Divisible by 3

This topic has expert replies
User avatar
Master | Next Rank: 500 Posts
Posts: 351
Joined: Mon Jul 04, 2011 10:25 pm
Thanked: 57 times
Followed by:4 members

Divisible by 3

by akhilsuhag » Tue Nov 18, 2014 1:10 pm
Is the tens digit of a three-digit positive integer p divisible by 3?

(1) p-7 is a multiple of 3

(2) p-13 is a multiple of 3
Please press "thanks" if you think my post has helped you.. Cheers!!

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Tue Nov 18, 2014 1:28 pm
akhilsuhag wrote:Is the tens digit of a three-digit positive integer p divisible by 3?

(1) p-7 is a multiple of 3

(2) p-13 is a multiple of 3
Note:
0 is considered divisible by EVERY POSITIVE INTEGER.

Statement 1: p-7 is a multiple of 3
Make a list of options for p-7:
p-7 = 93, 96, 99, 102, 105...

Adding 7 to each value in the list, we get the following options for p:
p = 100, 103, 106, 109, 112...

If p=100, then its tens digit is 0, which is divisible by 3.
If p=112, then its tens digit is 1, which is NOT divisible by 3.
INSUFFICIENT.

Statement 2: p-13 is a multiple of 3
Make a list of options for p-13:
p-13 = 87, 90, 93, 96, 99...
Adding 13 to each value in the list, we get the following options for p:
p = 100, 103, 106, 109, 112...

Both statements yield the same list of options for p.
Implication:
Even when the two statements are combined, it's possible that p=100 (in which case its tens digit is divisible by 3) or that p=112 (in which case its tens digit is NOT divisible by 3).
Thus, the the two statements combined are INSUFFICIENT.

The correct answer is E.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Tue Nov 18, 2014 1:29 pm
akhilsuhag wrote:Is the tens digit of a three-digit positive integer p divisible by 3?

(1) p-7 is a multiple of 3
(2) p-13 is a multiple of 3
Target question: Is the tens digit of a three-digit positive integer p divisible by 3?

IMPORTANT: When I scan the two statements, I see that they both tell me the SAME THING.
There's a rule that says: If N and K are both divisible by d, then N+K and N-K are also divisible by d.

OBSERVE: p-13 is 6 less than p-7. In other words, (p-7) - 6 = p-13

Statement 1 tells us that p-7 is divisible by 3, and we know that 6 is divisible by 3. So, by the above rule, p-13 must be divisible by 3.

When two statements provide the SAME information, we can conclude that the correct answer will be either D or E.

Statement 1: p-7 is a multiple of 3
There are several values of p that satisfy this condition. Here are two:
Case a: p = 139, in which case p-7 = 132, and 132 is divisible by 3. In this case, the tens digit of p (3) IS divisible by 3
Case b: p = 118, in which case p-7 = 111, and 111 is divisible by 3. In this case, the tens digit of p (1) is NOT divisible by 3
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: p-13 is a multiple of 3
We already learned that statements 1 and 2 provide the SAME information. So, if statement 1 is NOT SUFFICIENT, we know that statement 2 is NOT SUFFICIENT

If you're not convinced, check out these two conflicting cases:
Case a: p = 139, in which case p-13 = 126, and 126 is divisible by 3. In this case, the tens digit of p (3) IS divisible by 3
Case b: p = 118, in which case p-13 = 105, and 105 is divisible by 3. In this case, the tens digit of p (1) is NOT divisible by 3
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Since both statements are not sufficient, and since both statements provide the SAME information, the combined statements are NOT SUFFICIENT

Answer = E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image

Master | Next Rank: 500 Posts
Posts: 447
Joined: Fri Nov 08, 2013 7:25 am
Thanked: 25 times
Followed by:1 members

by Mathsbuddy » Wed Nov 19, 2014 7:47 am
Consider the value p = ABC where A,B,C are the three digits.
Working backwards, firstly, for B to be a multiple of 3, it can only be 3, 6 or 9.

So, these are possibilities:
p = A3C, A6C, A9C

If we subtract 7 (= -10 + 3)we get:
p = A2C+3, A5C+3, A8C+3
For p to be a multiple of 3, then the sum of all digits must be a multiple of 3
As A and C are both undefined, then Statement A is NOT SUFFICIENT

Instead if we subtract 13 (= -10 - 3)we get:
p = A2C-3, A5C-3, A8C-3
For p to be a multiple of 3, then the sum of all digits must be a multiple of 3
As A and C are both undefined, then Statement A is NOT SUFFICIENT

Combined statements will still leave A and C undefined. NOT SUFFICIENT>

Answer = (E)

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Wed Nov 19, 2014 8:52 am
Mathsbuddy wrote: Working backwards, firstly, for B to be a multiple of 3, it can only be 3, 6 or 9.
Hey Mathsbuddy,

You're absolutely right about the answer being E, but be careful. If B is a multiple of 3, then B can equal 0, 3, 6 or 9

Zero is a multiple of all integers. Likewise, zero is divisible by all integers.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image

Master | Next Rank: 500 Posts
Posts: 447
Joined: Fri Nov 08, 2013 7:25 am
Thanked: 25 times
Followed by:1 members

by Mathsbuddy » Wed Nov 19, 2014 10:05 am
Brent@GMATPrepNow wrote:
Mathsbuddy wrote: Working backwards, firstly, for B to be a multiple of 3, it can only be 3, 6 or 9.
Hey Mathsbuddy,

You're absolutely right about the answer being E, but be careful. If B is a multiple of 3, then B can equal 0, 3, 6 or 9

Zero is a multiple of all integers. Likewise, zero is divisible by all integers.

Cheers,
Brent
Thanks Brent, I did wonder if I should have included zero!
However, as it is not sufficient for the other values, it's sufficiency for zero becomes obsolete.
Nonetheless, I haven't totally convinced myself if 'working backwards' is 100% valid!