How many three-digit integers can be divided by 2 to produce

hi,
3 digit number divided by 2==>this will yield either a 2 digit number or a 3 digit number

case1:if it yields a 2 digit number
let three digit number =xyz
therefore 2 digit no.=yz===>as unit and tens digit are same
now
we can write :
100x+10y+z=2*(10y+z)
solving
100x = 10y+z
clearly this is not possible as LHS is a 3 digit number and RHS is a 2 digit number.

case 2: if it yields a 3 digit number lets say ayz
then we can write
100x+10y+z=2(100a+10y+z)
solving
100(x-2a)=10y+z
this is only possible if 10y+z =0...and x-2a=0
it means y=z=0..and x=2a
therefore possible values are
put a=1=>x=2=>number=200
a=2=>x=4=>number=400
a=3=>x=6=>number=600
a=4=>x=8=>number=800
a=5=>x=10=>number=1000(its a 4 digit hence not possible)
so only 4 numbers are possible.

Two conditions have to be met for the units digit of a 2 digit number to be the same when divided by 2:

1) The units digit of the original number must be 0.
2) The tens digit of the original number has to be even.

The same conditions can be extended to a 3 digit number if both the units digit and the tens digit of the new number have to be the same.

That is,

1) the units digit and tens digit of the original number have to be 0.
2) the hundreds digit of the original number has to be even

So these numbers are 200, 400, 600 and 800

Thus 4 numbers are possible.

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