Tag Info

Hot answers tagged

95

So in mathematics, as far as I know, you can't divide 100% by 3, without having that 0,1..% left.... No! we can in Math and also in real life. This is similar to ask can we divide $1$ into 3 parts? And the answer is again yes. $$1÷3=\frac{1}{3}$$ because adding $\frac{1}{3}$ three times give $1$. Consider 3 sticks of same length. Align these three ...


69

Meanwhile in ancient Greece... For quite a long time, greek (and not only) mathematicians described numbers as lengths of certain line segments. So, when asked "What is $\sqrt{2}$ equal to?" they'd draw a $1\times1$ square (nevermind the unit), draw it's diagonal and say "There it is! This diagonal's length equals exactly $\sqrt{2}$!". So to answer your ...


31

I think you have problems with this because you're thinking in base 10, and 10 (in base 10) doesn't divide evenly by 3. Think in base 3 instead. $100\%$ in base 3 is: $100\% = 10\% + 10\% + 10\%$ Which are trivially demonstrated to be equal parts, with no remainder. Addendum: although I have substituted its meaning analogously, $\%$ is the symbol for ...


24

One way to see this is to distinguish between definition and representation. I think you have began that. Definition: 1/3 is by definition exactly on third of 1. Representaion: "1/3" is a representation that is useful here. However there is no representaion of 1/3 using something like "0.3333333...". Unless you do not have an infinite paper or screen, of ...


20

From a false assumption you can derive anything. Answer what you want: it will be correct. For example: the answer is $\pi^2$, and I'll prove it. Suppose not. Then, by hypothesis, $5/2=3$, so $5=6$ and, substracting $5$ to each side of equation, $0=1$, a contradiction. So the answer is $\pi^2$.


19

I think this is more a question of language than of mathematics. (Indicated also by the fact that a "foreign country" is mentioned.) A possible understanding of "half" in this case would be that "half" is an operation that assigns integers to integers by splitting them in to parts as evenly as possible and then taking the largest part. In other words, by ...


18

The fundamental fallacy in your reasoning is that "natural == real". Just because a number never ends doesn't mean it isn't a "real" number with real application. You have three apples. Three is a "natural number"; it is positive and whole, used in counting and other "everyday math". Now, you are converting these three apples to "100%". "Percent" is from ...


11

It's been answered above, but I thought I should try for a consise answer. The fact that 100, in base 10, divided by 3 lacks a finite representation does not mean that it's impossible to divide 100 by 3. In fact most real numbers can not be represented by a finite decimal number.


8

Alright - With this sort of topic, we have to bring in a few ideas in order to explain this in full. The first one is: Infinity The concept of infinity is to have an endlessly large number. The fact is that you can't ever write out infinity as an actual number, because you can simply add another digit at the end, and you'll have a number larger than ...


6

I imagine this to be a problem caused by this foreign country not having the concept of the number zero. If you think about it as a number line, without a 0: If you were to divide this line into two equal parts, you would draw a line through the tick that corresponds to the number 3. Therefore, you could say that "half of 5 is 3" The same goes for a ...


4

A ratio of three is a rarity in coinage systems. They tend to use $1,2,5$ and things like that. I thought to check the old British coinage, because I knew it had funny values. Sure enough In the years just prior to decimalisation, the circulating British coins were the half crown (2/6), two shillings or florin (2/-), shilling (1/-), sixpence ...


3

Let $D$ be the check's actual number of dollars and $C$ be its actual number of cents. Then the actual amount of the check, expressed in pennies, is $$A=100D+C$$ The amount the man is given is $$G=100C+D$$ The pertinent equation is $$G-5=2A$$ Can you take it from there? Added later: A couple of people correctly admonished me for leaving the ...


3

You have made a good start. Your equation 2 is $9\cdot 10^{d-1}+x=70x+63$ or $x=\frac 9{69}(10^{d-1}-7)$ so you need $10^{d-1} \equiv 7 \pmod {69}$ Now you can just start looking. Increase $d$, multiply by $10$ and reduce the result $\pmod {69}$ It starts like this $$\begin {array}{c | c}d-1&10^{d-1}\\ \hline 1&10\\2&10\cdot 10 =100 \equiv 31 ...


3

N' := 7*N if the last digit of N = 9 then the last digit of N is like the last digit of 9*7, which is 3 (Remember 6 carry). The 3 is also the penultimate digit of N. 3*7+6 = 27 -> 7 is the next number (2carry) ... and so on, till the next Digit is 9 and carry is 0. N Ü N' 9 6 3 3 2 7 7 5 1 1 1 2 2 1 5 5 3 6 6 4 5 5 3 9 9 6 6 6 4 8 8 6 0 0 0 6 6 4 2 2 1 ...


3

In the real numbers, you can divide any quantity by 3 and get 3 equal parts. This is because of how we've defined the real numbers. You pose a very valid question, though, as real numbers aren't the only way to model quantities. For instance, we could also use the natural numbers (the ones we use for counting things discretely.) For instance, if you were ...


3

The answer is Mr.C. In detail following things happened: They competed in $5$ disciplines. The values of $x,y,z$ are $(5,2,1)$ respectively. Mr. A came first in $4$ disciplines and second in long jump, thus collecting $4\times5+2=22$ points. Note that this means that Mr. A won the 1/4 mile run. Mr. B came first in the long jump and then (necessarily) ...


2

Zubin's method is good, but still I will show my attempt. For first triplet ... $5*\color{red}{12}+7*\color{red}{2}=74$ For second triplet ... $5*\color{red}{26}+21*\color{red}{16}=466$ $12-2=10=26-16$ For first triplet we used bigger number for first multiplication, for second triplet we used smaller number for first multplication, so for third triplet ...


2

What you know is that $$ \text{The new income} = 120\%\text{ of the old income} $$ The new income is price per ticket, which is $(1-z)x$, times the number of tickets sold, $1.5y$. So your left hand side is good. The old revenue was the old price of tickets, $x$, times the old number of guests, $y$. Multiply by $1.2$ to get the increase of $20\%$. All in all ...


2

The $5_a$ must be interpreted as being half of $10_a$. So $\dfrac{5_a}2=3$ is equivalent to saying $10_a=12$, a third of which is obviously $4$. Imagine for instance counting from $1$ to $5$ on fingers, and using a clenched fist to represent the $6$. One could then easily count in duodecimal on two hands.


2

This is the Erdos-Szekeres theorem on monotonic subsequences. I'm pasting a proof from wikipedia: Given a sequence of length $(r − 1)(s − 1) + 1$, label each number $n_ i$ in the sequence with the pair $(a_i,b_i)$, where $a_i$ is the length of the longest monotonically increasing subsequence ending with $n_i$ and $b_i$ is the length of the longest ...


2

I guess it came out of confusion; Confusing the number 100 with the totality of something (100%, 1 unit) - However, the value 4 can be put into a unit and start being 100% of itself, you can divide 100% by 3 and get 1/3 and 0(»zero) left but you can't do it with the 4. -- I think, as @KeithS wrote I am making a switcheroo, although not intentional. As ...


2

$$ \left\{ \begin{align} A + B + C + D &= EF \\ E + F + G + H &= CJ \\ B + G + J &= xD \\ EF + CJ + xD &= By \\ A - H &\gt F \end{align} \right. $$ $\begin{align} \\ A &\in \{1,2,3,4,5,6,7,8,9\} \\ B &\in \{1,2,3,4,5,6,7,8,9\} \\ C &\in \{1,2,3,4,5,6,7,8,9\} \\ D &\in \{1,2,3,4,5,6,7,8,9\} ...


2

Let the centre of the circle be $O$, and let $\angle BOA=\theta$. Then our sum $AB+AC$ is $1+\cos\theta+\sin\theta$. It is enough to maximize $\cos\theta+\sin\theta$, or equivalently to maximize $(\cos\theta +\sin\theta)^2$, that is, $1+2\sin\theta\cos\theta$, which is $1+\sin(2\theta)$. This reaches a maximum when $\theta=\dots$.


2

You have that $$A\implies \neg B\\\neg A\implies B\\B\implies A$$ but we can stop right there because $A\implies \neg B$ means that the contrapositive is also true, so that $B\implies\neg A$, which contradicts the last statement. This world cannot exist, which is a little bit different than saying that the identies cannot be determined, because there is no ...


2

Whenever you talk about the probability of an event, you have to talk about the sample space (or equivalently, the measure) with respect to which you're taking the probability; otherwise, you run into potential paradoxes such as the Bertrand Paradox. Since you're talking about a Sudoku puzzle, the (IMHO) most natural sample space for determining your ...


2

Maybe I'm being obtuse and there is one 'obvious' answer, but it seems to me that we could generate infinitely many possible 'answers' through simply interpolating? That is, let $$f(x) = 11x - 12207$$ Then this function interpolates all of these points. (So maybe the 'obvious' answer is just to carry on interpolating along this line to get $1117 \mapsto ...


2

I found the answer by writing down the numbers 0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 on a piece of paper and writing down the numbers 7 10 1 3 6 11 8 4 5 9 0 2 on another piece of paper and sliding one against the other and counting the number of matches. 5 positions have 0 matches. 3 positions have 1 match. 3 positions have 2 matches. 1 ...


2

Here is my attempt at a purely geometric approach:                      Arrange a set of equal-size circles as in image 2, and position the top and bottom circles so that their centers form one side of a square. From the four-fold symmetry, it is seen that ...


2

The way to solve any problem of this type is to write some equations. Words are slippery and hard to work with; equations are much better. In this case, using $d$ for the number of dollars on the check and $c$ for the number of cents, you can write expressions for the total amount of money the check was worth, and the later process described: $$ 100c + d - 5 ...


2

Well let the number of Dollars in the amount he should have got be $D$ and the number of Cents $C$. He should have got $100D+C$. He actually got $100C+D$. After his spending spree he had $100C+D-5$ and this was equal to $200D+2C$ Let's assume for the moment that $D\ge 5$ and $2C\lt 100$ so there are no carries. Then we have $100C=200D$ or $C=2D$ equating ...



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