Question

Find a number which leaves remainders 2,3 and 2 when divided by 3,5 and 7 respectively.

Answer 1

15/2

Answer 2

Um...:-)

Answer 3

is it a correct answer?

Answer 4

After finding such a number , find the set of all such numbers..

Answer 5

is it a correct answer?

Not an integer..

Answer 6

the number would be an integer?

Answer 7

Number theory type question....

Answer 8

good, i always knew e-studier, u r 1000 times more intelligent than me

Answer 9

U looking for medals..:-)

Answer 10

no, this is the fact

Answer 11

Good, and the set?

Answer 12

e-studier, do you know any proof for the prime numbers greater than 10 to be of the form 6n+1? can you help me?

Answer 13

Yes, I will post it in a bit..

Answer 14

sorry,\[6n \pm1\]

Answer 15

105*n-82

Answer 16

Any integer must be 6m,6m+1,6m+2,6m+3,6m+4,6m+5.

Three are even and one is divisible by 3.

So, excepting 2 and 3, any prime must be 6m +1 or 6m+5 (this last is the same as the "next" m minus 1).

Answer 17

105*n-82

or 105n +23

Good

Care to tell everyone how u did it?

Answer 18

trial and error . All that i knew was the set would be related to LCM of 3,5,7 and i got first number to be equal to 23 so just made a relation bw 105 and 23

Answer 19

Fair enough, u can solve more "mathematically" with the assistance of linear congruences.

x = 2 mod 3, x = 3 mod 5, x=2 mod 7 solve simultaneously.

Answer 20

ah chinese remainder theorem
there is an algorithm of sorts

Answer 21

Yup, although u don't really need it for this one...

Answer 22

oh well i am going to use it anyway to refresh my memory with my morning coffee because i don't really remember it

Answer 23

This one u can say to satisfy x=2 m3 x is of form 3r+2 and to satisfy x=3 m5 (3r = 1 m5)

->r=2 m5 (r = 5s +2) ->

x= 3r+2 = 3(5s+2) +2 = 15s +8 which has to satisfy the last one

15s+8 = 2 m7 -> s+1 = s+1=2 m7 so s=1 m7 (s=7t+1 ->

x = 15(7t+1) + 8 = 105t+23 (x=23 m 105)

Answer 24

can't you always use something like this?

Answer 25

i just got one number using an algorithm and got 316, now i can subtract off 105 a couple times to get smaller ones. i wonder if my arithmetic is correct

Answer 26

maybe i messed up dang

Answer 27

Well, in principle, yes, the algorithmic way just keeps it a bit more organized.

Answer 28

clearly i messed up my arithmetic somewhere. damn. i am going to try again and see where i messed up