If log3N+log9Nlog3N+log9N is whole number, then how many numbers possible for N between 100 to 100?	Options:	A) 1, 9, 81	B) 1, 9, 89	C) 1, 9, 88	D) 1, 9, 22	E) 1, 9, 11
log3N+log9Nlog3N+log9N = log3N+log32Nlog3N+log32N = log3N+12log3Nlog3N+12log3N =32log3N32log3N   Now this value should be whole number.   Let 32log3N32log3N = w   ⇒log3N=23w⇒log3N=23w   N=3(23w)N=3(23w)   As N is a positive integer, So for w = 0, 3, 6 we get N = 1, 9, 81.   1, 9, 81nswer:1, 9, 81
<result>A</result>
If log3N+log9Nlog3N+log9N is whole number, then how many numbers possible for N between 100 to 100?