2

Is there a way to find the relation between numbers . I have 2 groups of numbers

Group a 

5 --> 20...
6 --> 32...
7 --> 112...
8 --> 192...
9 --> 576...
10 --> 1024...

And group b 

7 --> 56...
8 --> 80...
9 --> 432...
10 --> 672...

Is there a way to find relation between numbers and if there's no way to do that except mentally . What is the relation in the two groups

Thanks in advance

user3767289
  • 25

1 Answers1

2

Unfortunately, there is often no definitive way to determine a pattern within some given set of numbers, since patterns may be as arbitrary as you like. However, there are some ways of seeing patterns a little more clearly.

By way of example, I'm going to rewrite these relations in a slightly different way. Using that as a hint, try and find the relations yourself. I'll place the answer in a spoiler below.

Group A:

$5\to 2^2\cdot5$

$6\to 2^5$

$7\to 2^4\cdot7$

$8\to 2^6\cdot3$

$9\to 2^6\cdot9$

$10\to 2^{10}$

Group B:

$7\to 2^3\cdot7$

$8\to 2^4\cdot5$

$9\to 2^4\cdot3\cdot9$

$10\to 2^5\cdot3\cdot7$

Group A:

When $n$ is odd, $n\to n2^{n-2}$. When $n$ is even, $n\to2^{n-3}(n-2)$

Group B:

When $n$ is odd, $n\to 2^{n-6}\cdot n(n-3)$. When $n$ is even, $n\to 2^{n/2}(n-3)(n-7)$

Santana Afton
  • 6,758
  • thank you, jazzy. i have found a relation for all numbers in group a it's 2^n-3 (n-2+2(n mod 2)) but isn't there such one for group b without separating odd and even . – user3767289 Mar 14 '17 at 07:34
  • oh i found one for group b too : 2^((n-6(n mod 2)) / (-(n mod 2)+2)) x (n-7 + 7(n mod 2)) x (n-3) it's a bit long but it works . thanks – user3767289 Mar 14 '17 at 08:02
  • 1
    When you put "$n\bmod2$" into your formula, 3767289, you are separating odd and even. – Gerry Myerson Mar 14 '17 at 08:04