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## Hot answers tagged puzzle

18

Using only basic arithmetic: $$\frac{4\times 3}{.12} =100$$

15

How about: $1^3 \cdot 4 = 100_2$

12

And solutions with logarithms (here natural logarithm) and ceil function. $$\left\lceil \ln 2^{143}\right\rceil = 100 = \sqrt{\left\lceil \ln 2^{13}\right\rceil^4}$$ Edit: (I don't wan't to make other answer it's not that type of question) Or with binomial coefficient. $$\binom{4+1}{3}^2 = 100$$ How to get it? Note that binomial coefficient equal to ...

12

No it is not possible cause after each leap the area of the triangle spanned by the three frogs remains the same. Now the area of an equilateral triangle with sides $1$ obviously does not equal the area of a triangle with sides $2$. Thus this will not be possible. Edit: The reason the area doesn't change is if you have triangle ABC then the area equals ...

11

In your equation, if you solve for $a$ you get: $$a = \frac{\sqrt{M^2-60b^2}}{3} - 4b$$ If we're looking for integer solutions then this means that $M^2-60b^2$ must be a square. Also it must be divisible by $3$, so at least you only need search over (pseudo-)triples of the form: $$9x^2 + 60b^2 = M^2$$ Moreover you want $a$ to be positive so you require $x ... 10 Idea: You can simplify things a bit if you make the middle kid be$n$years old with a difference of$k$. Then the kids ages are$n-4k, n-3k, n-2k, n-k, n, n+k, n+2k, n+3k, n+4k$. The sum of the squares of the ages is$9n^2+60k^2$. Dad's age is$\sqrt{9n^2+60k^2}$. I think you can find reasonable values fairly quickly that give a realistic ... 9 What about$(1)(3!+4)^2$? Is factorial allowed? Is$2$allowed as an exponent? 8 Hint: Powers of$3$follow the last digit pattern$3,9,7,1,3,9,7,1,...$8$14=2^{2^2}-2$and$18=2^{2^2}+2$$13=\frac{22}2+2$$24=\frac{(2^2)!}{\frac22}$$20=\sqrt{{22}^2}-2=(2^2)!-2^2$$22=\sqrt{\left[(2^2)!-2\right]^2}$7 It seems that it is imposiible because in the following coloring of a triangle (with side 1) grid we see that grid vertices of each triangle with side$1$are 3-colored, grid vertices of each triangle with side$2$are 1-colored, and a jump doesn't change the color of the jumping frog place. 6 Though I suspect this may be pushing common functions..$2^2 + 2 + \Gamma(2) = 722/{\sqrt{2}^2} = 11\int_{2/2}^{22}dx =21\int_{2-2}^{22}dx =22(2+2)! + \sqrt{2+2} = 26$4 The following is a complete list of the rationals obtainable by the five binary operations you gave, from a list of four twos. There are numerous expressions for many of the results and several invalid expressions (like$(2-2)^{2-2}$), but only the first valid expression my program found is shown below. $$\begin{array}{ll} -4194302 & 2-2^{22} \\ -482 ... 4 0=2+2-2-2 uses four twos not three. 10=\cfrac {22-2}2 4 \left\lfloor \exp\left(2^2\right)\right\rfloor+2+2=11 4 This problem is one of my favorite problems also, I personally was not able to solve it the first time I saw it. There is an idea that is very hard to refute, if there are more than two elves with blue eyes Santa is really not giving any information to anyone. However lets look at the following idea. Suppose there was exactly one person with blue eyes, ... 4 No, they can't. The process is clearly reversible (at each step the frog can jump back over the frog it just jumped over), so if it was possible, it would be possible to go from a triangle with sides of length 2 to one of length 1. Consider the following grid of equilateral triangles with sides of length 2 and the frogs being the green dots (sorry for my ... 4 If factorial is allowed, then (3!+4)^2 could make the job (inspired by Nicholas). 4 How about (4!+1)(3!-2)=100 I can't see anyway of doing this without factorials. 4 Denote by M the optimal weight of the most heavy bag. Hint: At all times, the weigh of the least heavy bag has to be smaller than M (consider the total weight of items). The weigh of any item has to be smaller or equal than M (the most heavy item has to go somewhere). I hope this helps \ddot\smile 3 All the coefficients on the left are multiples of 3, so M must be a multiple of 3, let M=3N. Now we have a^2+8ab+\frac {68}3b^2=N^2and b must be a multiple of 3. It is pretty unreasonable for b to equal 6, as the sons would span 48 years, but you can try it if you want. We have still cut down the case work. Nothing more pops out at ... 3 The discriminat of the equation$$9a^2+72ab+204b^2-M^2=0$$is$$4(M^2-540b^2)$$so M^2-540b^2 must be a perfect square, say, n^2. Therefore,$$540b^2=M^2-n^2=(M+n)(M-n)$$The arithmetic mean of M+n and M-n is M, and it is greater that the geometric mean, \sqrt{540b^2}>23b. Since M is the age of a man, b\le 5. Since M+n and M-n have ... 3 This is a classic riddle, whose solution surprisingly requires the axiom of choice as discussed here. The solution, however, is short and clever. Encode the colors into 0 and 1, and define the equivalence relation on 2^\Bbb N, \langle x_i\rangle\sim\langle y_i\rangle if and only if there is some k such that for all n\geq k, x_n=y_n. Using the ... 3 The following Haskell program: import Control.Monad; remove1 x [] = [] remove1 x (h:t) = if (x == h) then t else h : remove1 x t remove xs ys = foldr remove1 ys xs digits = [0..9] numeral = foldl (\a b -> a * 10 + b ) 0 solutions = do s <- remove [0] digits; e <- remove [s,0] digits; v <- remove ... 3 I'm pretty sure that 7 is impossible. In fact, 11,13,15,17,19 and every number from 20 to 30 is imposssible, too. A=Possible results with two twos: 0,1,4 B=New possible results with three twos: 3,6,8,16 C=New results of an operation with a two and an element of B: 5,9,10,12,14,18,32,36,64,256 Operating two elements of A gives ... 3 this is for 24 and 26$$22+2\Gamma (2)=24(2+2)!+2\Gamma (2)=26$$3 A slightly more creative way for 12 and 17:$|2+2\sqrt{-2}|^2 = 12|(2+2)!+\sqrt{-2}|/{\sqrt{2}} = 17$3 We don't seem to have 15 yet:${2\cdot 2 + 2 \choose 2} = 15$If we're going to use$\Gamma$and the like, we can get$30$too:$2 \cdot {\Gamma(2+2) \choose 2} = 30$No$19$or$27$or$29$yet either (though this keeps getting more absurd):$\lfloor{2+\sqrt{2}}\rfloor^{\lfloor{2+\sqrt{2}}\rfloor} = 27\lfloor \mathrm{Exp}(2)\rfloor \cdot ...

2

Are you allowed $.\dot 2=\frac 29$? That opens up a host of possibilities - for example $2+2+\sqrt {\cfrac 2{.\dot 2}}=7$

2

Experimentally it appears that seems that something close to Hagen von Eitzen's comment of the construction using a regular pentagon works for all obtuse triangles. Here is an example, with two isosceles triangles cutting off the sharp angles, and then choosing a suitable point inside the remaining pentagon to give $7$ acute angled triangles in all. Not ...

2

Let's start with the comment Hagen wrote: Construct a regular pentagon, connect each vertex to its center, and prolong three of its edges (all but two non-adjacent edges). This gives you a $(108°,36°,36°)$ triangle partitioned into seven acute triangles. You can solve a wide range of cases by distorting this figure. Hopefuly, one can manage to solve ...

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