Hot answers tagged

10

Other Way $$10^k-1=(10-1)(10^{k-1}+10^{k-2}+\cdots+1)$$


5

Define the sequence given by $b_1=1,b_2=1$ and $b_{n+2}=3b_{n+1}-b_{n}$ for $n>0$. To prove that $a_n=b_n^2$ we just have to show: $b_{n+2}^2=7b_{n+1}^2-b_{n}^2-2$. Of course $b_{n+2}^2=9b_{n+1}^2-6b_{n+1}b_n+b_{n-1}^2$ So to finish we just need to prove $3b_{n+1}b_n=b_{n+1}^2+b_n^2+1$. Notice $b_{n+1}^2+b_n^2+1=b_{n+1}(3b_n-b_{n-1})+b_n^2+1=3b_{n+1}...


3

Your formula is essentially a computer program which iterates all integers in the range $[2,x−1]$. One can easily establish such formula, for example: $$F_n=\left\lfloor\frac{\left(\sum\limits_{k=2}^{n-1}\left\lceil\frac{{n}\bmod{k}}{n}\right\rceil\right)+2\cdot\left\lceil\frac{n-1}{n}\right\rceil}{n}\right\rfloor$$ The real challenge is to establish a ...


3

It's just that $10$ is $1$ modulo $9$, and modulo being compatible with multiplication. So $10^k \equiv 1^k \equiv 1 \mod 9$. More generally this shows that $(a+1)^k \equiv 1 \mod a$. Indeed, writing it this way suggests another proof: Expanding $(a+1)^k$ using the binomial theorem one see it is of the form $1+ a N$ for some integer $N$.


3

A brute-force approach $$a_{n} = 3a_{n-1} - a_{n-2}$$ Squaring on both sides, $$a_n^2 = 9a_{n-1}^2+a_{n-2}^2 -6a_{n-1}a_{n-2} $$ Similarly, $$a_{n-1}^2 = 9a_{n-2}^2+a_{n-3}^2 -6a_{n-2}a_{n-3} $$ Subtracting, $$a_n^2 - a_{n-1}^2 = 9a_{n-1}^2 - 8a_{n-2}^2 - a_{n-3}^2 -6a_{n-2}(a_{n-1}-a_{n-1}) $$ Since, $$3a_{n-2} = a_{n-1} + a_{n-3}$$ $$a_n^2 - a_{n-1}^2 = ...


2

I understand the thrust of your question. Today there is a huge body of knowledge in the general category of mathematics. But the bare bones of it were being worked out maybe even as far back as 10,000 years or even further back than that. So, if you think about it, some of the brightest minds of billions of humans have been hard at work for well over 10,...


2

Since you've tagged modular arithmetic: $$\begin{align*}10 \equiv 1 \pmod{9} \implies 10^k &\equiv 1^k \pmod{9} \\ 10^k &\equiv 1 \pmod{9} \\ 10^k -1 &\equiv 0\pmod{9}\end{align*}$$


2

For first, we may translate our sequence in order to have some sequence fulfilling $b_{n+2}=7b_{n+1}-b_n$. For such a purpose, it is enough to set $a_n=b_{n}+\frac{2}{5}$, leading to $b_0=b_1=\frac{3}{5}$. Now the characteristic polynomial of the sequence $\{b_n\}_{n\geq 0}$ is $$ p(x)=x^2-7x+1 $$ with roots given by $\frac{7\pm 3\sqrt{5}}{2}$, so the ...


2

Now I believe the correct answer is indeed of the order $\frac{n}{\ln n}$. I'll be using different estimates here, and I don't think the initially requested bound $S(n) < \frac{n+1}{2}$ falls out directly, so I'm posting this as a separate answer. Throughout the answer below, $p$ and $q$ will always be primes, so I will avoid writing this condition in ...


2

Just to provide more than a handful of data. 1) This is the code I used in Pari/GP : list=vectorv(1000);li=0; {forprime(p=2,11333, \\ 11333 is just a "idle-typing" value f=factor(p-1); for(k=1,rows(f), q=f[k,1]; if((q^q-1) % p == 0 , li++;list[li]=[p,q,(p-1)/q ] )) );} list=Mat(VE(list,li)) ...


2

An idea too long for a comment. Recall the Fibonacci sequence: $$1, 1, 2, 3, 5, 8, 13, \ldots$$ You have here the sequence: $$1^2, 2^2, 5^2, 13^2, \ldots$$ So: It looks to be the odd entries of the Fibonacci sequence, squared. Moreover, recall that the odd Fibonacci numbers $f(n)$ are generated by the recurrence relation: $$f(0) = f(1) = 1 \text{ } \...


1

Let us take the formula apart slowly. First we have a product over all the possible divisors of $x$. The product will be zero if just one factor is zero, so claiming that $H(x)$ is 0 only if $x$ is not prime, amounts to claiming that if $x$ is composite, one of the factors will be zero - my guess is that we'll get a zero if $i$ is a factor of $x$. Then let ...


1

I'd love to just add it as a comment, but 50400 (108 divisors), 55400 (112 divisors), 60480 (120 divisors), 65520 (112 divisors), and 69300 (108 divisors) suit the numbers you are looking for.


1

It's just a hand heuristic which is possible because the limits of the puzzle ... Having the 7 first primes power 1 is enough to get $2^7 > 100$ factors Their product is too high : $2.3.5.7.11.13.17 = 510510$ which is $> 70000$ Let's replace 17 by 16 = $2^4$ : $2^4 .3.5.7.11.13 = 240240$ // too high idem with 13 by 8 = $2^3$ : $2^7 .3.5.7.11 = ...


1

When a question involves specific number (in this case $10$) one has to use special property of that number. The special property of 10 relevant to our interest here is that it is one more than 9. Now we can use binomial theorem to expand $10^k =(9+1)^k$. In the binomial expansion on RHS except the final term $1^k$ all other terms are multiples of powers ...


1

By induction: When $k = 1$, $10^k - 1 = 9$ is divisible by $9$. Now suppose $k \geq 2$, and $10^{k-1}-1$ is divisible by $9$. By induction, we can write $10^{k-1} - 1 = 9a$ for some integer $a$. Then $$10^k - 10 = 90a$$ Add $9$ to both sides: $$10^k - 1 = 90a + 9 = 9(10a+1)$$ so $10^k - 1$ is divisible by $9$.


1

As $\frac{1}{n}>0$ for all natural $n$, all of $a,b,c$ have to be at least two. Let $a\leq b\leq c$. Unless $a=b=c=3$, $a=2$. Then we have $\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$, so similarly either $b=c=4$ or $b=3$. In the latter case, we get $c=6$. So, $(a,b,c)=(3,3,3),(2,4,4),(2,3,6)$ are all the solutions.


1

Hint Without loss of generality, let $a<b<c$ and set $t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. If $a=1$ then $t>1$ If $a\ge 3$ then $t<1$ Thus $a=2$


1

The key properties of xor are the following: If we have non-negative integers $a_1,a_2\dots a_n$ and we change exactly one of them then the xor always changes If we have non-negative integers $a_1,a_2\dots a_n$ and their xor is not $0$ then we can reduce exactly one of them, so that the xor of all of them is $0$. Try to prove these two properties. ...


1

The question seems to be: Why is $\underbrace{ 99\dots 99}_n$ a multiple of $9$? It is clear, since it is $\underbrace{ 11\dots 11}_n\times 9$



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