Contact
CoCalc Logo Icon
StoreFeaturesDocsShareSupport News AboutSign UpSign In
| Download

All published worksheets from http://sagenb.org

Views: 168733
Image: ubuntu2004

1.

mod(2^-1,67862310031)
33931155016

3.

mod(122^-1,343)
194

3(b)

(2^27-1)X=7^3mod(2^22-1)

can be split as product of two congruences,

(2^27-1)X=1mod(2^22-1)

1=7^3mod(2^22-1)

For Solvingfirst congruence,

(2^27-1)X=1mod(2^22-1)

or X=(2^27-1)^-1mod(2^22-1)

we first find inverse of 2^27-1

mod((2^27-1)^-1,2^22-1)
1353001

Thus X=7^3*((2^27-1)^-1) Mod (2^22-1)

or X=1353001*7^3 mod(2^22-1)

mod(1353001*7^3,2^22-1)
2706013

3(c)

mod(193707721^-1,761838257287)
119801956159

4(a)

a = 765355768, b = 76354890023, c = 863429

a = 765355768; b = 76354890023; c = 863429; GCD_1=xgcd(a,b); X_0= GCD_1[1]*c/gcd(a,b); X=X_0 k=1; while X<=0 : X=X_0+b*k/gcd(a,b) k=k+1; print X
39334375343
a = 765355768; b = 76354890023; c = 863429; xgcd(a,b)
(1, -11358092426, 113849703)

4(b)

a = 2^100-1; b = 2^102-1; c = 6442450941; GCD_1=xgcd(a,b); X_0= GCD_1[1]*c/gcd(a,b); X=X_0 k=1; while X<=0 : X=X_0+b*k/gcd(a,b) k=k+1; print X
1690200800304305868653681005913

4(c)

a = 3014774729910783238001; b = 15733624667337520130581; GCD_1=xgcd(a,b);
a = 3014774729910783238001; b = 15733624667337520130581; GCD_1=xgcd(a,b); X_0= GCD_1[1]*c/gcd(a,b); X=X_0 k=1; while X<=0 : X=X_0+b*k/gcd(a,b) k=k+1; print X
41811455202773941761/950017208401
a = 3014774729910783238001; b = 15733624667337520130581; xgcd(a,b);
(950017208401, 6489992021, -1243570020)

7.

41*41*40 mod(72549625,67240) mod(3^64905,41^3) 101^2*100 mod(72549625,1020100) 251*250 mod(72549625,62750) mod(3^122525,101^3) mod(3^10625,251^2) CRT_list([12549,796294,25854],[41^3,101^3,251^2])
3994608808930448

8.

euler_phi(1000)
400