Interchanging quotients and divisors

roses are red, violets are blue, putting a=6 and b=(6*n)-1 where n∈N ∩ n>1 would be the right clue.

If $a>b$ ,then $b=5$ ,but $a$ divided by $5$ gives a remainder $6$ ,which is impossible.

If $a<b$ ,then $a=6$ ,so $b$ divided by $6$ gives a remainder $5$ , we have $b=11,17,23...$

Hence there exists a family of solutions $(a,b)=(6,6n+5)$ ,where $n$ is a positive integer.

We have a solution if $a=b\times n+6$ and $b=a\times m+5$ , for integer quotients $n,m$ . Replacing one equation on the other one gets that $a(1-nm) = 5n+6 \\ b(1-nm) = 6m+5$ If $nm=0$ , then we have positive integer solutions given by $a = 5n+6\\ b= 6m+5.$ This is the case if either $n$ or $m$ are zero. Let's see what happens in each case:

• If $n=0$ then $a=6$ and $b=6m+5$ for any $m$ and therefore there are infinite solutions.
• If $m=0$ then $b=5$ which means that the remainder of $a/b$ is less than 5 and therefore cannot be 6. So this one case is not a solution.

We conclude that there are infinite solutions given by $a=6$ and $b=6m+5$ for $m=1,2,\dots$ .

$6/11$ has a remainder of $6$ and $11/6$ has a remainder of $5$ .

$a = 6$ and $b = 6n+5$ will always follow the given relationship for $n \in \mathbb{Z}^{+}$

Thus, there are infinitely many pairs $(a, b)$ .

p.s. Sorry, for the error before.

Write , $\textbf{a = bq + 6}$ and $\textbf{b = ap + 5}$ . Subtracting we get $\textbf{a-b = bq - ap + 1}$ which can be written as $\textbf{ax + by = 1}$ .where x,y,p,q are integers. Now from Bezout Identity there exist solution if $\textbf{ gcd(a,b) = 1}$ .Since there are infinite primes which can choose any 2.So there exist infinite solution.

A=NB+6 where n>=0.
B=MA+5 where m>=0 combining both A(1-M)=(N-1)B+11 Now m can only be 0 or 1 For M =0 N have infinite solutions , similarly for 1

A good way to try and solve problems like these is to start with the smallest possible solution and work your way up.

The smallest value that can have remainder of 5 is 5. Similarly, the smallest value that can have a remainder of 6 is 6.

If we try substituting $a$ with 6 and $b$ with 5, it doesn't work because 6 divided by 5 gives a remainder of 1.

No matter how many times we add 5 to $a$ , we will never get anything but a remainder of 1, since 5 goes into 6 once with a remainder of 1. Therefore, in order to have a remainder of 6 for $a$ divided by $b$ , $b$ must be greater than $a$ .

5 divided by 6 already has a remainder of 5, so we don't want to add anything extra. In this case, we should add 6, since 6 divides into 6 perfectly. It will make $b$ larger without adding anything to the remainder.

If we add 6 to $b$ , we get 11.

$(6,11)$ satisfies both conditions. Furthermore, we can keep adding 6 to $b$ to get the same result, thus meaning there are infinite solutions.

The solution is that you have to choose b = one of the multiples of 6 then substract that number by 1 ,then divide it by a = 6 and you'll have the remainder equals to 5, then the next step is to divide a = 6 by the number b you have choosen (the numbers are infinite but have to meet the first condition) then you'll get the result 0 ( because you can't divide a small number = 6, by a big number = the number you have chosen ) and the remainder equals to 6.