Can You Find A Four-digit Number Whose Digits Are Reversed When Multiplied By Four?

If the number is 4-digit, the least significant digit will not be 0, because that would require the most significant digit to be 0 as well. (4*0=0)
Suitable least-significant/most-significant pairs are
2/8, the number is 8xy2
4/6, the number is 6xy4
6/4, the number is 4xy6
8/2, the number is 2xy8
The most-significant digit will cause a carry into the 5th digit when any of the first 3 of these is multiplied by 4. So, the number must be of the form 2xy8.

We can write equations for x and y assuming n is the number carried over into the next place.
4y+3 = x+10n
4x+n = y (assuming no carry)

4(4x+n) +3 = x+10n    (substitute the second equation into the first)
16x + 4n +3 = x + 10n    (multiply it out)
15x + 3 = 6n    (subtract 4n+x from both sides)
5x + 1 = 2n    (divide both sides by 3)

x must be odd, so possible values are 1,3,5,7,9
x=1, n=3
x=3, n=8
Since n is the carry from the second column to the third, we're pretty sure it won't be 8 or higher. Thus, x=1 and y=(4*1+3)=7

The number you seek is 2178. When multiplied by 4, it is 8712.

Find two numbers whose difference is 79

X=4Y
=>1000a+100b+10c+d=4000d+400c+40b+4a
=>3999d+390c-60b-996a=0
=>1333d+130c-20b-332a=0......(1)
d!=0;d=3 Ydoes not remain 4 digit number . Again d is even 4a is even.
Therefore,d=2 ; a=4d or 4d+1 i.e 8 or 9 but as least significant digit of 4a is 2 ;a=8.
Putting a,d in (1) we get 1333-1328=10b-65c=>13c=2b-1=>13 divides (2b-1) => b=7 =>c=1
So,Y=2178;X=4Y=8712.