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33931155016
3.
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
1353001
Thus X=7^3*((2^27-1)^-1) Mod (2^22-1)
or X=1353001*7^3 mod(2^22-1)
2706013
3(c)
119801956159
4(a)
a = 765355768, b = 76354890023, c = 863429
39334375343
(1, -11358092426, 113849703)
4(b)
1690200800304305868653681005913
4(c)
41811455202773941761/950017208401
(950017208401, 6489992021, -1243570020)
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3994608808930448
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400