What is the remainder when the positive integer x is divided

[gphil]

Could please somebody explain how to approach the problem.
I saw some partial explanation of this problem in the previous posts, but, unfortunatelly, it wasn't too helpful for me. Thanks!!!

What is the remainder when the positive integer x is divided by 6.
1) When x is divided by 2, the remainder is 1; and when x is divided by 3, the remainder is 0.
2) when x is divided by 12, the remainder is 3.

The correct answer is D) Each statement alone is sufficient.

[nov1907]

(1) x divided by 2 leaves a remainder of 1. So x is odd and not divisible by 6 so remainder cannot be 0. x is a multiple of 3 and odd. Quick check shows 9 divided by 6 leaves a remainder of 3. 15 divided by 6 leaves a remainder of 3. Every odd multiple of 3 is 6 + the previous odd multiple. So any odd multiple of 3 will always leave the remainder 3. SO SUFFICIENT.

(2) x when divided by 12 leaves a remainder 3. so x-3 is divisible by 12. Means (x-3) is also divisible by 6. So x should leave the same remainder 3 when divided by 6. Mathematically:

x = 12a+3 = 6*(2a)+3= 6*b+3. So remainder when x is divided by 6 will always be 3. SO SUFFICIENT.

Answer is D.

[RonPurewal]

Yeah, that works.

This is also one of those problems where 'decoding' the statements by translating them into numbers works wonders: just sort through numbers by trial and error, and find the list of numbers that works for each statement.

(1) The numbers that work for this statement are 3, 9, 15, 21, etc. adding 6 each time.
(2) The numbers that work for this statement are 3, 15, 27, 39, etc. adding 12 each time.

If you make these lists, you'll find that either statement is sufficient on its own, so, D.

Incidentally, you should at least be able to translate '1 is the remainder upon dividing X by 2' into 'X is odd', IMMEDIATELY.

[Guest]

I had a question on this problem. I have no problem with calculating remainders, but what about when X = 3 in the first statement?

I was under the impression that a remainder only exists when the numerator is greater than the denominator? (i.e. 3/6 does not give us a remainder). Because of that impression I chose B, and outside of "3", I had no issue with how to calculate the first statement.

[RonPurewal]

I had a question on this problem. I have no problem with calculating remainders, but what about when X = 3 in the first statement?

I was under the impression that a remainder only exists when the numerator is greater than the denominator? (i.e. 3/6 does not give us a remainder). Because of that impression I chose B, and outside of "3", I had no issue with how to calculate the first statement.

see, that's the part they never told you about in grade school. if you divide 3 by 6, you darned well do get a remainder; it's 3.

think about it like this:
Remainder upon dividing N by 6
if you take N beer cans and put them into six-packs, then the remainder is the number of beer cans left over after you've made as many six-packs as you can.
...so, if you have 3 cans, you make 0 six-packs (which is 'as many as you can'), leaving ... 3 cans as your remainder.

general rule:
if N < M (both positive integers) and you divide N by M, then the remainder is N.

--

another way to justify the result is to realize that mathematical patterns don't spontaneously change. to wit: take a list of the remainders gotten by dividing certain #s by six. let's start with 20 and count downward.
20 --> remainder = 2
19 --> remainder = 1
18 --> 0
17 --> 5
16 --> 4
15 --> 3
14 --> 2
13 --> 1
12 --> 0
11 --> 5
10 --> 4
9 --> 3
8 --> 2
7 --> 1
6 --> 0
the pattern is obvious, and it must continue (as patterns are wont to do):
5 --> 5
4 --> 4
etc.

[taposh_dr]

Hello,

I am unable to understand this general rule:
if N < M (both positive integers) and you divide N by M, then the remainder is N.

If I try to Plug in numbers : let N=8 and M=10 i.e. N<M => 8 <10

Now when you divide N by M i.e. 8/10 i donot see how the remainder is N i.e. 8.

Can someone please explain.

Regards

T

[akhp77]

Try this way

Statement 1:
N = 2a + 1
N = 3b = (3b - 1) + 1

(3b - 1) is divisible by 2 if b = 1 and min of N is 3

So if N is divided by LCM of 2 & 3 the remainder would be 3

N = 6c + 3
Remainder = 3
Sufficient

Statement 2:
N = 12d + 2 = 6 *(2d) + 3
Remainder = 3
Sufficient

Ans: D

[StaceyKoprince]

if N < M (both positive integers) and you divide N by M, then the remainder is N.

If I try to Plug in numbers : let N=8 and M=10 i.e. N<M => 8 <10

Now when you divide N by M i.e. 8/10 i donot see how the remainder is N i.e. 8.

The remainder is whatever is left over AFTER you've divided evenly as many times as you can.

You can't divide 8 by 10 evenly at all (that is, you can't get any integer values when you do this division) because 8 is smaller than 10. Therefore, the amount "left over" is the amount you started with: 8.

Contrast this with dividing, say 18 by 10. First, 10 goes into 18 once, so we get the integer "1" and then we have 8 left over. The 10 can't go into that 8 any more, so the 8 is the remainder.

You may want to write out both of the above examples in terms of long division in order to see what I'm saying.

[wholf09]

What is the correct answer to this problem? When i have the pratice test show me what the answer is it says that it is B not D.( statement 2 alone but not statement 1 alone) Can someone please explain this problem.

[jnelson0612]

wholf,
The correct answer to the problem in the original post is D. If you have a different version of this problem please post and we will advise.

Please read through the explanation in this thread. If you then still have specific questions please let us know.

Thank you,

[vicksikand]

Hello,

I am unable to understand this general rule:
if N < M (both positive integers) and you divide N by M, then the remainder is N.

If I try to Plug in numbers : let N=8 and M=10 i.e. N<M => 8 <10

Now when you divide N by M i.e. 8/10 i donot see how the remainder is N i.e. 8.

Can someone please explain.

Regards

T

By defn
Dividend = Divisor x Quotient + Remainder
Ex: 4/6 ---->Dividend = 4, Divisor = 6, and quotient = 0(Because Dividend < Divisor)
thus Dividend = Remainder (4 for the example I am using)

[jnelson0612]

Agreed vicksickand.

[jeevan13]

akhp77
 Post subject: Re: What is the remainder when the positive integer x is divided
Try this way

Statement 1:
N = 2a + 1
N = 3b = (3b - 1) + 1

(3b - 1) is divisible by 2 if b = 1 and min of N is 3

So if N is divided by LCM of 2 & 3 the remainder would be 3

N = 6c + 3
Remainder = 3
Sufficient

Statement 2:
N = 12d + 2 = 6 *(2d) + 3
Remainder = 3
Sufficient

Ans: D

Hey akhp77,

Can u pls explain hw u arrived at
N = 6c + 3 frm

(3b - 1) is divisible by 2 if b = 1 and min of N is 3

So if N is divided by LCM of 2 & 3 the remainder would be 3

I mean i understand why the formula is correct bt I jst wanted to know so that I can get a general idea for other questions too.

Thnx alot,
Jeevan

[RonPurewal]

don't forget, amid all this complicated theoretical discussion, that this is a problem on which number-picking is very easy -- conservatively, i'd say 100,000 times easier than using theory. refer to this post (which is earlier in this same thread):
post5248.html#p5248

[jeevan13]

Hey Ron,

I understand it is easy to do these kind of questions by just substituting a number. But it would be really nice if you could explain to me the method for:

"
N = 2a + 1
N = 3b = (3b - 1) + 1

(3b - 1) is divisible by 2 if b = 1 and min of N is 3

So if N is divided by LCM of 2 & 3 the remainder would be 3

N = 6c + 3
Remainder = 3
"

I get the part whr
N = 2a + 1
N = 3b = (3b - 1) + 1

bt wht I dont get is

How u get

N = 6c + 3

from

(3b - 1) is divisible by 2 if b = 1 and min of N is 3

So if N is divided by LCM of 2 & 3 the remainder would be 3

Basically I want to know:
1. What is this third N, i.e N = 6c + 3?
2. How do you arrive at it?

Thnx a lot,
Jeevan