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 Divisible? Yes and No. (Posted on 2011-08-15)
Prove that 3^105+4^105 is divisible by 13*49*181*379 but is not divisible by either 5 or 11.

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If a and b are odd integers with ab = 105, and if  3a + 4a = 0 (mod m), then:

3a = - 4a = (-4)a  (mod m).          Therefore:         3ab = (-4)ab = - 4ab

giving : 3105 + 4105 = 0    (mod m).     So  3105 + 4105 is divisible by m.

Using this result in each case:

33 + 43 = 27 + 64 = 91  = 0 (mod 13),     so 13 divides  3105 + 4105

37 + 47 = 2187 + 16384 = 18571 = 0 (mod 49) so 49 divides  3105 + 4105

35 + 45 = 243 + 1024 = 1267 = 0 (mod 181) so 181 divides  3105 + 4105

37 + 47 = 2187 + 16384 = 18571 = 0 (mod 379) so 379 divides  3105 + 4105

It follows that  3105 + 4105  is divisible by 13*49*181*379, since 13, 49, 181

and 379 have no common factors,

Now on a different tack:  3105 = (3)(34)26 = 3*8126 = 3*126 = 3 (mod 5)

and:                              4105 = (4)(42)52 = 4*1652 = 4*152 = 4 (mod 5)

So  3105 + 4105 = 2 (mod 5) and therefore is not divisible by 5.

Also:                             3105 = (35)21 = 24321 = 121 = 1 (mod 11)

and:                              4105 = (45)21 = 102421 = 121 = 1 (mod 11)

So  3105 + 4105 = 2 (mod 11) and therefore is not divisible by 11.

 Posted by Harry on 2011-08-16 23:11:17

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