Question

Find the number of identical terms in the following two series:
a) 3,7,11,15,...,367
b) 2,9,16,23,...,709

Answer 1

find the intersection of 2 lines ...

Answer 2

y = 3+4n
y = 2+7n

Answer 3

what's the answer?

Answer 4

thats not a very good question ...

Answer 5

we are not a "whats the answer" website

Answer 6

i mean that kindly give me a solution

Answer 7

i have given you information which you should be capable of working out .. if you have questions about what ive offered, then as for clarification .. not for "the answer". We can work towards the solution, but I need you to give me something to work with.

Answer 8

the common terms will represent the intersection of the respective rules of each .. and they are not series, they are sequences.

a series is a sum of a sequence, whereas a sequence is a list of terms seperated by commas

Answer 9

thanks...is the answer 102 if i did it right?

Answer 10

no.. is the answer 13

Answer 11

hmm, i have to admit that they will not have the same nth term , so my original idea, no matter how spectaular ...is flawed lol

Answer 12

hey... i found that it's an A.P. and then i got answer 13

Answer 13

AP is correct, 13 may be one value they have in common, but is it the only value they share?

Answer 14

other value i got is 102...but it's not possible...as sequence b has 102 terms

Answer 15

i cant get a 13, canyou show me how you got to that?

Answer 16

yeah

Answer 17

a) 3,7,11,15,...,367
^^
13 is missing


b) 2,9,16,23,...,709
^^
13 is missing

Answer 18

by trial and error, i get 51 to be a common term between them

Answer 19

sequence 'a' has 92 terms and 'b' has 102
let pth term of a =qth term of b
3+(p-1)4=2+(q-1)7
4p-1=7q-5
4(p+1)=7q
(p+1)/7=q/4=k (say)
so p=7k-1 and q=4k
now p<=92 and q<=102
so k<=93/7 and k<=51/2
so k<=13+2/7
so k=13

Answer 20

can there be a quicker way to solve this?

Answer 21

that looks similar to what i just developed on the paper in front if me :)

let a = 3+4k , for some integer k
let b = 2+7n , for some integer n

when a=b we have:

3+4k = 2+7n or written another way

4n - 7k = -1

Answer 22

we can either view this as a euler number theory problem, or simply as a line that we can plot and compare integer solutions with

Answer 23

if we do a euler run on it; we need to find the gcd of 4 and 7 ...

7 = 4(1) + 3
4 = 3(1) + 1 <-- well, the gcd is already 1 so we dont have to modify it any

Answer 24

maybe modify it by a negative :) does this look like the content of your material? or shold we approach it another way?

Answer 25

phew... finally got my internet working

Answer 26

if we use a line approach:

4n - 7k = -1 , we can define n in terms of k

n = (7k-1)/4

k=0, n=-1/4 not an integer
k=1, n=6/4 not an integer
k=2, n=13/4 not an integer
k=3, n=20/4 = 5, thats good
k=4,5,6, n is not an integer
k=7, n=48/4 = 12 , thats good
seems to me that every 4th k produces an integer result

so when k = 3+4t, for t=0,1,2,3,4,....

n = (7(3+4t)-1)/4

n = (21+28t -1)/4

n = (20+28t)/4

n = 5+7t

Answer 27

the question is now, how many ts can we fit into 91 and 101 terms?

Answer 28

might be a simpler approach

3+4k = 367 ; k = 91

91 /4 = 22.75 so im thinking there are at best 22 or 23 terms they have in common

Answer 29

im really hoping some if this makes sense :)

Answer 30

any way to check it?

Answer 31

make a tables in excel and compare it side by side

Answer 32

maybe i'll try a java program

Answer 33

there are 25 as is, but i havent accounted for the limit of n and k in the number of terms

Answer 34

k can at best go to 101 ... n=55, k=96 is the best we can accomplish with this, and that adds up to 14 terms for me

Answer 35

maybe taking it simply as an AP gives 13, and maybe that's fine

Answer 36

i think i have my n and k bacwards ... let me rechk

Answer 37

yeah, i had them backwards :/ k on the left, n on the right,

i got to 13 eventually:)

Answer 38

so 13 is alright
THANKS amistre64

Answer 39

yep, let me see if i can make that word out a little better :) condense the beast!

let a = 3+4k , for some k = 0,1,2,3, ... , 3+4k = 367 ; 91
let b = 2+7n , for some n = 0,1,2,3,..., 2+7n = 709 ; 101

we can work it up to a setup for some t = 0,1,2,3,4,...

k = 3+4t
91 = 3+4t
22 = t

n = 5+7t
101 = 5+7t
13.714 = t

notice that this is only valid for the smaller value of t, so when t = 13

Answer 40

thanks

Answer 41

good luck :)