Hey again mate.

We have 362X215
1)3+6+2+X+2+1+5=4y insufficient, as x can be anything, so y can be anything.
2)if x differs from any of digits then it may be - 4,7,8,9, insufficient.

1+2)
19+x=4y,
plug in all 4 numbers from the stmt 2. The right X is X, when added to 19, will be divisible by 4.
Try to solve it, *=9.

NikolayZ wrote:Hey again mate.

We have 362X215
1)3+6+2+X+2+1+5=4y insufficient, as x can be anything, so y can be anything.
2)if x differs from any of digits then it may be - 4,7,8,9, insufficient.

1+2)
19+x=4y,
plug in all 4 numbers from the stmt 2. The right X is X, when added to 19, will be divisible by 4.
Try to solve it, *=9.

I agree with everything Nikolay said, the only thing I would add to make it slightly more clear is this.

When looking at statement 1 consider all of the possibilities 1-9, you will find that 19+x=4(y) can only work with the numbers 1,5,9.

So when combining statements 1&2, you know the answer needs to be found in both statements. 9 is the only number that works for both statements. (1,5,9 and 4,7,8,9)

NikolayZ wrote:Hey again mate.

We have 362X215
1)3+6+2+X+2+1+5=4y insufficient, as x can be anything, so y can be anything.
2)if x differs from any of digits then it may be - 4,7,8,9, insufficient.

1+2)
19+x=4y,
plug in all 4 numbers from the stmt 2. The right X is X, when added to 19, will be divisible by 4.
Try to solve it, *=9.

Just a point on statement (1) - x can't be anything, we can actually narrow it down to 3 possibilities.

19 + x must be a multiple of 4 and x is a digit from 0 to 9. So, 19 + x could be 20, 24 or 28, giving us possible x values of 1, 5 and 9.

Now, we didn't actually need to calculate that if we understood that since x varies between 0 and 9, there will be at least two values within that range that give us a multiple of 4 (if you have 4 consecutive integers, you have 1 multiple of 4; 8 consecutive integers guarantees 2 multiples of 4; we have 10 consecutive integers, so there will be either 2 or 3 multiples of 4 in the set).