 1 = 3b, where b is a nonnegative integer.
Solving for k, we get:
k = 3b - 1.

Thus:
3b - 1 = 57, 58, 59, 60, 61, 62, 63, 64, 65
3b = 58, 59, 60, 61, 62, 63, 64, 65, 66
b = 58/3, 59/3, 20, 61/3, 62/3, 21, 64/3, 65/3, 22.
Since b must be an integer, we get:
b = 20, 21, 22.

If b=20, then k = 3*20 - 1 = 59.
If b=21, then k = 3*21 - 1 = 62.
If b=22, then k = 3*22 - 1 = 65.
Since k can be different values, INSUFFICIENT.

Statements combined:
Statement 1: k = any odd integer between 56 and 66.
Statement 2: k = 59, 62, 65.
Implication:
When the statements are combined, it's possible that k=59 or that k=65.
Since k can be different values, INSUFFICIENT.

The correct answer is E.
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by g3lo » Thu Jan 28, 2016 10:11 am
Hi GMATGuruNY,

Thank you kindly for the detailed answer.

The logic makes sense, I guess I am trying to find shortcuts into the logic as I doubt I will have time to write everything out as said.

Best,
g3lo
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