Ans 1)Let the number be 10x+y
Reversed number=10y+x
Original number-Reversed number=10x+y-(10y+x)
=10x+y-10y-x
=9x-9y
=9(x-y)
Hence,we observe that the result is a multiple of 9 or to be more specific 9 multiplied to the difference of the ten's and the unit's digit.
Example:92-29=63 which is 9 multiplied to 7[9-2]
Ans 2)Let the number be 10x+y
Sum of digits=x+y
Original number-Sum of digits=10x+y-(x+y)
=10x+y-x-y
=9x
Hence,we observe that the result is a multiple of 9 or to be more specific 9 multiplied to the ten's digit of the number.
Example:92-11=81 which is 9 multiplied to 9(the ten's digit of the number)
Ans 3)Let the number be 100x+10y+z
Reversed number=100z+10y+x
Original number-Reversed number=100x+10y+z-(100z+10y+x)
=100x+10y+z-100z-10y-x
=99x-99z
=99(x-z)
Hence,we observe that the result is a multiple of 9,11 and 99 or to be more specific 9 multiplied to the difference of the hundred's's and the unit's digit.
Example:321-123=198 which is 99 multiplied to 2[3-1]
Ans 4)Let the number be 10000x+1000y+100z+10a+b
Reversed number=10000b+1000a+100z+10y+x
Original number-Reversed number=10000x+1000y+100z+10a+b
- (10000b+1000a+100z+10y+x)
=10000x+1000y+100z+10a+b
-10000b-1000a-100z-10y-x
=9999x+990y-990a-9999b.
=99(101x+10y-10a-101b)
Hence,we observe that the result is a multiple of 9,11 and 99 .
Example:54321-12345=41976 which is a multiple of 9,11,99