Remainder Differs

Since X gives a remainder of 24 when divided by Y, it is logical that 2X gives a remainder of 48 when divided by Y, but the remainder is actually 11.

Therefore, Y is more than 24 but less than 48, so we can calculate Y by subtracting the remainders:

48 - 11 = 37

See that 37 is prime, so it only has 2 factors: 1 and 37 respectively.

As we have said just now that Y is more than 24 but less than 48, only one of the factors fit this requirement, which is 37.

We have the following equations for $X,Y \in \mathbb{N}$ :

$\frac{X}{Y} = q_{1} + \frac{24}{Y}$ ; (i)

$\frac{2X}{Y} = q_{2} + \frac{11}{Y}$ (ii)

where $q_{1}, q_{2}$ are the respective quotients. Solving (i) for $X$ in terms of $Y$ gives $X = q_{1}Y + 24$ , and substituting this expression into (ii) yields:

$2( q_{1}Y + 24 ) = q_{2}Y + 11 \Rightarrow 37 = (q_{2} - 2q_{1})Y$ (iii)

Since 37 is prime, the only allowable values for $Y$ are 1 and 37. If $Y = 1$ , then NO remainders exist in (i) and (ii)! Hence $\boxed{Y=37}$ is our desired result.

From the given information we can write

$X=nY+24 \\ 2X=mY+11$ where m and n are integers.

Subtracting the bottom equation from twice the top one eliminates X to give

$0=(2n-m)Y+37$

which is easily solved to give

$Y=\frac{37}{m-2n}$

If we assume (I think we are expected to!) that Y is a natural number then we see that, because 37 is a prime number, this can only happen if the denominator (m-2n) equals one. In this case we find the answer

$\boxed{Y=37}$

Let X = kY + 24, 2X = mY + 11, but 2X = 2kY + 48, 2kY + 48 = mY + 11, so 37 = Y(m - 2k), and Y divides 37. Then Y = 37, since it is a prime. Ed Gray

If mod(X, Y) = 24, then mod(2X, Y) is congruent by 24*2 mod(Y), this one is congruent to 11 too. It means that 48-Y=11. What give us the answer Y = 37

Using modular arithmetic, we get the following:

$X \equiv 24 \mod Y$ , $2X \equiv 11 \mod Y$

$\implies (2X - X) \equiv (11 - 24) \mod Y$

$\implies X \equiv -13 \mod Y$

$\implies 24 \equiv -13 \mod Y$

$\implies 37 \equiv 0 \mod Y$

$\implies Y \mid 37$

37 is prime, so either $Y = 1$ or $Y = 37$ . If $Y = 1$ , then every number would be congruent to 0 mod $Y$ , which is false, so $Y = 37$ .

Since X when divided by Y gives remainder 24, implies that the divisor Y is greater than 24 and when 2X is divided by Y the reminder is not 2*24(i.e. 48) which means that the divisor Y is less than 48.

But in the latter case it leave the remainder 11 instead of 48 which indicates that some part of 48 is being consumed leaving out 11, so 48-11=37 is the value of the divisor Y.

The following equations would be according to the formula :-

Dividend / Divisor = Quotient + Remainder

Hence, x/y = a + 24

Multiplying y on the opposite side, x = ay + 24y...... (2)

Putting it into the other equation which is,

2x / y = a + 11 Putting (2) in X,

2ay + 48y / y = a + 11

y (2a+48) / y = a + 11

Dividing y by both sides, 2a + 48 = a + 11,

a = 37

There's a flaw in the solution given below though which is, the sign before the a or 37 will be - (subtraction). I found the other solutions hard to understand therefore I wrote this for reference 😅 I shall take it down if it gets flagged

We know that y should be greater than 24&11 otherwise y will divide them also. While doing rough work I found a pattern. 49=25×1+24.
2×49=25×3+(23). 50=26×1+24 .2×50=26×3+(22) In this way second remainder was decreasing. Therefore If 25 gives 23 remainder For 11as remainder 23-x=11.x=12 To get it we have to move 12 steps Forward 25+12 =37

Let us say that:

1) X = Z1 + 24

and

2) 2X = Z2 + 11

Where Z1 and Z2 are multiples of Y.

(1) into (2) gives us:

2(Z1 + 24) = Z2 + 11

2Z1 + 48 = Z2 + 11

Z2 = 2Z1 + 48 - 11

Z2 = 2Z1 + 37

Since Z1 (and by extension, 2Z1), and Z2 are multiples of Y, 37 must be divisible by Y as well.

As 37 is prime, Y = 37.

X congruent to 24 (mod Y) So, 2X congruent to 48 (mod Y), which is given congruent to 11(mod Y) Therefore, 48 congruent to 11(mod Y) Or Y divides 48 -11=37 Since 37 is a prime, Y =37..

I SOLED IT WITH PYTHON

>> [(x,y) for x in range(100) for y in range (1,100) if x%y==24 if (2*x)%y==11] [(24, 37), (61, 37), (98, 37)]

If x/y=p+24 and 2x/y=2p+11 then we see that y=[p+24=2p+11] which we can solve. We double the first side. 2p+48=2p+11 then cancel the p's and then swap 11 to the other side and get 48-11 or 37.

n and m are integers:

X=nY+24

2X=mY+11

2X=2nY+48

solving gives:

(2n-m)Y=-37

m-2n=37/Y

m and n are integers so m-2n is an integer and 37 divided by Y is an integer, 37 is prime so Y has to be 1 or 37 but it can't be 1 because then the remainder can't be 0

(X-24) = Ym => 2X-48 = 2Ym

(2X-11) = Yn

Subtracting first from second, we get:

37 = Y (n-2m)

Since 37 has factors 1 and 37, and it cannot be 1 (else there wouldn't be any remainders):

Y = 37