can you solve this ?

Let d be the divisor and x be the number that us being divided. Let z and y be the answers we get by dividing by the divisor. Hence x/d=y remainder 23 and 2x/d= z remainder 11. These can be written as dy+23=x and dz+11=2x. Using simultaneous equations we find that 2dy-dz+35=0. This can be written as d(2y-z)=-35. Now 35 has the factors 5,7,35 and 1. d cannot be 5 or 7 as they will not leave remainders of 23 and 11. d cannot be 1 as 1 does not leave a remainder. So, d=35.

One could write the first sentence as $x \equiv 23 (\operatorname{mod} y)$ , while second sentence as $2x \equiv 11 (\operatorname{mod} y)$ .

By multiplying the first congruence by 2, one would have $2x \equiv 46 (\operatorname{mod} y)$ . After using both new and the second congruence, we could conclude that the $y$ is equal to $35$ as $2x \equiv 46 \equiv 11 (\operatorname{mod} y)$ .

If the remainder has 2 come as 23 then the divisor should b atleast 24 .so by dividing 24 with double of dividend the remainder s 22. like wise i we divide 2(dividend)with 25 we get 21 as remainder . So this sequence goes like this. 26-20 27-19 28-18 29-17 30-16 31-15 32-14 33-13 34-12 35-11 So the ans s 35