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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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minhphamvn wrote:
Hi Karishma,

Thanks for the post. I just want to ask, whether we can form a general formula of a number which satisfies 2 equations: divided by "a", remainder b (x=a.n + b) and divided by "c", remainder d (x=c.m + d). Then decide the remainder of x when divided by a*b.

For example, without plugging in number, how to solve the problem: find remainder when x divided by 32 if x = 8n + 3 = 6m + 4?

I find 1 solution but it can only apply for a-b=1 (for eg: x=a.n +b = (a+1)m +d)
Then we have:
(a+1)x = a(a+1)n +(a+1)b
ax = a(a+1)m + ad

Minus 2 equation => x = a(a+1) (n-m) + (a+1)b - ad => remainder when x divided by a(a+1) is (a+1)b-ad.


Did you check out the last link provided above? Here it is again: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/05 ... s-part-ii/

With a case such as this: x = 8n + 3 = 6m + 4, for the first number of this form, you need to use hit and trial.
Note that this example is incorrect since 8n+3 is an odd number and 6m+4 is an even number. But check the post to see how to apply the method to correct questions.
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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vikasbansal227 wrote:
Dear All,


Its difficult that on what basis we took "First common integer in two patterns" as a remainder for the formula?


Can someone please explain,

Thanks



You want a number, n, that satisfies two conditions:
"leaves a remainder of 4 after division by 6" and
"a remainder of 2 after division by 8"

So you find the first such number by writing down the numbers. You get that it is 10.
10 satisfies both conditions.

What will be the next number?
Any number that is a multiple of 6 more than 10 will continue to satisfy the first condition. e.g. 16, 22, 28, 34 etc
Any number that is a multiple of 8 more than 10 will continue to satisfy the second condition. e.g. 18, 26, 34 etc

So any number that is a multiple of both 6 and 8 more than 10 will satisfy both conditions.
LCM of 6 and 8 is 24. 24 is divisible by both 6 and 8.

So any number that is a multiple of 24 more than 10 will satisfy both conditions.
e.g. 34, 58 .. etc
These numbers can be written as 10 + 24a.

Also, I think you did not check out the 4 links I gave here: positive-integer-n-leaves-a-remainder-of-4-after-division-by-93752-40.html#p1496547
They are very useful in understanding divisibility fundamentals.
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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bchekuri wrote:
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30?

A. 3
B. 12
C. 18
D. 22
E. 28

Solution:

Notice that 4 is 2 less than 6 and 3 is also 2 less than 5; thus, a number that has both requirements (i.e., a remainder of 4 after division by 6 and a remainder of 3 after division by 5) is 2 less than the LCM of 6 and 5. Since the LCM of 6 and 5 is 30, that number will be 28. Although we are given that n is greater than 30, we can modify the number by adding 30 to it. That is, n = 28 + 30 = 58, and the remainder when 58 is divided by 30 is 28 (note that 58/6 = 9 R 4 and 58/5 = 11 R 3, satisfying the requirements).

Answer: E
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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bchekuri wrote:
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30?

A. 3
B. 12
C. 18
D. 22
E. 28


Since n leaves a remainder of 4 (which is even) after division by 6 (which is even), we know n is even. A is wrong.
Since n leaves a remainder of 3 after division by 5, we know the ones digit of n is either 3 or 8. B and D are wrong.
Let's just try something for C. We need a remainder of 18 after dividing by 30. How about 48. If we divide by 6, we don't get a remainder of 4. That didn't work.
Let's try something for E. We need a remainder of 18 after dividing by 30. How about 58. If we divide by 6, we get a remainder of 4. If we divide by 5, we get a remainder of 3. Giddy up.

Answer choice E.
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
Bunuel wrote:
To elaborate more.

Suppose we are told that:
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?

The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: \(n=6p+4\). Thus according to this particular statement \(n\) could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ...

The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: \(n=8q+2\). Thus according to this particular statement \(n\) could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...

The above two statements are both true, which means that the only valid values of \(n\) are the values which are common in both patterns. For example \(n\) can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer \(q\).

So we should derive general formula (based on both statements) that will give us only valid values of \(n\).

How can these two statement be expressed in one formula of a type \(n=kx+r\)? Where \(x\) is divisor and \(r\) is a remainder.

Divisor \(x\) would be the least common multiple of above two divisors 6 and 8, hence \(x=24\).

Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\).

Therefore general formula based on both statements is \(n=24k+10\). Thus according to this general formula valid values of \(n\) are: 10, 34, 58, ...

Now, \(n\) divided by 12 will give us the reminder of 10 (as 24k is divisible by 12).

Hope it helps.


if n divided by 24 gives a remainder of 10, then why must n divided by 12 also give a remainder of 10?
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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Schachfreizeit wrote:
Bunuel wrote:
To elaborate more.

Suppose we are told that:
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?

The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: \(n=6p+4\). Thus according to this particular statement \(n\) could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ...

The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: \(n=8q+2\). Thus according to this particular statement \(n\) could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...

The above two statements are both true, which means that the only valid values of \(n\) are the values which are common in both patterns. For example \(n\) can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer \(q\).

So we should derive general formula (based on both statements) that will give us only valid values of \(n\).

How can these two statement be expressed in one formula of a type \(n=kx+r\)? Where \(x\) is divisor and \(r\) is a remainder.

Divisor \(x\) would be the least common multiple of above two divisors 6 and 8, hence \(x=24\).

Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\).

Therefore general formula based on both statements is \(n=24k+10\). Thus according to this general formula valid values of \(n\) are: 10, 34, 58, ...

Now, \(n\) divided by 12 will give us the reminder of 10 (as 24k is divisible by 12).

Hope it helps.


if n divided by 24 gives a remainder of 10, then why must n divided by 12 also give a remainder of 10?


n divided by 24 gives a remainder of 10:

n = 24k + 10.

24k is divisible by 12, so the remainder upon division n by 12 will be from 10 divided by 12. 10 divided by 12 gives the reminder of 10.

I think you should go through some theory to brush up fundamentals.

6. Remainders



For other subjects:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
Bunuel wrote:
bchekuri wrote:
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30?
(A) 3
(B) 12
(C) 18
(D) 22
(E) 28

How to approach this Problem?

Positive integer n leaves a remainder of 4 after division by 6 --> \(n=6p+4\) --> 4, 10, 16, 22, 28, ...
Positive integer n leaves a remainder of 3 after division by 5 --> \(n=5q+3\) --> 3, 8, 13, 18, 23, 28, ...

\(n=30k+28\) - we have 30 as lcm of 5 and 6 is 30 and we have 28 as the first common integer in the above patterns is 28.

Hence remainder when positive integer n is divided by 30 is 28.

Answer: E.

P.S. n>30 is a redundant information.

­What if the question asked for the remainder after division by 60? Would you have to multiply 30k + 28 by 2 to get 60k + 56 and therefore the answer would be 56? 
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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kayze wrote:
Bunuel wrote:
bchekuri wrote:
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30?
(A) 3
(B) 12
(C) 18
(D) 22
(E) 28

How to approach this Problem?

Positive integer n leaves a remainder of 4 after division by 6 --> \(n=6p+4\) --> 4, 10, 16, 22, 28, ...
Positive integer n leaves a remainder of 3 after division by 5 --> \(n=5q+3\) --> 3, 8, 13, 18, 23, 28, ...

\(n=30k+28\) - we have 30 as lcm of 5 and 6 is 30 and we have 28 as the first common integer in the above patterns is 28.

Hence remainder when positive integer n is divided by 30 is 28.

Answer: E.

P.S. n>30 is a redundant information.

­What if the question asked for the remainder after division by 60? Would you have to multiply 30k + 28 by 2 to get 60k + 56 and therefore the answer would be 56? 

­
No. If we multiply \(n=30k+28\) by 2 we get \(2n=60k+56\), not \(n=60k+56\). Th remainder when n is divided by 60 cannot be determined because from \(n=30k+28\), n could be 28, 58, 88, 118, ... Hence, the remainder upon division by 60 could be 28, or 58.­
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