1. ## Modulo Stuff...

'=' means 'congruent to'

(Uses Chinese Remainder Theorem)

x=0(mod4) --> 3600*Y1=1(mod4)
x=1(mod9) --> 1600*Y2=1(mod9)
x=2(mod16) --> 900*Y3=1(mod16)
x=3(mod25) --> 576*Y4=1(mod25)

Knowing that M1=3600, M2=1600, M3=900, and M4=576

Is it possible to 'simplify' the congruences to the right? I know I couldn't... or is there a different approach to solving the system of congruences to the left?

Any pointers will do.
Thanks!

2. Originally Posted by AgentNXP
'=' means 'congruent to'

(Uses Chinese Remainder Theorem)

x=0(mod4) --> 3600*Y1=1(mod4)
x=1(mod9) --> 1600*Y2=1(mod9)
x=2(mod16) --> 900*Y3=1(mod16)
x=3(mod25) --> 576*Y4=1(mod25)

Knowing that M1=3600, M2=1600, M3=900, and M4=576

Is it possible to 'simplify' the congruences to the right? I know I couldn't... or is there a different approach to solving the system of congruences to the left?

Any pointers will do.
Thanks!
There is a problem because you can't have x=0(mod4) and x=2(mod16) in the same system.

3. Originally Posted by AgentNXP
'=' means 'congruent to'

(Uses Chinese Remainder Theorem)

x=0(mod4) --> 11025*Y1=1(mod4)
x=1(mod9) --> 4900*Y2=1(mod9)
x=2(mod25) --> 1764*Y3=1(mod25)
x=3(mod49) --> 900*Y4=1(mod49)

Knowing that M1=11025, M2=4900, M3=1764, and M4=900

Is it possible to 'simplify' the congruences to the right? I know I couldn't... or is there a different approach to solving the system of congruences to the left?

Any pointers will do.
Thanks!
Crap! Thanks for noticing that. Fixed above.

I'll try it again now--cheers!

4. x=0(mod4) --> 11025*Y1=1(mod4) --> Y1=1(mod4)
x=1(mod9) --> 4900*Y2=1(mod9) --> Y2=7(mod9)
x=2(mod25) --> 1764*Y3=1(mod25) --> Y3=9(mod25
x=3(mod49) --> 900*Y4=1(mod49) --> Y4=43(mod49)

Knowing that M1=11025, M2=4900, M3=1764, and M4=900

So after re-doing the problem...

I end up getting x=39053(mod 44100)

Now, what I am trying to get: Four consecutive numbers none of which is “square-free” (that is, each of which can be divided by a perfect square). It's not working so well for me.