Find the sum of all positive integers n, each of which has precisely 16 positive divisors and a successor n+1 that has precisely 3 divisors.
Only 120 and 168 satisfy all the given conditions as:
120 = (2^3)*3*5 -> # divisors = 4*2*2=16
121 has precisely 3 divosors as; 1,11, and 121
168= (2^3)*3*7 -> # 4*2*2= 16, and:
169 has precisely 3 divisors as 1,13, and 169
Consequently, the required sum= 120+168=288
*** Interesting puzzle. I will certainly try to post an independent analytic solution of my own.
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