Question

If k is an interger and numbers 1+k, 1+2k and 1+8k are divisible by 3,5 and 7 respectively, then the smallest positive value of k lies in the interval
a) 0 to 20
b) 21 to 40
c) 41 to 60
d) 61 to 80
e) none of these

Answer 1

I think you can use the Chinese Remainder Theorem here. See here for a good tutorial on this: http://www.youtube.com/watch?v=3PkxN_r9up8

Answer 2

I have worked out the answer. Let me know what answer you get and I'll tell you if it matches the one I found using this theorem.

Answer 3

i think i get the chinese remainder theorem but i am not sure how to apply it in this problem

Answer 4

ok, lets take each part that you have been given.
1+k is divisible by 3 means:\[1+k=0\text{ mod }3\]\[\therefore k=-1\text{ mod }3=2\text{ mod }3\]similarly 1+2k is divisible by 5 means:\[1+2k=0\text{ mod }5\]\[\therefore 2k=-1\text{ mod }5=4\text{ mod }5\]\[\therefore k=2\text{ mod }5\]and lastly, 1+8k is divisible by 7 means:\[1+8k=0\text{ mod }7\]\[\therefore 8k=-1\text{ mod }7=6\text{ mod }7\]\[\therefore 4k=3\text{ mod }7\]\[\therefore -3k=3\text{ mod }7\]\[\therefore l=-1\text{ mod }7=6\text{ mod }7\]
so we end up with:\[k=2\text{ mod }3\]\[k=2\text{ mod }5\]\[k=6\text{ mod }7\]Now just use the theorem to find the general solution for k.

Answer 5

i got 452 for k

Answer 6

you need to get the general solution, something like: k = a + nb
where a and b are integers and 'n' is a variable multiplier that can range from \(-\infty\) to \(+\infty\). you can then use that formula to work out the minimum positive value for k.

Answer 7

452 is not one of the valid values for k.

Answer 8

k = 260 +/- n105

Answer 9

the n105 is correct but the 260 is wrong.

Answer 10

I need to leave for a wedding now - but if you write out the steps you took here then when I get back I can check them to see where you made a mistake.

Answer 11

o i find my mistake
now i get k = 272 +/- 105 and the smallest value is 62

Answer 12

thanks asnaseer for that info - interesting stuff

Answer 13

yociyoci - that is correct - well done for pursuing this.
jimmyrep - yw

Answer 14

Thank you very much for showing this new theorem to me and showing me the steps to this problem. Thank you!!!!

Answer 15

yw

Answer 16

asnaseer, i have a problem posted earlier, and it was left unfinish, can you help me out? Thanks.
http://openstudy.com/study#/updates/4f3bcab2e4b0fc0c1a0f1d10

Answer 17

I haven't seen a problem like that one before - let me think over it - I'll get back to you.

Answer 18

okay, thank you.