If the least common multiple of a positive integer x, 4^3 and 6^5

Question

If the least common multiple of a positive integer  x ,  4^3  and  6^5  is  6^6 , then how many values can  x  take?

A. 1
B. 6
C. 7
D. 30
E. 36

Bunuel · Accepted Answer

If the least common multiple of a positive integer  x ,  4^3  and  6^5  is  6^6 , then how many values can  x  take?

A.  1 
B.  6 
C.  7 
D.  30 
E.  36

We are given that  6^6=2^{6}*3^{6}  is the least common multiple (LCM) of three numbers:
 
 x ;
 
 4^3=2^6 ;
 
 6^5 = 2^{5}*3^5 ;
 
First notice that  x   cannot include any prime factors other than 2 and/or 3  as the LCM is exclusively composed of these primes.
 
Next, since the exponent of 3 in the LCM is greater than the exponent of 3 in the other two numbers,  x  must have  3^6  as a factor. Otherwise, the factor  3^{6}  wouldn't appear in the LCM.
 
Furthermore,   x  can include the prime number 2 raised to any power from 0 to 6, inclusive  (the LCM limits the power of 2 in  x  to 6, so it cannot be any higher).
 
Therefore,  x  can take on a total of 7 distinct values, namely:
 
 3^6 ;
 
 2*3^6 ;
 
 2^2*3^6 ;
 
 2^3*3^6 ;
 
 2^4*3^6 ;
 
 2^5*3^6 ;
 
 2^6*3^6 .

Answer: C

manpreetsingh86 · Answer

4^3=2^6 ,  6^5= 2^5.3^5

now if the power of 3 in x is less than 6, then l.c.m will have  3^5 in it.
if the power of 3 in x is more than 6, then l.c.m will have a term more than  3^6 .

thus x must have  3^6  in it, and x cannot have any prime number other than 2 and 3. why ?? because in that case l.c.m won't be  6^6 .
Also, maximum power of 2 that x can contain is 6. thus power of 2 in x can vary from 2^0 to 2^6. this gives us 7 possible cases. which are  2^0.3^6,2^1.3^6,2^2.3^6,2^3.3^6,2^4,3^6,2^5.3^6,2^6.3^6

hence answer is C

TarPhi · Answer

x=3^6 * 2^0, 3^6 * 2^1, 3^6 * 2^2, 3^6 * 2^3, 3^6 * 2^4, 3^6 * 2^5, 3^6 * 2^6

Hence 7

bo12345 · Answer

I came accross this question during a Focus test and I think the wording in this question is ambigous. Please fix it or correct me if I am wrong. IMO the clear wording should be something like:

"The least common multiple of x ,4^3 and 6^5 is 6^6. How many positive integers x satisfy this?"

Bunuel · Answer

This is standard GMAT wording, even if it feels unusual at first. The structure is intentional and precise. It’s important to get familiar with how such questions are phrased.

ManishaPrepMinds · Answer

we need to find the values of x where the least common multiple (LCM) of x, 4^3, and 6^5 equals 6^6.

Step 1: Find prime factorizations  
 
 4^3 = 2^6 
 6^5 = (2 × 3)^5 = 2^5 × 3^5 
 6^6 = (2 × 3)^6 = 2^6 × 3^6  
  Step 2: Apply LCM rule   
For the LCM of numbers, we take the highest power of each prime factor that appears in any of the numbers.

Currently, without x: 
LCM(2^6, 2^5 × 3^5) = 2^6 × 3^5 
But we need the LCM to be 2^6 × 3^6.

Step 3: Determine constraints on x   This means x must contribute the factor 3^6 to make the LCM equal to 2^6 × 3^6.

Since we already have 2^6, x cannot contribute any higher power of 2, or the LCM would exceed 2^6 × 3^6.

So x must be of the form:  x = 2^a × 3^6  where:
 
 a ≤ 6 (so as not to exceed the power of 2 in the target LCM) 
 The power of 3 must be exactly 6

Step 4: Count possible values   This gives us x = 2^a × 3^6 where a can be 0, 1, 2, 3, 4, 5, or 6.
That's  7 possible values  for x:
 
 x = 3^6 
 x = 2^1 × 3^6  
 x = 2^2 × 3^6 
 x = 2^3 × 3^6  
 x = 2^4 × 3^6 
 x = 2^5 × 3^6  
 x = 2^6 × 3^6   
 Answer: C. 7

The Key Insight Students Miss

The main trap is not recognizing that  x can have different powers of 2  (from 2^0 to 2^6) while maintaining the same LCM.

Many students find one valid value and assume it's unique, and choose A(1) which is a  trap answer  or they don't systematically consider all possibilities.

The correct approach requires understanding that  x = 2^a × 3^6  where a ∈ {0, 1, 2, 3, 4, 5, 6}, giving exactly  7 values .

nobodyaal · Answer

Hey Bunuel! Can you please share why x cannot be 6^6 itself? Like if LCM for 3, 4 & x is 12; cant x be 12? Please let me know if I am getting this wrong, thanks!

Bunuel · Answer

x  can  be 6^6. There is no restriction against that.

The LCM is allowed to be equal to one of the numbers involved. Just like LCM(3, 4, 12) = 12, here LCM(x, 2^6, 2^5*3^5) can equal x if x already contains the highest required powers. That is why 2^6*3^6 (i.e., 6^6) is included among the valid values of x.

If the least common multiple of a positive integer x, 4^3 and 6^5

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If the least common multiple of a positive integer x, 4^3 and 6^5

Prep Toolkit

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