Number Theory. LCM
Number Theory
Level
3
For how many 'k' is lcm(6^6, 8^8, k) = 12^12
24
26
27
23
25
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1 solution
(6^6) = (2^6) ∗ (3^6).
(8^8) = (2^24)
Now, we know that the least common multiple of the above two numbers and k is:
(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.