number properties

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sud21: Master | Next Rank: 500 Posts; Posts: 344; Joined: Sat Nov 12, 2011 3:21 amFollowed by: 2 members; Thanked: 1 times

by sud21 » Sat Jan 28, 2012 11:14 pm

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

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Source: Beat The GMAT — Problem Solving | Sat Jan 28, 2012 11:14 pm

rijul007: Legendary Member; Posts: 588; Joined: Sun Oct 16, 2011 9:42 amFollowed by: 9 members; Location: New Delhi, India; Thanked: 130 times; GMAT Score:720

by rijul007 » Sat Jan 28, 2012 11:26 pm

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

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Abhishek009: Master | Next Rank: 500 Posts; Posts: 359; Joined: Wed Mar 11, 2009 4:37 amFollowed by: 2 members; Location: Kolkata, India; Thanked: 50 times

by Abhishek009 » Sun Jan 29, 2012 10:03 am

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...

Abhishek

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ronnie1985: Legendary Member; Posts: 626; Joined: Fri Dec 23, 2011 2:50 amFollowed by: 10 members; Location: Ahmedabad; Thanked: 31 times

by ronnie1985 » Sun Jan 29, 2012 10:08 am

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.

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