- #1
- 239
- 0
4
5
11
34
65
111
175
260
505
540
671
what is the next line
5
11
34
65
111
175
260
505
540
671
what is the next line
Yet another number pattern
- Thread starter ArielGenesis
- Start date
Answers and Replies
- #2
- 239
- 0
really no one got an idea ?
- #3
- 1
- 0
Is the next number 802.
- #4
- 147
- 1
ArielGenesis, maybe you can give any hint?
- #5
- 239
- 0
my fault acually, but it is suppose to be
1
5
15
34
65
111
175
260
369
505
671
not much diffrent isn't it.
no neveza, it's not 802
the hint is that it involve listing a group of number and doing sum.
1
5
15
34
65
111
175
260
369
505
671
not much diffrent isn't it.
no neveza, it's not 802
the hint is that it involve listing a group of number and doing sum.
- #6
- 1,110
- 20
1 = 1
5 = 2 + 3
15 = 4 + 5 + 6
34 = 7 + 8 + 9 + 10
65 = 11 + 12 + 13 + 14 + 15
111 = 16 + 17 + 18 + 19 + 20 + 21
175 = 22 + 23 + 24 + 24 + 26 + 27 + 28
260 = 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36
369 = 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45
505 = 46 + 47 + 48 + 49 + 50 + 51 + 52 + 53 + 54 + 55
671 = 56 + 57 + 58 + 59 + 60 + 61 + 62 + 63 + 64 + 65 + 66
870 = 67 + 68 + 69 + 70 + 71 + 72 + 73 + 74 + 75 + 76 + 77 + 78
- #7
- 222
- 0
4
5
11
34
65
111
175
260
505
540
671
1381
Is that anywhere close? You really don't want to know how my messed up mind found that answer though. It might make you go crazy...
Oopss! You posted another one...
1
5
15
34
65
111
175
260
369
505
671
870
Looks like that should be the answer for this second set of numbers.
5
11
34
65
111
175
260
505
540
671
1381
Is that anywhere close? You really don't want to know how my messed up mind found that answer though. It might make you go crazy...
Oopss! You posted another one...
1
5
15
34
65
111
175
260
369
505
671
870
Looks like that should be the answer for this second set of numbers.
Last edited:
- #8
- 222
- 0
Wow! We got the answers a different way; but I'm sure that they equate.
I did it this way:
C = 6 9 12 15 18 21 24 27 30 ? [ Difference in #'s from B ]
B = 4 10 19 31 46 64 85 109 136 166 ? [ Difference in #'s from A]
A=1, 5, 15, 34, 65, 111, 175, 260, 369, 505, 671, ?
In C, the ? should, of course, be 33, which makes the ? in B = 199, which makes the ? in A = 870
I did it this way:
C = 6 9 12 15 18 21 24 27 30 ? [ Difference in #'s from B ]
B = 4 10 19 31 46 64 85 109 136 166 ? [ Difference in #'s from A]
A=1, 5, 15, 34, 65, 111, 175, 260, 369, 505, 671, ?
In C, the ? should, of course, be 33, which makes the ? in B = 199, which makes the ? in A = 870
- #9
- 1,110
- 20
Rahmuss, your approach is systematic and works in a large number of cases of this kind of puzzle.
- #10
- 239
- 0
sorry, it is my fault. and i yup, jimmy (as usual) got it right
can we form a formula based on rahmuss asnwer. as i also had not notice it. and how could it happen to be 3 if he is going to make D = 3,3,3,3,3,3,3 [diffrence in #'s from c]
can we form a formula based on rahmuss asnwer. as i also had not notice it. and how could it happen to be 3 if he is going to make D = 3,3,3,3,3,3,3 [diffrence in #'s from c]
- #11
- 1,110
- 20
One possible formula is this difference formula:ArielGenesis said:
[itex]A_n = 3 \times A_{n-1} - 3 \times A_{n-2} + A_{n - 3} + 3;[/itex]
Another would be a cubic polynomial in n, but I haven't figured out the coefficients yet.