Question

what is the largest natural number that can divides the numbers 1723, 2010, and 5741 and each remainder is 1 ?

Answer 1

write a computer program for that one

Answer 2

@ganeshie8 , @waterineyes
any idea ?
i need a way

Answer 3

factor

Answer 4

u wanna find a natural number that divides all with remainder 1?

Answer 5

yes divides that numbers and always gives remains 1 respectively

Answer 6

ok well

Answer 7

then just look at umm

Answer 8

1724, 2011, and 5742

Answer 9

factor these out and see largest divisor

Answer 10

wait 2011 is prime

Answer 11

that wud mean ur answer is 1... that cant be it lemme see

Answer 12

i was represent all number as :
1723 = nk + 1
2010 = np + 1
5741 = nr + 1
with n is divisor, and how to get the largest value of n ?

Answer 13

oh i know lower by 1?

Answer 14

gcd(1722, 2009, and 5740)?

Answer 15

287?

Answer 16

im not sure --"

Answer 17

ok look

Answer 18

umm

Answer 19

ifff

Answer 20

g=GCD of 1723,2010,5471, which leaves rem 1 then

1723/r = k+1/r
1723= rk+1
1723-1 =rk
therefore r must divide 1722 evenly

Answer 21

similarily r will divide all the other numbers -1 evenly

Answer 22

so we simply need to find the GCD of the numbers shifter down by 1

Answer 23

shifted*

Answer 24

GCD(1722, 2009, and 5740)= 287

Answer 25

ok?

Answer 26

where it says g=GCD, it shud say r=GCD

Answer 27

yes, it is 287.i checked it by calculator...
i think i can undestand now, just by using find the GCD
thnks, @dan815

Answer 28

how you can be fast to get GCD(1722, 2009, and 5740)= 287 ??
by using a program comptr ?

Answer 29

umm

Answer 30

ya its not that hard to write a factoring program though especially for such small numbers

Answer 31

sieve of erathonesis

Answer 32

says there mut be a factor that is between 2 to sqrt(1722)

Answer 33

so u check primes between those numbers that divide one u divide then u are checking for 2 to floor sqrt(1722/p1)

Answer 34

primes between there

Answer 35

its pretty fast factoring system

Answer 36

or go to wolfram and type
GCD(n1,n2,n3) :P

Answer 37

or just factor like a good boy

Answer 38

oh, ok... thanks again :)

yea, i checked on wolfram:
http://www.wolframalpha.com/input/?i=GCD+of+%281722%2C+2009%2C+and+5740%29&dataset=&equal=Submit

i dont like do that like a good boy ;P

Answer 39

1722=2*861
=2*3*287

Answer 40

8+6+1 = 15 therefore 3 must factor

Answer 41

if have
n1*100+n2*10+n3*1
then
(n1*100+n2*10+n3*1)/3 = n1*(33+1/3)+n2(3+1/3)+n3/3
=33n1+3n2+n3/3+n1/3+n2/3
=33n1+3n2+ (n1+n2+n3)/3
therefore as long as n1+n2+n3 add upto a multiple of 3

Answer 42

u can see how this same pattern follows infinite digits

Answer 43

1723 = nk + 1
2010 = np + 1
5741 = nr + 1

n | (1723-1)
n | (2010-1)
n | (5741-1)

Answer 44

yes, i caught that @ganeshie8 . thanks :)

Answer 45

euclid algorithm is superior to prime factorization for finding gcd

Answer 46

what is that algoritm

Answer 47

2009 = 1722*1 + 287
1722 = 287*6 + 0

so gcd(2009,1722) = 287

Answer 48

:O coool

Answer 49

since 287 | 5743, we are done.