a and b are positive integers less than or equal to 9. If a

Dear Bunnel,

What about if I take a=4 b=5.

It's not divisible by 3 please assist.

It works even when you use 4 and 5.
4=a
5=b
so the number is 454545.
Adding up all 3(4) + 3(5) = 3(9) = 27.
This is divisible by 3.
The of the question essence lies the data point that ab is repeated thrice.
hence should be divisible by 3
Hope this helps

454545 when divided by 3 gives the result 151515

\(\frac{454545}{3} = 151515\)

CONCEPT: Rule of Divisibility of 3 is "If sum of the digits of the Number is divisible by 3 then the number will be divisible by 2

SUm of the digits of No. 454545 is 27 which is divisible by 3 hence the Number 454545 will be divisible by 3

I hope it helps!

(A) 3: ababab is divisible by 3 because the sum of its digits is 3(a + b), a multiple of 3 for any
integers a and b.
(B) 4: An integer is divisible by 4 if its last two digits represent a two-digit number that is itself
divisible by 4. It is uncertain whether the two-digit integer ab is divisible by 4.
(C) 5: An integer is divisible by 5 if the last digit is 0 or 5. It is uncertain whether the positive
integer b is 5.
(D) 6: An integer is divisible by 6 if it is even and divisible by 3. We already established that ababab is divisible by 3, but it is uncertain whether the last digit b is even, a requirement for ababab to be even.

Alternatively, we can tackle this problem by thinking about the place values of the unknowns. If we had a two-digit number ab, we could express it as 10a + 1b. By similar logic, ababab can be expressed as follows:
ababab = 100,000a + 10,000b + 1,000a + 100b + 10a + b
If we combine like terms, we get the following:
ababab = 101,010a + 10,101b

At this point, we can spot a common term: each term is a multiple of 10,101. If we factor 10,101 from each term, the expression can be written as follows:
ababab = 10,101(10a + b), where a and b are individual digits.

Or simply:

ababab = 10,101(ab), where ab is a two-digit number.
Since we don't know the value of the two-digit number ab, we cannot know what its factors are. To find a known factor of ababab, our only option is to find a factor of 10,101.
At this point, we can recognize that 10,101 is a multiple of 3 (the sum of the digits is 3). Therefore, ababab must be a multiple of 3.

The correct answer is A.

the sum of the digits of ababab is 3(a+b) which is divisible by 3. Hence the answer is 3.

I am sorry but I cannot deduct from the question that you have to sum up the digits!!! Where did you get that from the question??

You can try the worst case scenario to reduce the options. Setting a and b to be both prime numbers, this will eliminate all of the even answers.

Next, knowing that one of the options has a 5, and knowing that numbers having 2-digits or more ending in 0 and 5 are divisible by 5,you can eliminate 5 by setting "b" as a prime number that is not 5.

Example: a=5, b=7

575757

Now you are down to 2 options, a and e. Starting with A, you will see that 3 is a factor of 575757

Next try another number, let's say 737373. 3 also works.

I'm now leaning towards A.

This is how I solved this,

We are told that a and b are positive and <=9
So the maximum value for ababab can be 999999 and the minimum value can be 111111

We have to find a number that MUST be a factor, so we take the minimum value-111111=3*11*37

Hence, 3 is the answer.

Please let me know if my approach is incorrect.
Bunuel KarishmaB

Sure you can take a number of the form and check (as you did) but the issue here is that it has an option 'None of these.' So what works for one example, may not work for all such numbers and hence, you need to establish it conceptually.
e.g. ababab = ab * 10101
Since 10101 is divisible by 3, no matter what ab is, ababab will be divisible by 3.

That said, GMAT does not give this option "None of these" so normally you don't have to worry about it.

Thank you for your response.
I get the conclusion that 3 can divide 10101, but can you explain how do we arrive at 10101*ab?

Also if I chose 101010 as the minimum value instead of 111111, and reduced it to its primes, would it then be correct to determine that 3 will always be a divisor and hence, a factor.
Eg: 101010=2*3*5*7*13*37

You could take any example - it needn't be the smallest such number. (As an aside, note that in this case, since a and b are positive integers, 101010 is not the smallest such number since b cannot be 0.)
A number which must be a factor of all numbers of the form ababab will be a factor of your example too. Hence you can pick any number of the form ababab. Note that it is not necessary that if the smallest such number has that factor then every greater number will also have that factor. For example, 11 is a factor of 111111 but it will not be a factor of 636363.

The point is that if we are looking for a factor of ababab such that this factor is a factor of all such numbers, then it will be a factor of 111111, 131313, 878787, 929292 etc. So I could pick any number, find its factors and then try to match it with the given options.
If I had picked 929292, then I would get 3, 4 and 6 as factors. So I would need to pick another number such as 878787 to confirm that 4 and 6 are not necessarily factors of all numbers of the form ababab.

As for how to figure out that ababab = ab * 10101, it is all about pattern recognition.

We see that aaa = a * 111 (try the multiplication)
Then if we have 2 digits getting repeated, ababab = ab * 10101 (again, try the multiplication by taking values for a and b)
Similarly, abcabc = abc * 1001

I suggest you to check out these posts and videos on factors and factorisation:

Factors:
https://anaprep.com/number-properties-f ... -a-number/
https://anaprep.com/number-properties-r ... e-factors/
https://youtu.be/DxIH8rjhpKY

Factorisation
https://youtu.be/Kd-4cH4cqHw

a and b are positive integers less than or equal to 9. If a and b are assembled into the six-digit number ababab, which of the following must be a factor of ababab?


Sum of digits of ababab = 3a +3b = 3(a+b) i.e. multiple of 3 therefore, Option A.

NOTE :- We can not comment about a+b therefore, ababab being multiple of 6 is not a must be true