Question

Find the smallest integer K which when divided by 6, 5, 4, 3, 2 successively leaves remainders 5, 4, 3, 2 and 1 respectively.

Answer 1

@ajprincess

Answer 2

chinese remainders can take awhile :)

Answer 3

R u familiar with chinese theorem? @goformit100

Answer 4

No Madam I am not familiar with it.

Answer 5

i recall it as\[\sum aMy\]

Answer 6

http://mathworld.wolfram.com/ChineseRemainderTheorem.html.
Pls take a look at this and see if u can understand it:)

Answer 7

does the CRT find the smallest? or just "a" number?

Answer 8

ok let me check it out.

Answer 9

it depends on the values of \[\bar n_i\] we find. @amistre64

Answer 10

I did the question as


LCM of 6, 5, 4, 3, 2, 1 = 60
So Number is 60 – 1 = 59


is it correct ?

Answer 11

I got 59 as the required K.

Answer 12

the CRT a=remainder, M=product of divisors, divided by the required divisor; and y is a little complicated, My = 1 (mod divisor)

5 6!/6 ; y 6!/6 = 1 (mod 6)
+ 4 6!/5 ; y 6!/5 = 1 (mod 5)
+ 3 6!/4 ; y 6!/4 = 1 (mod 4)
+ 2 6!/3 ; y 6!/3 = 1 (mod 3)
+ 1 6!/2 ; y 6!/2 = 1 (mod 2)

Answer 13

if 59 fits the bill, then it would be simple enough to chk

Answer 14

Thank you Sir and Madam.

Answer 15

59 = 60 +-1
59 = 60 +-2

these remainders dont seem to be +1, +2, +3, ...

Answer 16

but i might have done that a little off

Answer 17

59/6 = .9 + 5
59/5 = .11 + 4
59/4 = .14 + 3

it looks promising

Answer 18

Thankyou Sir