Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

S1 is sufficient. \(1+5+4+3+&+2\) must be divisible by 9. This is possible only when \(& = 3\) .

S2 is not sufficient. \(&2\) must be divisible by 4. This is possible when \(& = 1\) or \(& = 3\) .

I realize this is a pretty rudimentary question but would somebody please explain why all the digits are added together to determine divisibility by 9 for statement 1, but for statement 2, only the first two digits are added to determine divisibility. The answer is just not obviously clear to me.

Hey answer is pretty simple. !! there are few rules for division you need to understand before answering such kind of questions. Rule I. divisible by "2" - any no. will be considered divisible by two if the last digit of that no. is divisible by 2.

ex XXX4 - this no. will always be divisible by 2

Rule II. for numbers "3" and "9"

if sum of the digits of number is divisible by "3" or "9" then the number is considered divisible by "3" or "9" consider no. 123456 , sum of digits = 1+2+3+4+5+6=21 , now 21 is divisible by 3 only not by 9. therefore 123456 is divisible by 3 only.

Rule III divisible by "4" for any no. to be divisible by 4 , its last two digits should be divisible by 4. eg. X32 will always be divisible by 4 for any value of X. You can try this exercise yourself

@meetnaren Thanks for the valuable tip, but it would be great if u could explain it a little more, so that it is easy to understand. At first, I didn't get but then I used some random numbers & divided them to verify and it did work ( maybe I've got dull a bit )
_________________

"I choose to rise after every fall" Target=770 http://challengemba.blogspot.com Kudos??

2: last digit should be divisible by 2. 3: sum of the digits should be divisible by 3. 4: last 2 digits should be divisible by 4. 5: last digit should be 0 or 5. 6: last digit should be divisible by 2 and sum of the digits should be divisible by 3. 7: given below 8: last 3 digits should be divisible by 8. 9: sum of the digits should be divisible by 9.

Divisibility by 7:- Take the last digit in a number. Double and subtract the last digit in your number from the rest of the digits. Repeat the process for larger numbers. Example: 287 (Double the 7 to get 14. Subtract 14 from 28 to get 7 which is divisible by 7 and we can now say that 287 is divisible by 7.

1 - A is divisible by 9. That means 1+5+4+3+2+& must be divisible by 9. Now 1+5+4+3+2 is 15. Although, in general & can any integer value from 0 - 9, but it can only take a value that makes A a multile of 9. When &=3, A = 18 --> A is divisible by 9 we need 9 more to reach next multiple of 9, but since we have already used 3 to reach 18 (15 + 3 = 18), we need 12 to reach next multiple of 9, i.e 27. (15 + 12 = 27) Because & = 12 is not possible, we can only use & = 3. Statement 1 is sufficient.

2 - For a number to be divisible by 4, its last two digits must be divisible by 4 (i.e a multiple of 4). Considering this rule, 154312 should be divisible by 4. 154352 should also be divisible by 4. Because we have found two values for &, this statement is not sufficient.

@meetnaren or @AtifS - can you explain the rule? seems like a rule that could be really helpful. Thks

Divisibility Rules

2 - If the last digit is even, the number is divisible by 2.

3 - If the sum of the digits is divisible by 3, the number is also.

4 - If the last two digits form a number divisible by 4, the number is also.

5 - If the last digit is a 5 or a 0, the number is divisible by 5.

6 - If the number is divisible by both 3 and 2, it is also divisible by 6.

7 - Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.

8 - If the last three digits of a number are divisible by 8, then so is the whole number.

9 - If the sum of the digits is divisible by 9, so is the number.

10 - If the number ends in 0, it is divisible by 10.

11 - If you sum every second digit and then subtract all other digits and the answer is: 0, or is divisible by 11, then the number is divisible by 11. Example: to see whether 9,488,699 is divisible by 11, sum every second digit: 4+8+9=21, then subtract the sum of other digits: 21-(9+8+6+9)=-11, -11 is divisible by 11, hence 9,488,699 is divisible by 11.

12 - If the number is divisible by both 3 and 4, it is also divisible by 12.

25 - Numbers ending with 00, 25, 50, or 75 represent numbers divisible by 25.

@meetnaren or @AtifS - can you explain the rule? seems like a rule that could be really helpful. Thks

Divisibility Rules

2 - If the last digit is even, the number is divisible by 2.

3 - If the sum of the digits is divisible by 3, the number is also.

4 - If the last two digits form a number divisible by 4, the number is also.

5 - If the last digit is a 5 or a 0, the number is divisible by 5.

6 - If the number is divisible by both 3 and 2, it is also divisible by 6.

7 - Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.

8 - If the last three digits of a number are divisible by 8, then so is the whole number.

9 - If the sum of the digits is divisible by 9, so is the number.

10 - If the number ends in 0, it is divisible by 10.

11 - If you sum every second digit and then subtract all other digits and the answer is: 0, or is divisible by 11, then the number is divisible by 11. Example: to see whether 9,488,699 is divisible by 11, sum every second digit: 4+8+9=21, then subtract the sum of other digits: 21-(9+8+6+9)=-11, -11 is divisible by 11, hence 9,488,699 is divisible by 11.

12 - If the number is divisible by both 3 and 4, it is also divisible by 12.

25 - Numbers ending with 00, 25, 50, or 75 represent numbers divisible by 25.

S1 is sufficient. \(1+5+4+3+&+2\) must be divisible by 9. This is possible only when \(& = 3\) .

S2 is not sufficient. \(&2\) must be divisible by 4. This is possible when \(& = 1\) or \(& = 3\) .

I realize this is a pretty rudimentary question but would somebody please explain why all the digits are added together to determine divisibility by 9 for statement 1, but for statement 2, only the first two digits are added to determine divisibility. The answer is just not obviously clear to me.

\(@\) represents the tens digit in integer \(A = 1543@2\). What is the value of \(@\) ?

(1) \(A\) is divisible by 9. In order an integer to be divisible by 9, the sum of its digit must be divisible by 9. Hence \(1+5+4+3+@+2=15+@\) must be divisible by 9, and since @ is a digit then it can only equal to 3. Sufficient.

(2) \(A\) is divisible by 4. In order an integer to be divisible by 4, its last two digits must be divisible by 4. Hence \(@2\) must be divisible by 4, which is possible if \(@=1\), \(@=3\), \(@=5\), \(@=7\), or \(@=9\). Not sufficient.

We will have to check the divisibility rules for both statements: For S1 to be true &=3. hence S1 is sufficient For S2 to be true & =1/3 (as last two digits have to be divisible by 4). hence S2 is not sufficient. Thus answer is A

@meetnaren or @AtifS - can you explain the rule? seems like a rule that could be really helpful. Thks

Divisibility Rules

2 - If the last digit is even, the number is divisible by 2.

3 - If the sum of the digits is divisible by 3, the number is also.

4 - If the last two digits form a number divisible by 4, the number is also.

5 - If the last digit is a 5 or a 0, the number is divisible by 5.

6 - If the number is divisible by both 3 and 2, it is also divisible by 6.

7 - Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.

8 - If the last three digits of a number are divisible by 8, then so is the whole number.

9 - If the sum of the digits is divisible by 9, so is the number.

10 - If the number ends in 0, it is divisible by 10.

11 - If you sum every second digit and then subtract all other digits and the answer is: 0, or is divisible by 11, then the number is divisible by 11. Example: to see whether 9,488,699 is divisible by 11, sum every second digit: 4+8+9=21, then subtract the sum of other digits: 21-(9+8+6+9)=-11, -11 is divisible by 11, hence 9,488,699 is divisible by 11.

12 - If the number is divisible by both 3 and 4, it is also divisible by 12.

25 - Numbers ending with 00, 25, 50, or 75 represent numbers divisible by 25.

@meetnaren or @AtifS - can you explain the rule? seems like a rule that could be really helpful. Thks

Divisibility Rules

2 - If the last digit is even, the number is divisible by 2.

3 - If the sum of the digits is divisible by 3, the number is also.

4 - If the last two digits form a number divisible by 4, the number is also.

5 - If the last digit is a 5 or a 0, the number is divisible by 5.

6 - If the number is divisible by both 3 and 2, it is also divisible by 6.

7 - Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.

8 - If the last three digits of a number are divisible by 8, then so is the whole number.

9 - If the sum of the digits is divisible by 9, so is the number.

10 - If the number ends in 0, it is divisible by 10.

11 - If you sum every second digit and then subtract all other digits and the answer is: 0, or is divisible by 11, then the number is divisible by 11. Example: to see whether 9,488,699 is divisible by 11, sum every second digit: 4+8+9=21, then subtract the sum of other digits: 21-(9+8+6+9)=-11, -11 is divisible by 11, hence 9,488,699 is divisible by 11.

12 - If the number is divisible by both 3 and 4, it is also divisible by 12.

25 - Numbers ending with 00, 25, 50, or 75 represent numbers divisible by 25.

As 0 is the first multiple of every number than will it not be divisible by every number.

thanks Sid

Yes, zero is divisible by EVERY integer except zero itself, since 0/integer=integer (or, which is the same, zero is a multiple of every integer except zero itself).
_________________

1) A is divisible by 9 -> 1+5+4+3+2+& = 15+& is divisible by 9 (0=<&=<9) -> &=3 -> sufficient 2) A is divisible by 4 -> &2 is divisible by 4. &=3 and &=1 are both ok -> insufficient