acid_burn wrote:

How many three-digit integers can be divided by 2 to produce a new integer with the same tens digit and units digit as the original integer?

A. None

B. One

C. Two

D. Three

E. Four

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.

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