If positive integer x is divided by 5, the result is p and

Bunuel, I tried using this method below as described in

https://gmatclub.com/forum/manhattan-remainder-problem-93752.html#p721341

I got stuck. Please help

X=5P+3 , x can be 8 13 18 23...58
X=11Q+3, x can be 14,25,....58

To form the equation n=kx+r
n=55K+58

Not sure how to proceed.

Since X-3 is divisible both by 5 and by 11,which are prime numbers, so P/11 or X-3/11 will always be with remainder 0

First of all you don't need to use that approach to solve the problem.

Next, you are making a mistake while deriving a general formula.

Positive integer x is divided by 5, the result is p and the remainder 3: \(x=5p+3\) --> \(x\) can be: 3, 8, 13, ... Notice that the least value of \(x\) for which it gives the remainder of 3 upon division by 5 is 3 itself: 3 divided by 5 yields remainder of 3.

Positive integer x is divided by 11, the the remainder 3: \(x=11q+3\) --> \(x\) can be: 3, 14, 25, ... Th same here the least value of \(x\) is 3: 3 divided by 11 yields remainder of 3.

General formula would be: \(x=55k+3\).

Check the problems below for which you can use this approach:
positive-integer-n-leaves-a-remainder-of-4-after-division-by-93752.html
if-n-is-a-positive-integer-greater-than-16-is-n-a-prime-129829.html
when-positive-integer-x-is-divided-by-5-the-remainder-is-128470.html
when-n-is-divided-by-5-the-remainder-is-2-when-n-is-divided-82624.html
when-positive-integer-n-is-divided-by-5-the-remainder-is-90442.html
when-the-positive-integer-a-is-divided-by-5-and-125591.html
what-is-the-value-of-length-n-100-meter-of-wire-126500.html
a-group-of-n-students-can-be-divided-into-equal-groups-of-126384.html
when-the-positive-integer-a-is-divided-by-5-and-7-the-104480.html
positive-integer-n-leaves-a-remainder-of-4-after-division-by-93752.html
when-positive-integer-n-is-divided-by-3-the-remainder-is-86155.html

Hope it helps.

If the remainder is same in both the cases,
x = 5p + 3
x = 11q + 3

then x = 55a + 3

Since 5p has 55 as a factor, p must be divisible by 11. So remainder is 0

x=5p+3;
x=11q+3

So, 5p+3=11q+3..
5p=11q
p=11(q/5)

P should be a multiple of 11... & p divided by 11 should give R=0

Using the same approach, we know that at p=11 the value of X=58, for both the expressions. Hence p is a multiple of 11 so the remainder is 0.
Though this is still a more time consuming approach that the ones stated above.

Please correct me if I am wrong.

Nityam

Hi I have a quick question on this problem. How are you getting to 55 in the combined equation? Why can't X be 3? If you divide 3 by both 5 and 11, the remainder is 3 so I'm not sure what I am missing. Thanks for any help you can give.

I have discussed the general case there.

Given that:
x = 5p + 3
x = 11q + 3

We can say that x = 55a + 3
i.e. when we divide x by 55 (the LCM of 5 and 11), the remainder will be 3 in that case too.


Sure, the number x can be 3 too. In that case p = 0, q = 0 and a = 0. When you divide p by 11, the remainder will be 0.

When we got 5p = 11k, since 5 and 11 is prime number -> k must be divisible by 5 and p must be divisible by 11 -> A is correct

I think I took the long road approaching this problem, not sure whether the correct one or not (after looking at the very simple and logical solution given by Bunuel), but this is the way I did it.

x= 5p+3
x=11q+3
p=11c+r

I substituted for p which lead to:

5(11c+r)+3=11q +r --> 5(11c+r)-11q=0 --> 55c+5r-11q=0 --> 11(5c-q)+5r=0. Given that r must be non negative, I concluded that in order for this equation to be 0, r must be equal to 0, as well. Please let me know if this conclusion is faulty.

I also had the same line of thinking. thanks for all the guidance.

I have done via number estimation.
I have taken list of all multiple values of 11 [ 22,33,44,55,66, ..] and added 3 to each. That results to [25, 36, 47, 58, 69, ...]
58 satisfy with first rule provided in question stem. 58 / 5 = 11 * 5 + 3. Therefore when we divide 55 with 11, remainder is 0.

So 'result' = 'Quotient'?

x= 5p+3
x=11q+3

first number that’s common to both is 3 and therefore p=0
now, 0/11 will give a remainder of 0.

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