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53

All numbers of this kind are divisible by $11$. Consider prime $\mathit{abcdefg}$ where each letter is a digit. The number $\mathit{gfedcbaabcdefg}$ is divisible by $11$ because the alternating sum of digits is always zero: $$g - f + e - d + c - b + a - a + b - c + d - e + f - g = 0.$$


44

As John's answer notes, every palindromic number with an even number of digits is divisible by $11$, because the alternating sum of its digits is zero (and thus a multiple of $11$). For example, for $81233218$, we have: $$8-1+2-3+3-2+1-8 = 0,$$ and so $81233218$ is divisible by $11$. The reason why this divisibility rule works is most easily seen using a ...


26

As has been pointed out in the comments, this is not a special property of the primes. Rather it's true that whenever you reverse a number and append the result to itself, the result is always divisible by 11. To see why this is true, consider the number $A = 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + 10^0 a_0$, where $a_0, \ldots, a_{n-1}$ are all ...


11

Hint $\ $ If in radix $b,\,$ we have $\,n = f(b) = f_0\! + f_1 b +\,\cdots+ f_n b^n,\,$ then reversing the digits of $\,n\,$ yields the integer $\,\bar n = \bar f(b)\,$ for $\,\bar f(b) = b^n f(b^{-1})$ the reversed polynomial. Appending yields $$ {\rm mod}\,\ b\!+\!1\!:\,\ b\equiv -1\,\Rightarrow\, b^{n+1} f(b) + b^n f(b^{-1})\,\equiv\, (-1)^n ...


10

The reasoning here is very simple. $2^n$ divides $2^m$ for $n,m$ positive integers if and only if $n \le m$ ; this is just because you multiply a bunch of $2$'s together, so if there are less multiples of $2$ on one side than on the other, then the side with less $2$'s divides the other side. Now $n=2^{2011} \le 2^{2012}=m$ shouldn't be hard to see. Hope ...


10

More specifically, $2^{2^{2012}}=2^{2^{2011}\times2}=(2^{2^{2011}})^{2}$


9

Every number obtained by this construction is divisible by $11$. Here is a simple test for divisibility by $11$: if the number is $a_r\cdots a_0$, it is divisible by $11$ iff $\sum_{i=0}^r (-1)^ia_i = 0$. This follows from the fact that $10^{2i+1}=-1\bmod{11}$ and $10^{2i} = 1\bmod{11}$. Thus, $a_r\cdots a_0 = \sum 10^ia_i = \sum (-1)^i a_i\bmod{11}$. Now, ...


7

${\rm mod}\ 7\!:\,\ \overbrace{55\cdots 55}^{1+3n\rm\,\ fives}\, =\, \dfrac{5(10^{1+3n}\!-1)}9\, \equiv\, \dfrac{-2\,(3^{1+3n}-1)}2 \,\equiv\, -3(\color{#c00}{3^3})^{n}\!+1 \equiv 4\ $ by $\ \color{#c00}{3^3\equiv -1},\ n$ odd


6

After noting $555555$ is divisible by $7$, note further that $555555\times 10^r$ is divisible by $7$ for any positive integer $r$. So you can cast out groups of six $5$s starting at the most significant digit, without changing the remainder on division by $7$. This gets rid of $996$ of the $5$s, leaving $5555$. Then $4949$ is obviously divisible by $7$ ...


6

you have it, but all you need is your observation that: $$ a^3-a=(a-1)a(a+1) $$ since exactly one of any three consecutive integers is divisible by 3 addendum PROOF (for Git Gud) let $a$ be the smallest number for which $3$ does not divide $a(a+1)(a+2)$ clearly $a \gt 0$ since $3 | 0$ write $k=(a-1)a(a+1)$, so $k$ is divisible by $3$ since $3$ doesn't ...


6

Note that $n^2$ is even if and only if $n$ is even. Since $m^2+n^2$ is even, you see that either both $m,n$ are even, or both are odd. If both $m$, $n$ are even, then $mn$ is divisible by $4$. If both $m$, $n$ are odd, you have $m=2k+1$, $n=2l+1$ so $m^2+n^2=4k^2+4k+4l^2+4l+2$ is not divisible by $4$. This gives the result.


5

Hint:- $6\equiv1 \pmod 5\implies 6^n\equiv1\pmod 5\tag{1}$ $-5(n-1)\equiv 0\pmod 5\tag{2}$ Solution:-


5

Directly: $$6^{n}-5n+4=\left(6-1\right)\left(6^{n-1}+6^{n-2}+\cdots+6^{1}+6^{0}\right)-5n+5=$$$$5\left(6^{n-1}+6^{n-2}+\cdots+6^{1}+6^{0}-n+1\right)$$


5

The part about $p,q$ being prime is a bit of a red herring. I did the same thing without primality required, got this with the larger number up to 46,000. Should not be too difficult to prove that every other Fibonacci number is involved, the ones that are also Markov numbers. http://en.wikipedia.org/wiki/Markov_number From the outcome and context, it is ...


5

True because $$2011 +1 = 2012 \Longrightarrow 2^{2012} = 2 \cdot 2^{2011} \Longrightarrow 2^{2^{2012}}= (2^{2^{2011}})^2$$


5

I've seen these same arguments in quite a few books, it all looks pretty standard issue to me. But what you seem to be missing is the motivation for all of this, the why do we care. That motivation is this famous fact: $$6 = 2 \times 3 = (1 - \sqrt{-5})(1 + \sqrt{-5}).$$ It's also true that $$6 = -2 \times -3 = (-1 - \sqrt{-5})(-1 + \sqrt{-5}).$$ We've got ...


5

You first find $2^2$, $5^2$ and $7^2$, $\mod 39$, to be $4$, $25$ and $10$ respectively. These add to $39$. Their squares are $16$, $1$, and $22$, which also add to $39$. This is power $2^2$. The squares of these are $22$, $1$, $16$, which is the same as before, Thus if it is true for $2^n$, it's true for $2^{n + 1}$. therefore $2^a + 5^a + 7^a$ is a ...


4

Yes, you are on the right track. All your reasoning makes sense to me. On your question about the associates By the properties of the norm, associates have the same norm. So the only possible associates in your list are $1 + \sqrt{-5}$ and $1 - \sqrt{-5}$. Now determine all units of $\mathbb{Z}[\sqrt{-5}]$ by finding all integer solutions of $N(a + ...


4

There are some great answers here, but If your doing mental math, this might be the easiest if you basic modular arithmetic properties. First look at the pattern 5, 55, 555, 555.... If you notice that this can be written as $a_1 = 5$ and $a_n = 10a_{n-1}+5$ This produces the pattern. Now if we perform the modulo we get $a_n \equiv 3a_{n-1}-2~\mod 7$. ...


4

Note that $111111$ is divisible by $7$. This follows either from $111111 = \frac{10^6-1}{9}$ where $10^6-1$ is divisible by $7$ by Fermat's little theorem, or from the factorization $111111 = 111 \cdot 1001 = 111 \cdot 7 \cdot 11 \cdot 13$. This implies that $555555$ is also divisible by $7$, hence $$ \underbrace{555 \cdots 5}_{k \text{ times } 5} \mod 7 $$ ...


4

(Nearly complete) hint: The only perfect squares $\mod 3$ are $0$ and $1$.


4

Hint: $\text{lcm}(5,7,9)=315$: $$315\times 1=315$$ $$315\times 2 =630$$ $$315\times 3 =945$$ $$315\times 4 = 1260$$ Well?


4

The crucial observation is that $\gcd(u,v)=\gcd(v,u)=\gcd(u,u\pm v)$. From this follows Euclid's algorithm for computing gcd. The significance of the gcd in your problem is that the crucial observation implies that all points visited by allowed moves have the same gcd. So, you can divide their coordinates by that gcd and reduce the problem to a grid with ...


4

Note $\let\phi\varphi$if $\phi=\frac{1+\sqrt5}2$ and $\psi=-\phi^{-1}$ are the roots of $x^2-x-1$, then $F_n=\frac1{\sqrt5}(\phi^n-\psi^n)$ (also for negative $n$). Observe ...


4

We are to prove that $x^5-x$ is divisible by 10. Also $x^5-x=x(x^4-1)=x(x-1)(x+1)(x^2+1)$, therefore it is the product of two consecutive numbers, $x(x-1)$, hence divisible by $2$.(recall product of $r$ consecutive numbers is divisible by $r!$) also, $n^p-n$ is divisible by a prime $p$ (Fermat's little theorem). Plug $p=5$ to show it is divisible by $5$. ...


4

Consider the numbers $6,10,15$. Their $\gcd$ is $1$, but the $\gcd$ of any pair is greater than $1$. More generally, we can produce a list $x_1,x_2,\dots,x_n$ of integers with $\gcd$ $1$ such that the $\gcd$ of any $n-1$ of them is greater than $1$. Let $p_1,p_2,\dots,p_n$ be distinct primes, and let $N$ be their product. Let $x_k=\frac{N}{p_k}$.


4

$7$ is the largest integer less than the square root of $50$.


4

Since $4\cdot(n+2)+(-1)\cdot(4n+1)=7$, we can say that either $\gcd(4n+1,n+2)=1$ or $\gcd(4n+1,n+2)=7$. Both cases can happen: for $n=1$ the gcd is $1$, for $n=5$ the gcd is $7$.


4

They are the squares so the answer is $\lfloor \sqrt{1000} \rfloor = 31$. To see this notice that the number of divisors of a number is the product of each exponent plus one, i.e. $n = \prod{p_i^{e_i}}$ and $\tau(n) = \prod{(e_i+1)}$. If $\tau(n)$ is odd then all $e_i$ are even which means $n$ is a square.



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