I need help I'm taking a test for an interview...
2006-07-03 09:24:31 · 10 answers · asked by mashc10 2 in Science & Mathematics ➔ Mathematics
The right answer is not 4
2006-07-03 19:09:44 · update #1
One could find innumerable patterns here. The one that I found is multiply by 2, add 1, multiply by 2, subtract 1, multiply by 2, add 1, multiply by 2, subtract 1, and so on.
1*2=2, 2+1=3, 3*2=6, 6-1=5, 5*2=10
The next number is 10.
2006-07-03 09:32:56 · answer #1 · answered by rar4000 2 · 0⤊ 0⤋
4
2006-07-03 16:52:18 · answer #2 · answered by LAB811 1 · 0⤊ 0⤋
4
2006-07-03 16:32:17 · answer #3 · answered by Edward A 1 · 0⤊ 0⤋
4
2006-07-03 16:30:53 · answer #4 · answered by Anonymous · 0⤊ 0⤋
4
2006-07-03 16:28:42 · answer #5 · answered by Commensalism 2 · 0⤊ 0⤋
One solution is -14.
If you plug x=6 into this formula,
f(x) = 1 + 1(x-1) + 0(x-1)(x-2)/2 + 2(x-1)(x-2)(x-3)/6 + -8(x-1)(x-2)(x-3)(x-4)/24, you will get f(6)=-14. That formula also gives f(1)=1, f(2)=2, f(3)=3, f(4)=6, f(5)=5.
Here is my solution to your problem. There are undoubtedly other solutions.
We know that given n ordered pairs { (x1, y1), (x2, y2), …, (xn, yn) }, there must be an (n-1) degree polynomial f(x) that contains those ordered pairs. If we can calculate that polynomial, then we can calculate f(x) at x=x subscript n+1.
There are certainly many ways to calculate that polynomial. One way would be solving n equations in n unknowns.
There is a topic in Numerical Analysis called "differences", and Newton's difference
formulas provide an easy way to calculate this polynomial. The formula is:
f(x) = f(x1)/0! + Df(x1)(x-x1)/1! + D2f(x1)(x-x1)(x-x2)/2! + D3f(x1)(x-x1)(x-x2)(x-x2)/3! + …
where Df(x1) = [f(x2) - f(x1)] / [ x2 - x1], etc. In our case the xi are successive integers.
f(x) = f(x1)/0! + Df(x1)(x-x1)/1! + D2f(x1)(x-x1)(x-x2)/2! + D3f(x1)(x-x1)(x-x2)(x-x3)/3! + …
f(x) = 1 + 1(x-1) + 0(x-1)(x-2)/2! + 2(x-1)(x-2)(x-3)/3! + -8(x-1)(x-2)(x-3)(x-4)/4!
Check:xf(x)
11
22
33
46
55
And then the next term is:
xf(x)
6-14
However, -14 is just one possible answer. Any number can be the next answer. For example,
I can find another polynomial that works for the same first five numbers, but also contains, say
-99 as the next number.
2006-07-03 16:53:59 · answer #6 · answered by fcas80 7 · 0⤊ 0⤋
I hate questions like these, because there are several possible answers. One answer I thought of was that this could be two series put together. All the odd terms could just be odd numbers, and each even term could be 3 * the last even term. If that is the case, then the next term would be 18. Another possibility is that the odd terms are two apart from each other, and the even terms are 4 apart from each other. If that is the case, then the next term is 10. If I were you, I'd note how ambiguous the question is and mention several possible solutions.
2006-07-03 16:38:12 · answer #7 · answered by anonymous 7 · 0⤊ 0⤋
30
2006-07-03 16:27:55 · answer #8 · answered by Anonymous · 0⤊ 0⤋
I quess 4 follows the pattern. 1+1=2, 2+1=3, 3+3=6, 6-1=5,5-1=4.
I just outh out of geo. though.
2006-07-03 16:34:12 · answer #9 · answered by wingnut3.1415 2 · 0⤊ 0⤋
4
what ? what interview? is that the exact question form the interview?
2006-07-03 16:32:01 · answer #10 · answered by rufus_t 2 · 0⤊ 0⤋