For How Many Values of k is 12^12 the Least Common Multiple GMAT Problem Solving

Question: For how many values of k is 12^12 the least common multiple of the positive integers 6^6, 8^8 and k?

1. 23
2. 24
3. 25​
4. 26
5. 27

“For how many values of k is 12^12 the least common multiple”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "GMAT Official Guide Quantitative Review". To solve GMAT Problem Solving questions a student must have knowledge about a good number of qualitative skills. GMAT Quant section consists of 31 questions in total. The GMAT quant topics in the problem-solving part require calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1:
The problem statement states that
12^12 = 2^24∗3^12 is the least common multiple of the following three numbers:
6^6 = 2^6 * 3^6
8^8 = 2^24
And k;
we can first notice that:
k cannot have any other primes other than 2 or/and 3, because LCM contains only these primes.
Now, since the power of 3 in LCM is higher than the powers of 3 in either the first number or in the second, than
k must have 3^12 as its multiple (else how 3^12 would appear in LCM?).
Hence, we get:
12^12 = (2*2*3)^12
=> (2^24) * (3^12)
Thus, k will also be in the form of : (2^a) * (3^b)
Now, b has to be equal to 12 since in order for (2^24) * (3^12) to be a common multiple, at least one of the numbers must have the terms (2^24) and (3^12) as
its factors. (not necessarily the same number).
We can see that 8^8 already takes care of the 2^24 part.
Thus, k has to take care of the 3^12 part of the LCM.
This means that the value k is (2^a) * (3^12) where a can be any value from 0 to 24 (both inclusive) without changing the value of the LCM.
Thus K can have 25 values.

Correct Answer: C

Approach Solution 2:

According to the problem statement,
The least common multiple of the following three numbers is 1212 = 224312.
6^6 = 2^6 * 3^6
8^8 = 2^24
And k;
We can see that:
Since LCM only contains the primes 2 or/and 3 there can be no more primes for k.
Considering that the power of 3 in LCM is more than the powers of 3 in either the first or second number, then
If k didn't have 312 as its multiple, how would 312 show up in the LCM?
Thus, we obtain:
12^12 = (2*2*3)^12
=> (2^24) * (3^12)
As a result, k will also take the form of: (2a) * (3b).

Now, b must be equal to 12 since at least one of the numbers must include the terms (224) and (312) as part of it for (224) * (312) to be a common multiple.
its components (maybe not the same quantity).
We can see that the 224 component is already taken care of by 88.
K must thus handle the 312 component of the LCM.
This indicates that the value of k is (2a) * (312), where a can have any value between 0 and 24 (inclusive) without affecting the LCM's value.
K therefore has a possible 25 values.

Correct Answer: C

Approach Solution 3:

The problem statement states that
1212 = 224312 is the least frequent multiple of the next three digits.
6^6 = 2^6 * 3^6
8^8 = 2^24
And k;
We notice that:
There can be no further primes for k because LCM only includes the prime numbers 2 or/and 3.
Given that the LCM's power of three exceeds the powers of 3 in either the first or second number,
How would the number 312 appear in the LCM if k didn't contain 312 as its multiple?
Therefore, we get:
12^12 = (2*2*3)^12
=> (2^24) * (3^12)
K will thus likewise have the form (2a) *. (3b).

Now, for (224) * (312) to be a common multiple, b must equal 12. This is because at least one of the numbers must contain the terms (224) and (312) in its whole.
its components (maybe not the same quantity).
We can see that 88 has already taken care of the 224 component.
K must thus handle the LCM's 312 component.
Thus, it can be concluded that the value of k is (2a) * (312), where a can be any positive integer between 0 and 24 (inclusive) without changing the value of the LCM.
K therefore has a range of 25 potential values.

Correct Answer: C

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