Seemingly Difficult but too easy - Part 4!

First, note that a fraction $\frac{m}{k}$ with coprime positive integers $m$ and $k$ has a terminating decimal representation if and only if $k=2^a5^b$ . Otherwise, there would exist no $c$ , such that $\frac{m}{k}\cdot 10^c$ is an integer.

Then, rewrite $N=\frac{2n+1}{n(n-1)}=\frac{2n+(n-n)+1}{n(n-1)}=\frac{3}{n-1} - \frac{1}{n}.$ Both $\frac{3}{n-1}$ and $\frac{1}{n}$ have to terminate in order for $N$ to be terminating. If $n$ would have a factor that is not a multiple of $2$ or $5$ , then the second term would not terminate. If $n$ would be a multiple of $2$ and $5$ , $n-1$ would be coprime to $2$ and $5$ and the first term would not terminate. Thus $n=2^k$ or $5^k$ such that (for some $l$ ) either $2^k-1=5^l,\ \ \ 5^k-1= 2^l,\ \ \ 2^k-1= 3\cdot 5^l\ \ or \ \ 5^k-1= 3\cdot 2^l.$

Case 1: $n=2^k$ and $n-1=5^l \Leftrightarrow 2^k=5^l+1$ . Since $5^l+1\equiv 1^l+1 \equiv 2 \pmod 4$ , the only solution to this equation is $k=1, l=0$ and therefore $n=2$ .

Case 2: $n=5^k$ and $n-1= 2^l$ . Assume $k=2b$ . Then $2^l=5^{2b}-1=(5^b-1)(5^b+1)$ . We have already shown that $5^b+1=2^m$ can only hold true for $b=0$ . Now assume $k=2b-1$ . Note that $\sum_{i=0}^d a^i=\frac{a^{d+1}-1}{a-1}$ . Using this, consider $2^l=5^{2b-1}-1=4(\sum_{i=0}^{2b-2} 5^i).$ This only holds true if both factors are a power of $2$ . Since the sum is always odd, the only possible value for the second factor is $1$ and thus $b=k=1$ and therefore $n=5$ .

Case 3: $n=2^k$ and $n-1= 3\cdot 5^l$ . Note that, for $l>0$ , this expression's last digit is $5$ . Therefore $k=4b$ and $3\cdot 5^l=2^{4b}-1=(2^{2b}-1)(2^{2b}+1),$ which we have shown in case 1 and case 2 to only be true for $b=1, k=4$ and therefore $n=16$ . For $l=0$ we find $2^2-1= 3\cdot 5^0$ and $n=4$

Case 4: $n=5^k$ and $n-1=3\cdot 2^l$ . This can be proven almost identically to case 2. It yields $k=2$ and therefore $n=25$ .

Summing all of them together makes $2+4+5+16+25=\boxed{52}$ .

Note: These are special cases of Catalan's conjecture .

2n+1/n(n-1) = (-1/n)+(3/n-1) For it to be terminating decimal both the terms must be terminating , For 1<n<=10 (n not equal to 1 as it will make denominator zero) 1/n is terminating only for n=2,4,5, 8 and 10 but 3/n-1 is terminating for n=2,4,5 only. Now for n>10 , for 1/n to be terminating it has to some multiple of either of 1/2 , 1/4 and 1/5 (we reject 1/8 and 1/10 as it will make 3/n-1 non terminating) Now only for 1/n = (1/4)^2 and 1/n = (1/5)^2 , 1/n and 3/n-1 are simultaneously terminating . hence all possible values of n are 2,4,5,16 and 25 . Therefore answer is 52.

Let the number n be of the form 2^a 3^b 5^c..... so, basically, the number n would have prime factors other than 2 and 5... but if that were the case, 2n+1 would not have other prime factors and the fraction would be non terminating... so, n has to be of the form 2^a 5^b... so, given this, 1/n would be a terminating decimal... let's consider (2n+1)/(n-1)... this can be re-written as 2+3/(n-1)... so, 2^a 5^b-1 should be of the form 2^l 5^m 3 or 2^l 5^m... if there were higher powers of 3 in it, it would be non-terminating... but, 2^a 5^b-1 can be divisible by either some power of 2 or some power 5 but not both at the same time... hence n is of the form 2^a or 5^b... hence two possibilities here: 1. 2^a-1 = 5^k or 3 5^k for the above equation, other than a = 1,2, for any other solution, clearly, a has to be a multiple of 4... hence 2^a-1 can be re-written as some 16^l-1 which can be factored as (2^l-1) (2^l+1)(2^(2l)+1) and would clearly have prime factors other than 3 & 5, except when a = 4 2. 5^b-1 = 2^p or 3 2^p 5^b-1 = (5-1) (5^(b-1)+5^(b-2)+.....+5+1) so, clearly, the second part has to have 2 and 3 (either 3^0 or 3^1) as its only factors... but the second part of the product ends in 31, 06, 156, 656, 81... clearly, these would require prime factors other than 2 and 3 (or probably more powers of 3) except when b = 1, 2... so, the only n's satisfying the above condition are 2, 4, 5, 16, 25 and hence the sum is 52...