# Tag Info

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$10easy inductions show that \begin{align*} &1^n+9^n\equiv\begin{cases} 0,&\text{if }n\text{ is odd}\\ 2,&\text{if }n\text{ is even}\;, \end{cases}\\ &2^n+8^n\equiv\begin{cases} 0,&\text{if }n\text{ is odd}\\ 8,&\text{if }n\equiv 2\pmod4\\ 2,&\text{if }n\equiv 4\pmod4\;, \end{cases}\\ &3^n+7^n\equiv\begin{cases} 0,&... 6 Given that the order of any of the components here will divide 4, since that is the Carmichael function value for 10, it is only necessary to check the values for n=\{1,2,3, 4\}. The results follow a pattern across the range of values (all \bmod 10):\begin{array}{c|c} n & 1^n & 2^n & 3^n & 4^n & 5^n & 6^n & 7^n & 8^... 5 Taking\phi = \frac{1 + \sqrt 5}{2},$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)\...

Only top voted, non community-wiki answers of a minimum length are eligible