I don't have any idea about a "systematic" way to solve this one. So I just tried the lowest primes.
2 x 2 x 2 x 2 x 2 x 2 = 32
From this I realized that if I stick with 2s and add a 3, I will get 96, anything above that will give me numbers with more than 2 digits.
Therefore, you have just 32 and 96. Choose C.
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AleksandrM
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I did it pretty much the same way guys ... dont know if there is another approach to solve this one ...
The smallest prime number is 2, so I just worked with:
2*2*2*2*2*2 = 2^6 = 64. So this is one of the 2-digit numbers with "length" of 6.
Next prime number is 3, so you have:
2*2*2*2*2*3 = 2^5 * 3^1 = 96
There is no need to carry on, since anything else will take you over 100 ... and hence will no longer be a 2-digit number.
So we only have two 2-digit numbers with a length of 6.
any other approaches out there to solve this ?
The smallest prime number is 2, so I just worked with:
2*2*2*2*2*2 = 2^6 = 64. So this is one of the 2-digit numbers with "length" of 6.
Next prime number is 3, so you have:
2*2*2*2*2*3 = 2^5 * 3^1 = 96
There is no need to carry on, since anything else will take you over 100 ... and hence will no longer be a 2-digit number.
So we only have two 2-digit numbers with a length of 6.
any other approaches out there to solve this ?
