Computers

Any of the major manufacturers; AMD processors such as a Sempron of the Athlon. Intel Pentium 4 etc..

The performance of any computer and hence it's software, benefit from having adequate memory installed. This is almost as important as the processor that's used.

Go for at least 512MB of memory. Most motherboards these days only support DDR memory, which is fast than SD RAM.

i think stick with p4 for that one because amd works allot better with games than large programs like image editing and mp3 encoding and do't listen to the rumours that the p4 fans die out all fans die out eventually

Fromj 20gig

AMD Athlon 64. Go for 2+gHz, x2 (dual core). AMD processors are cheaper but just as good if not better than Intel. You just pay for the cheap 1 cent Intel sticker if you go for Intel. If you want Intel, that's okay too.

FAN IT TO THE MAX!!! Fans are cheap! Buy them!

Frankly, I would go for one of the pentiums, maybe the Pentium 4 CPU. It's cheap, you can get 20 on eBay for $50 dollars, it's common, so you can just find a cheap motherboard that has a 775 socket. It's not as bad as Celeron or Atom, and it's not as expensive or ultra-powerful as the i3/i5/i7 CPUs. But, if you want one that is also cheap, but is more in the 64-bit range, go for a cheap 604 socket Xeon. SeanHolshouser

This is a good question.

By experiment, 1 is not, 2 is not, 4 is not, 8 is not, 16 is not. So there are some counting numbers that are not the sum of two or more counting numbers.

It turns out that a counting number can be expressed as the sum of two or more consecutive counting numbers if and only if it is not a power of 2.

Let us do some algebra (sorry about the lack of nice symbols).

Consider n consecutive counting numbers, starting at a (so n >= 2, a >= 1).

The numbers are a, a+1, ... , a + (n-1).

If we add these up, we get the sum s = na + n(n-1)/2.

For example, five consecutive numbers starting at 7 are 7, 8, 9, 10, 11.

Here n = 5 and a = 7. The sum of the five numbers is s = 5x7 + 5x(5-1)/2 = 45.

Fact: Every sum of at least two consecutive counting numbers has an odd factor.

Proof: We analyse s = na + n(n-1)/2. There are two cases.

Case 1: n is odd. Then s = n(a + (n-1)/2). Since n is odd, n-1 is even, so (n-1)/2 is a whole number, and a + (n-1)/2 is also a whole number.

Then s has a factor of n, and n is odd.

Case 2: n is even. Write m = n/2, so m is a whole number.

We find s = 2ma + 2m(2m-1)/2. so

s = 2ma + m(2m-1).

Thus s = m(2a + 2m - 1).

Whatever a and m are, 2a + 2m must be even, so 2a + 2m - 1 is odd.

Thus s has an odd factor, namely 2a + 2m - 1.

Consequence: If a counting number is a power of 2, it cannot be expressed as the sum of two or more consecutive counting numbers, because a number that is a power of 2 does not have any odd factors.

That is half the story. As far as the Fact above tells us, there might be some numbers that do have odd factors and cannot be expressed as the sum of two or more consecutive counting numbers.

Actually, there are not.

Any counting number can be written as a product pq, where p is a power of 2, and q is an odd number (which could be 1). If the number has an odd factor, q is at least 3.

For example, 80 = 16 x 5, so p = 16, q = 5 for the number 80.

Fact: Every counting number that has an odd factor can be expressed as the sum of two or more consecutive counting numbers.

Proof: Write our counting number k as the product pq; since k is supposed to have an odd factor, q is at least 3. We note that 2p = q is impossible, because 2p is even and q is odd. There are two cases.

Case 1: 2p < q.

Set n = 2p and a = (q - 2p + 1)/2.

Since 2p < q, q - 2p + 1 is positive, and since q is odd, q - 2p + 1 is even. So a = (q - 2p + 1)/2 is a counting number.

Let us work out s = na + n(n-1)/2. We get

s = (2p)(q - 2p + 1)/2 + (2p)(2p - 1)/2

So s = p(q - 2p + 1) + p(2p-1), which works out to pq, as it should.

Thus in Case 1 k = pq is the sum of nconsecutive counting numbers starting at a.

Case 2: 2p > q.

Set n = q and a = p - (q-1)/2.

Since 2p > q, certainly 2p > q - 1, so 2p - (q - 1) is positive. Also, since q is odd, p - (q - 1) is even. So a = p - (q-1)/2 is a counting number.

Let us work out s = na + n(n-1)/2. We get

s = q( p - (q-1)/2) + q(q-1)/2. This works out to pq.

Thus in Case 2 k = pq is the sum of nconsecutive counting numbers starting at a.

In both cases, k = pq is the sum of nconsecutive counting numbers starting at a, and our proof is finished.

For an example, try k = 68. Since 68 = 4 x 17, p = 4 and q = 17. Since 2 x 4 < 17, this is an example of Case 1.

Set n = 2p = 8, and a = (q - 2p + 1)/2 = 5. Then 68 should be the sum of 8 consecutive numbers starting at 5.

Check: 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 68.

For another example, try k = 80. We have 80 = 16 x 5, so p = 16, q = 5. Since 2 x 16 > 5, this is an example of Case 2.

Set n = q = 5 and a = p - (q-1)/2 = 14. Then 80 should be the sum of 5 consecutive numbers starting at 14.

Check: 14 + 15 + 16 + 17 + 18 = 80.

The two Facts above give us a complete solution to the problem: a counting number can be expressed as the sum of two or more consecutive counting numbers if and only if it is not a power of 2.

Comment: We have shown that a number that has an odd factor can be expressed as the sum of two or more consecutive counting numbers, and the proof of the second Fact shows a way to do it. But there may be more than one way of expressing a number as the sum of two or more consecutive counting numbers.

For example, 60 = 4 x 15, so it fits under Case 1, and we find that n = 8, and a = 4.

So 60 = 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11.

But also 60 = 10 + 11 + 12 + 13 + 14, which doesn't come out of our proof.

elements

APEX :)