Problem Solving

An integer when divided by 6 has remainder 2 and when divided by 8 gives remainder 4. What is the remainder when the integer is divided by 48?

1. 0

2. between 1 and 6

3. between 7 and 12

4. between 13 and 19

5. <= 20

[spoiler]OA: E[/spoiler]

I'm not sure if my approach is right ....
When I use this method I get remainders >= 20....


20 is the least possible integer which satisfies the given divisibility conditions.

So the series of numbers which will satisfy the condition can be written as
20 + xLCM (6,8 ) where x = 0,1,2,..
20+24x

=>Series 20, 44, 68 , 92....

When we divide the above series by 48, we get remainders 20, 44,20, 44

cramya wrote:An integer when divided by 6 has remainder 2 and when divided by 8 gives remainder 4. What is the remainder when the integer is divided by 48?

1. 0

2. between 1 and 6

3. between 7 and 12

4. between 13 and 19

5. <= 20

[spoiler]OA: E[/spoiler]

Beautiful question Cramya.

Lets go back to our fundemantals, before we solve this problem:

LCM, Lowest Common Multiple:

In arithmetic and number theory, the least common multiple or lowest common multiple (lcm) or smallest common multiple of two integers a and b is the smallest positive integer that is a multiple of both a and b.

Let's say the remainder was zero when the integer was divided by 6 and 8.

Then, LCM(6,8 ) = 24 will be the lowest number that can satisfy this question and obviously diving by 48 would give us either 24 or 0 as a remainder.

But this is Alice's wonderland. In GMAT we will have 2 as a remainder when divided by 6, and 4 as a remainder when divided by 8. We need to find this NUMBER and find it its multiples or series and test it with 48.

Integer, I , can be defined as:

I = 6a+2 , if we use the remainder equation

I = 8b+4 if we use the remainder equation

If we insert the first equation into the second

6a+2-4 = 8b

6a-2 =8b ( which means that 6a-2 is divisible by 8 )

b=0,1,2,3,4,...etc

for b=0, a=not integer
b=1, a=not integer
b=2, a=3 BINGO

So I=6a+2 =20

you can actually find the same number by sorting the numbers like:

2,8,14,20,26,...etc ( 6a+2)
4,12,20,28,...etc (8b+4)


Now we have to find out when this pattern will repeat itself:

Our star point is 20, and our LCM is 24

just like y=ax+b

Integer = 24X+20

So integer set will lookl like as below:

20, 44, 68, 92, ...

if we divide these numbers by 48 the remainder set will be:

20, 44, 20, 44, 20, 44

Hence the answer is:

F) 20 or 44

Sris and Logitech,
Thats is the easiest way to go about this problem...

Good luck guys!

Regards,
Cramya

Great work Logitech. I did it another way that was not as sophisticated as your answer. But I guess, I am that type of guy :

We know that
a) 6/x = remainter of 2 &
b) 8/x = remainder of 4.

Now what would be the remainder if X were divided by 48? So we know that X has to be larger than 48.

I listed the multiples of 6 and 8 (and larger than 48 ) to find a number that would fit the criteria of a) and b):

Multiples of 6:
48 54 60 66 72...etc

Multiples of 8:
48 56 64 72 80 88 96

Using the equations above, I plugged in numbers to find an answer that would fit the criteria of a) and b). So starting with the multiples of 6's, I began with 54. a) 54+2 = 56. b) 56 - 4 = 52. Looking at the Multiples of 8 there isn't a 52. So I moved to the next number- a) 60 + 2 = 62. b) 62 - 4 =58. There isn't a 58 in the Multiples of 8 column. Now when you get to 66....a) 66 + 2 = 68. b) 68 - 4 = 64. Is there a 64 in the Multiples of 8 column? YES. So the number is 68! 68 divided by 48 leaves a remainder of 20.

Thus E.

jnellaz wrote:Great work Logitech. I did it another way that was not as sophisticated as your answer. But I guess, I am that type of guy :

Sweet.

I know that the last thing GMAT cares is how sophisticated our answers are

cramya wrote:Sris and Logitech,
Thats is the easiest way to go about this problem...

Good luck guys!

Regards,
Cramya

I would just add one thing.

The way the question is worded, you don't need to look for patterns or anything but just one remainder. I mean, it says you have an integer that fits the 6a + 2 and 8b + 4 constraints and the remainder has to fit one of the five answers. So as soon as you find one remainder, for example 20, you are done. It only fits one answer and that answer has to be the right answer.

Done.

As divisors and remainders are all even numbers, we can simplify the problem by halving all values:
I = 3x + 1 = 4y + 2 = 24z + 0.5R

Test possible values of x:
For even values of x, y is unachievable
If x = 1, I = 4, y is unachievable
If x = 3, I = 10, y = 2, z is unachievable
If x = 5, I = 16, y is unachievable
If x = 7, I = 22, y = 5, z is unachievable
If x = 9, I = 28, y is unachievable
If x = 11, I = 34, y = 8, z = 1, 0.5R = 10
Therefore R minimum = 20
So R >= 20

cramya wrote:An integer when divided by 6 has remainder 2 and when divided by 8 gives remainder 4. What is the remainder when the integer is divided by 48?

1. 0

2. between 1 and 6

3. between 7 and 12

4. between 13 and 19

5. <= 20

[spoiler]OA: E[/spoiler]

I guess the easiest is to find such a number first

divided by 6 has remainder 2 = 14, 20, 26, 32, 38, 44, 50, 56 etc

divided by 8 has remainder 4 = 12, 20, 28, 36, 44, etc

We have two such examples 20 and 44

Divide them by 48 and see the remainder they are >or= 20

Answer: Option E

Note that Answer 5 should be: >=20 (not <=20)

cramya wrote:An integer when divided by 6 has remainder 2 and when divided by 8 gives remainder 4. What is the remainder when the integer is divided by 48?

1. 0

2. between 1 and 6

3. between 7 and 12

4. between 13 and 19

5. <= 20

[spoiler]OA: E[/spoiler]

I like your method... very quick, but 20 and 44 are both less than 48

Note that as we need an integer bigger than 48, we could save more time by only looking at values that are >=48

Divided by 6 has remainder 2 = 50, 56, 62, 68, 74, etc...

Divided by 8 has remainder 4 = 52, 60, 68, e

Our lowest common value is 68

Divide this by 48 and see the lowest remainder is 20

Mathsbuddy wrote:I like your method... very quick, but 20 and 44 are both less than 48

Note that as we need an integer bigger than 48, we could save more time by only looking at values that are >=48

Divided by 6 has remainder 2 = 50, 56, 62, 68, 74, etc...

Divided by 8 has remainder 4 = 52, 60, 68, e

Our lowest common value is 68

Divide this by 48 and see the lowest remainder is 20

You are mistaken about it here mathbuddy.

Nowhere is it mentioned that the number itself is lesser than 48.

Therefore
1) The values of number 20 and 44 are well examples to be acceptable for the given question

2) 22 and 44 when divided by 48 leave remainders 20 and 44 respectively which is only the reflection that when such numbers (satisfying to the given condition in question) are divided by 48 then remainders can be either 20 or 44

3) You may cross-check with the example that you are obtaining to solve the question i.e. 68 when divided by 68 leaves remainder 20 and if you find the next such number and divide by 48 then remainder will be 44 (you certainly might want to check that aswell)

4) If it's a new observation for you then you might not want to leave this post without pressing "Thanks"

All the best!!!

GMATinsight wrote:

Mathsbuddy wrote:I like your method... very quick, but 20 and 44 are both less than 48

Note that as we need an integer bigger than 48, we could save more time by only looking at values that are >=48

Divided by 6 has remainder 2 = 50, 56, 62, 68, 74, etc...

Divided by 8 has remainder 4 = 52, 60, 68, e

Our lowest common value is 68

Divide this by 48 and see the lowest remainder is 20

You are mistaken about it here mathbuddy.

Nowhere is it mentioned that the number itself is lesser than 48.

Therefore
1) The values of number 20 and 44 are well examples to be acceptable for the given question

2) 22 and 44 when divided by 48 leave remainders 20 and 44 respectively which is only the reflection that when such numbers (satisfying to the given condition in question) are divided by 48 then remainders can be either 20 or 44

3) You may cross-check with the example that you are obtaining to solve the question i.e. 68 when divided by 68 leaves remainder 20 and if you find the next such number and divide by 48 then remainder will be 44 (you certainly might want to check that aswell)

4) If it's a new observation for you then you might not want to leave this post without pressing "Thanks"

All the best!!!

Thank you for pointing this out. I overlooked quotients of zero with a remainder.
Based on this, I might then add to your previous entry:

divided by 6 has remainder 2 = 2, 8, 14, 20, 26, 32, 38, 44, 50, 56 etc
divided by 8 has remainder 4 = 4, 12, 20, 28, 36, 44, etc

Although not used, I think it is only prudent to include all values, just in case.
Cheers! [/u]