How do I determine what the highest power of 2 is that divides 89,275,744?

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- Jan 28th 2007, 07:50 PMIdeasmanDivisibility Rules
How do I determine what the highest power of 2 is that divides 89,275,744?

- Jan 28th 2007, 09:04 PMCaptainBlack
- Jan 29th 2007, 07:02 AMThePerfectHacker
The theorem says, and follow the pattern....

N is divisible by 2 if and only if the last digit is even.

N is divisible by 4=2^2 if and only if the last two digits are divisible by 4.

N is divisible by 8=2^3 if and only if the last three digits are divisible by 8.

And thus on.....

Thus, given 89,275,744

1)We see it is divisible by 2 because 2 divides 2.

2)We see it is divisible by 2^2 because 4 divides 44.

3)We see it is divisible by 2^3 because 8 divides 744.

And so on....

Find the point where it stops being divisible and you have an answer. - Jan 29th 2007, 10:20 AMSoroban
Hello, Ideasman

Quote:

How do I determine what the highest power of 2 is that divides 89,275,744?

Convert $\displaystyle 89,275,744$ to binary and we get:

. . $\displaystyle 101,010,100,100,011,110,101,100,000_2$

. . . . . . . . . . . . . . . . . . . - - - $\displaystyle \uparrow$

. . . and it is obvious that $\displaystyle \overbrace{100000_2 = 32}$ is the greatest divisor.

Just kidding!