jade3 wrote:
For how many values of k is 12^12 the least common multiple of the positive integers 6^6, 8^8 and k?
A. 23
B. 24
C. 25
D. 26
E. 27
Lets use a quick example
What is the LCM of 2,4,9,12,
Factorise all the numbers one by one and write them in prime numbers raised to exponent form \((Prime1)^m\) X \((Prime2)^m\)...
\(2=2^1\)
\(4=2*2==>2^2\)
\(9=3*3==>3^2\)
\(12=4*3==>2*2*3==>2^2 * 3^1\)
NOW LCM OF THESE NUMBERS WILL TAKE THE HIGHEST POWER OF EACH PRIME FROM EACH NUMBER (ONE TIMES ONLY)
So LCM = \(2^2 * 3^3\)==> 4*9=36
Notice how \(2^1\) and \(3^1\) are not contributing towards the LCM at all.
Now apply the same logic to your question
You already know LCM is = \(12^{12}=(4*3)^{12}\)==> \((2^2*3)^{12}\) ==> \(2^{24}*3^{12}\)
Similarly \(6^6= (2*3)^6==>2^6*3^6\) So we know \(6^6\) is neither contributing 2's or 3's towards the LCM
\(8^8= (2^3)^8==> 2^{24}\) , So we know 8 is contributing all the \(2^{24}\)towards our LCM
Now we need a \(3^{12}\) to reach the LCM
Since K is the only remaining digit therefore K must contribute \(3^{12}\)
but it is also possible K can or cannot have \(2^m\) in it also and the values of \(2^m\) can vary from \(2^0 to 2^{24}\)
Remember for LCM we take the highest power, so \(2^{24}\) can be common in \(8^8\) as well as K
therefore total values of 2 in k = \(2^0 to 2^{24}\) (Total=25) and one compulsory value of \(3^{12}\) (total= 1)
Total=26 values
Answer is D
Why am in overshooting by 1?
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