### Brain Teasers

# Ascending Order

Arrange the following numbers in ascending order without using a calculator.

2^6666 , 3^3333 , 6^2222

2^6666 , 3^3333 , 6^2222

### Answer

3^3333<6^2222<2^6666

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## Comments

Try to solve this without using log or a calculator

Nice one. Hint: if you take the 1,111th root

of each number, you don't change the order.

of each number, you don't change the order.

umm all i did was take the base and the first power number into consideration and ignoring the ongoing repeating power. ex. 2^6=64, 3^3=27, and 6^2= 36 ...then i arranged it without a calculator! =)

I like hairemo's approach, which is the same as dewtell's. Instead of converting to x^1111, I converted all to x^6666:

2^6666 > 6^(1/3)^6666 > 3^(1/2)^6666

Basically 2 > cube root of 6 > square root of 3

Not quite as elegant, but almost as easy.

2^6666 > 6^(1/3)^6666 > 3^(1/2)^6666

Basically 2 > cube root of 6 > square root of 3

Not quite as elegant, but almost as easy.

I had to make silly theorems using logarithms. Nevertheless they worked.

1.) 2^x > 3^y iff (x/y) > logarithm to the base 2 of 3

2.) 2^x > 6^y diff (x/y) > logarithm to the base 2 of 6

knowing that log base 2 of 3 is less than 2

and log base 2 of 6 is less than 3. We can see that 2^6666 is the biggest. Lastly, you can compare 6^2222 and 3^3333 with the same sort of theorem. I'm too lazy to mention how I did it

1.) 2^x > 3^y iff (x/y) > logarithm to the base 2 of 3

2.) 2^x > 6^y diff (x/y) > logarithm to the base 2 of 6

knowing that log base 2 of 3 is less than 2

and log base 2 of 6 is less than 3. We can see that 2^6666 is the biggest. Lastly, you can compare 6^2222 and 3^3333 with the same sort of theorem. I'm too lazy to mention how I did it

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