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General Forum: Education | | ******MATHEMATICS****** | |
| now urs turn....evarainaa question ivvandi....
Posted by: Raj Sekhar At: 13, Oct 2003 3:13:24 PM IST
correct giri gaaru....
Posted by: Raj Sekhar At: 13, Oct 2003 3:12:57 PM IST
okay thankyou raaj...
Posted by: Mrs. Aruna At: 13, Oct 2003 3:03:05 PM IST
series choosukunte kashtam lendi...
15^80 = (13 + 2)^80
expand chesthe anni 13 powers vasthaayi except the last term which is 2^80
2^80 = [ 2^2] [ 2^6^13 ] =4*[65 - 1]^13
which contains all powers of 65 except the last term (-1)^13
hence 15^80 mod 13 = 2^80 mod 13 = 4*(-1)^13 mod 13 =-4 mod 13 = 9
Posted by: Raj Sekhar At: 13, Oct 2003 2:52:42 PM IST
yes true....but how ante theleedhu...
Posted by: Mrs. Aruna At: 13, Oct 2003 2:47:27 PM IST
konchem explainc cheyyandee...mundhu question ans....
naaku2,4,8,3,6,12....ila series vachchindhi....
Posted by: Mrs. Aruna At: 13, Oct 2003 2:44:57 PM IST
next question :
(any odd square -1) is a multiple of "8"
Is it true? how?
Posted by: Raj Sekhar At: 13, Oct 2003 2:42:00 PM IST
right giri gaaru......
Posted by: Raj Sekhar At: 13, Oct 2003 2:27:49 PM IST
8??? guess cheyyadame theleeledhu raj....cheppandee ans..
Posted by: Mrs. Aruna At: 13, Oct 2003 2:05:08 PM IST
ok....
what is the remainder when 15^80 is divided by 13?
Posted by: Raj Sekhar At: 12, Oct 2003 2:12:39 PM IST
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