sud21 wrote:the unit' digit of a number is twice the hundredth' digit. And the number is divisible by 3. Which of the following must be true?
I. the tenth digit is nonzero
II. the tenth digit is odd
III. the tenth digit is a multiple of 3

I. the tenth digit is nonzero
consider the number 603
this statement cannot be always correct


II. the tenth digit is odd
603, 864,..

not always true

III. the tenth digit is a multiple of 3
let us say units digit is x
hundredth digit is 2x
tenth digit is a
the number is multiple of 3
hence, 2x+a+x = 3n
a = 3(n-x)

tenth digit is a multiple of 3

True

Option III is correct

sud21 wrote:the unit' digit of a number is twice the hundredth' digit. And the number is divisible by 3. Which of the following must be true?
I. the tenth digit is nonzero
II. the tenth digit is odd
III. the tenth digit is a multiple of 3

The structure of the number will b as follows -

H T U

Now it's given -

The Unit' digit of a number is twice the hundredth' digit..


So , the digits will be -

H T 2H

Summation of the digits will be -

3H + T

Now in order to make the number divisible by 3 T must be a multiple of 3 , or we can write T as a multiple of 3 ie , 3k


So now the summation of the number stands as follows -

3H + 3k



Now let's consider the problem statements given -

I. the tenth digit is nonzero

Now the units digit can't b 0 , coz then the Hundredth's digit must have to b zero and it's not possible ...

II. the tenth digit is odd

Now this can not be possible , coz anything multiplied by 2 makes it an even number...

III. the tenth digit is a multiple of 3

Now this has just been proved above , so it's true...

Let the no be "xy(2x)"
The no is then 100x+y+2x = 102x+y which is divisible by 3. Thus, 3x+y is also a multiple of 3, which implies y must be a multiple of 3.