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30

Divisibility rules generally rely on the remainders of the weights of digits having a certain regularity. The standard method for divisibility by $3$ in the decimal system works because the weights of all digits have remainder $1$ modulo $3$. The same is true for $9$. For $11$, things are only slightly more complicated: Since odd digits have remainder $1$ ...


17

When reading roman numerals, I prefer to think in the following way: Read from left to right, and if at any point the value of a character decreases, put a comma between the decrease. Then, add each block together. MCMXCVI $\mapsto$ M,CM,XC,V,I $\mapsto$ $1000+900+90+5+1=1996$ MDCCCLXXIV $\mapsto$ M,D,CCC,L,XX,IV $\mapsto$ $1000+500+300+50+20+4=1874$ ...


9

Add the digits, but multiply the even digits by 2. This works because $5 \equiv 2 \mod 3$, $5^2 \equiv 1 \mod 3$, etc.


9

\begin{align} abcde_{10} &= 10000a + 1000b + 100c + 10d + e \\ &= (9999a + a) + (1001b - b) + (99c + c) + (11d - d) + e \\ &= (9999a + 1001b + 99c + 11d) + (a-b+c-d+e) \\ &= 11(101a + 91b + 11c + d) + (a-b+c-d+e) \\ &\equiv a-b+c-d+e \pmod{11} \end{align} Another way to look at this problem is to note that $10^{...


7

Our sum is odd, so all we need to do is to compute it modulo $5$. Note that the congruence class of $k^n$ modulo $5$ is the same as the congruence class of $k^{n+4}$ modulo $5$. This is obvious if $k$ is divisible by $5$. And if $k$ is not divisible by $5$ then $k^4\equiv 1\pmod{5}$. So to find the last digit for any $n$, it is enough to know the last ...


7

Hint: This is equivalent to asking what $$1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n$$ is modulo $10$. But $m=m-10$ mod $10$, so modulo 10 the above is the same as $$1^n+2^n+3^n+4^n+5^n+(-4)^n+(-3)^n+(-2)^n+(-1)^n.$$ What does this equal if $n$ is odd versus even?


7

Modulo $10$ easy inductions show that $$\begin{align*} &1^n+9^n\equiv\begin{cases} 0,&\text{if }n\text{ is odd}\\ 2,&\text{if }n\text{ is even}\;, \end{cases}\\ &2^n+8^n\equiv\begin{cases} 0,&\text{if }n\text{ is odd}\\ 8,&\text{if }n\equiv 2\pmod4\\ 2,&\text{if }n\equiv 4\pmod4\;, \end{cases}\\ &3^n+7^n\equiv\begin{cases} 0,&...


6

Given that the order of any of the components here will divide $4$, since that is the Carmichael function value for $10$, it is only necessary to check the values for $n=\{1,2,3, 4\}$. The results follow a pattern across the range of values (all $\bmod 10$): $$\begin{array}{c|c} n & 1^n & 2^n & 3^n & 4^n & 5^n & 6^n & 7^n & 8^...


5

Taking $\phi = \frac{1 + \sqrt 5}{2},$ your ratio is exactly $$ \frac{\log {10} - 4 \log \phi}{5 \log \phi - \log {10}} \approx \frac{0.377737792}{0.103474033} \approx 3.650556386 $$ The exact formula (Binet) for the Fibonacci numbers does not matter, since taking logarithms makes any constant coefficient disappear in the limit. All that matters is that the ...


5

When you are reading Roman numerals, start from the left-most character. Read rightward until the value of the character increases. Then, section those two characters off, and repeat. That sounds really complicated, and I wrote it somewhat poorly, so here are some examples. In $XIX$, we start with the left $X$ which is $10$. Then we move to the $I$ which ...


4

If $n$ is odd, then as in the answer by @Semiclassical, $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n\equiv5\pmod{10}.$ If $n$ is even, say $n=2m,$ then first observe $4^m\equiv 5+(-1)^m\pmod{10}$ and $9^m\equiv(-1)^m\pmod{10}.$ So we have $\begin{align}1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n&\equiv2(1+2^n+3^n+4^n)+5^n\\&\equiv2(1+(5+(-1)^m)+((-1)^m)+(5+(-1)^{2m}...


3

Yes, this is correct. There is no infinite decimal between the infinite decimals $0.\overline9$ and $1.\overline0$; they follow each other in the lexicographic ordering of the infinite decimals; whereas between any two real numbers there are further real numbers. Thus no bijection between the infinite decimals and the reals can respect their orders. Put ...


3

You can apply the following logics: the units are denoted I, II, III, IV, V, VI, VII, VIII$^*$, IX; the tenths are denoted X, XX, XXX, XL, L, LX, LXX, LXXX, XC; the hundredths C, CC, CCC, CD, D, DC, DCC, DCCC, CM; the thousands, M, MM, MMM. numbers are written in thousands, hundredths, tenths and units from left to right. no other pattern is allowed. ...


3

The $n=(d_m \cdots d_0)_5$ then $n$ is divisible by $3$ iff $d_0-d_1+d_2-d_3+\cdots$ is divisible by $3$. This follows from $5^k \equiv 1 \bmod 3$ if $k$ is even and $5^k \equiv -1 \bmod 3$ if $k$ is odd.


2

XIX is read left to right, the "I" is always applied to the final X. XIX = X + IX = 10 + 9 XXI = X + XI = 10 + 11


2

Notice that this is $0.\overline{110}$, so it corresponds to $\frac{110_2}{1000_2-1}=\frac 6 7$. This is how repeating decimals in different bases work: The repeating part can be written as the part that repeats over the difference between the power of the base and $1$. We need to find where we can do this for $\frac 6 7$ in base $3$. We need to find: $$\...


1

One can represent finite trees by a Dyck word, that is a word of the context-free language $D^*$ generated by the grammar $S \to (S) + SS + 1$. The following article gives several possible enumerations for $D^*$: Zoltan Kasa, Generating and ranking of Dyck words, Acta Univ. Sapientiae, Informatica, 1, 1 (2009) 109–118 See also this paper: Yu. S. Medvedeva,...


1

Let $a$ be a whole number. We'll write $a$ as $a=a_0+a_1\cdot10+_{\cdots}+a_k\cdot 10^k$. Let's check when $a (mod 11)=\overline{a}=0$ ($\overline a$ is just to make the writing more comfortable. $\overline a=\overline{a_0+a_1\cdot10+_{\cdots}+a_k\cdot 10^k}=\overline{a_0+a_1\cdot(-1)+_{\cdots}+a_k\cdot (-1)^k}=\overline{a_0-a_1+_\cdots+a_k\cdot(-1)^k}$ ...


1

A) is just fine. B) $15*16^4+11*16^3+7*16^2+12*16+5=$ $15*2^{16}+11*2^{12}+7*2^8+12*2^4+5=$ $15*2*8^5+11*8^4+7*4*8^2+12*2*8+5=$ $30*8^5+(8+3)*8^4+28*8^2+24*8+5=$ $(3*8+6)*8^5+8^5+3*8^4+(3*8+4)*8^2+3*8^2+5=$ $3*8^6+7*8^5+3*8^4+3*8^3+7*8^2+5=$ $3733705$.


1

The rule for b) is to first convert into decimal base(i.e. multiply by powers of 10) and then convert them to Octal (remainders when dividing by 8)


1

Hint $ $ Radix notation has Polynomial form $\,n = d_0\! + d_1 5 + d_2 5^2\! +\cdots + d_k 5^k\! = P(5)\,$ so ${\rm mod}\ 3\!:\ \color{#c00}5\equiv \color{#c00}{-1}\,\Rightarrow\ n = P(\color{#c00}5) \equiv P(\color{#c00}{-1}) \equiv d_0 - d_1 + d_2 - \cdots + (-1)^k d_k\, $ by applying the Polynomial Congruence Rule, i.e. $\,a\equiv b\,\Rightarrow\,P(a)\...



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