# Non calc paper question

1. Mar 31, 2007

### thomas49th

Hi, I have a question

tThe number 40 can be written as $$2^m$$ x n, where m and n are prime numbers

I managed to guess that m = 3 and n = 5 by trials and error by using low prime numbers. Is this the proper/best way to work out the answer. If not can someone show me a proper way

Thx

2. Mar 31, 2007

### Mindscrape

Your way works, because it got you the right answer, but consider what if the number was larger. The way I would do it, which is just one way of many, would be to write out the factors of 40 = {1x40, 2x20, ... , 40x1}. Since you know that 2^m can at most be 40, and also must be prime (meaning the first factor can only be 2^2, 2^3, 2^5 (which is not in the set of factors so it can be discarded)), that will narrow down the options of m, and from there you can find which of the 2 options of n will give the right factors. 4x10 (no!), or 8x5 (yes!)

My way would only work for numbers relatively small though because you don't to do a bunch of factors for numbers in tens of thousands. There could be something better out there with logarithms or systems of equations.

Last edited: Mar 31, 2007
3. Mar 31, 2007

### Gib Z

Yes there are better ways using Modular Arithmetic i think, but this is a GSCE or something paper, and considering as don't learn logs, I think they'll be fine with the guess and check for now, which is probably why they've been given a small number.

4. Mar 31, 2007

### Gib Z

Scratch last comment, I found an easier, yet systematic method that doesn't require guess and check.

$$40 = 2^m \cdot n$$.

Repeated divide by 2.
$$20=2^{m-1}\cdot n$$
$$10=2^{m-2} \cdot n$$
$$5=2^{m-3}\cdot n$$

Since 5 can not be divided by 2 any further, this is the lowest you can reduce it. Since 2 is obviously not a factor, m-3 must be zero, or m=3. And its also obvious to see n then must be 5. HAPPY SUNSHINE :D:D:D