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Mathematics 13 Online
OpenStudy (anonymous):

5,9,1,2,2,3,5,9,1,2,2,3 How would you find out what would the 620th number would be?

OpenStudy (amistre64):

by recognizing a pattern

OpenStudy (agreene):

write an equation to model it, then plug in 620 for your variable.

OpenStudy (amistre64):

lol ... this isnt equatable; its just see-able

OpenStudy (amistre64):

every 6th term you repeat the pattern

OpenStudy (agreene):

everything can be taylored. TAYLOR I SAY!

OpenStudy (amistre64):

so id say this is mod6 at least

OpenStudy (anonymous):

Is it 2?

OpenStudy (amistre64):

what is 620 divided by 6?

OpenStudy (anonymous):

you find the rule of the data set, and as x you plug in 620, so it would be y=mx+b, y is what you look for, m is what they change by each time, x is the number in the sequence, and b is the y intercept (when x=0)

OpenStudy (amistre64):

and the remainder is important here

OpenStudy (anonymous):

so it's 1 then?

OpenStudy (amistre64):

it looks to be 1 yes

OpenStudy (anonymous):

So the remainder is all it is?

OpenStudy (amistre64):

err... let me chk that first

OpenStudy (amistre64):

6|620 13R2 right?

OpenStudy (anonymous):

bah... now i have to work it :P, lol 1 sec leme check

OpenStudy (amistre64):

forgot how to divide lol

OpenStudy (anonymous):

its 9, right?

OpenStudy (anonymous):

its 9

OpenStudy (anonymous):

I'm so confused now :p

OpenStudy (amistre64):

103 ---- 6|620 6 ---- 20 18 ----- 2 we have a remainder of 2

OpenStudy (amistre64):

so its the 2nd term

OpenStudy (amistre64):

5,9,1,2,2,3, 5,9,1,2,2,3, 5,9,1,2,2,3, 5,9,1,2,2,3, .... mod6

OpenStudy (anonymous):

therefore the second number in the data set

OpenStudy (amistre64):

we have to start at 0 with the mods tho

OpenStudy (amistre64):

so its the 012 term

OpenStudy (amistre64):

id say 1 :)

OpenStudy (anonymous):

ah I see ;)

OpenStudy (anonymous):

With using simple logic you know that any number divisible by 6 is the sixth term. So the 618th term would be the same as the 6th term, then the 619th would be the same as the 1st and then the 620th would be the same as the 2nd term, which is 9.

OpenStudy (amistre64):

1,2, 3, 4, 5, 6 7,8,9,10,11,12 6/12 = 2 R 0, is the last term hmmm 6/8 = 1 R 2 is the second term

OpenStudy (anonymous):

620th term, 6 numbers to restart (5,9,1,2,2,3,) divide 620 by 6 as many even times as possible, so 1st 100, then 3 (complete sets) now you can take out 618 (6 times 103) from 620, you get 2 then you go back to the set of numbers and pick the second in the line which is 9 9 is your answer

OpenStudy (amistre64):

right, the answer is the 2nd term, a 9. final offer lol

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