Math Is Fun Forum

Ethereal

Guest

Elementary, yet challenging problem...

I am constructing a kind of a game.. It is a quest of some sort and there are clues that one should follow. Clues are numbered 1, 2, 3, 4.. etc. Thus those playing should do what the clue under number "1" is saying, then number "2".. etc..

Now, at that point the game is too easy.. i want to make them sweat a bit.. need a bit of chaos in all that... so my idea is to represent the numbers as formulas and scatter them around (not following logically one after another). For example "1" can be anything on the power of 0, "2" can be root of 4.. etc.

Problem is I have about 200 different clues.. so you can imagine the amount of number substitutes I have to come up with..

That's what i need some help with. There is no limit to imagination here.. the more complex it is, the better (only restriction is formula length, it should be short because of the limited space on the sheets I will print the game on)
Any branch of mathematics can be used.. logarithms, trigonometry, complex numbers, lim, rot, div, integrals and derivatives etc... just anything that would give a numerical result.

I would really appreciate any help you guys can provide in coming up with the substitution formulas.
Thanks in advance,
Alex

bossk171

Member

Registered: 2007-07-16

Posts: 305

Re: Elementary, yet challenging problem...

Any branch of mathematics can be used.. logarithms, trigonometry, complex numbers, lim, rot, div, integrals and derivatives etc... just anything that would give a numerical result.

Counting Down:

200 = 5²(2³-1)(2-1)
199 = 200 + arccos(pi)
198 = √(39204)
197 = 200 - log(1000)
196 = (x-6)²(x-1)²  [x = 2³]
195 = x²-390x+38020 (solve for x)
194 = Area of rectangle with sides 2 and 97
193 = 44th prime
192 =  f'''(π/4)   [f(x) = 3*sin(4x)]
191 = x intercept of y=2x-382
190 = 1+2+3+4+...+19
189 = distance (in meters) traveled at 7 meters/sec for 27 sec
188 = Re[(10+3i)(20+4i)]
187 = area of triangle with base 22 and height 17
186 = number of seconds in 3.1 minutes
185 = distance from (5,5) to (65,180)
184 = 5²√(16) + 80 + |6-10|
183 = x²-136x+3783 [x = 100]
182 = (100*cos(x)+80*sin(x)+2*cos(x+2π))*(√2) [x=π/4]
181 = largest prime factor of 25340
180 = 6∫(x²+2x)dx (evaluate from 2 to 5)
179 = lim (x⇒∞) (179x+3)/x
178 = x-value of the min of f(x) = (x-300)*(x-56)
177 = 175 + ∫sin(x)dx [0<x<pi]
176 = LCM(16,22)
175 = 5² + 5² + 5³
174 = 200 + (-4+6i)(2+3i)
173 = distance from the origin to (52,165)
172 = volume of block with side lengths 2,2,43
171 = n(n+1)/2 [n=18]
170 = 10*(∫(-17/x²)dx) [1<x<∞]
169 = area of circle with radius 13/√π
168 = 10*2^4 + 2^3
167 = 10100111 (base 2!)
166 = 369 mod 203
165 = 11 Choose 3
164 =The shortest possible perimeter of an isosceles triangle with two of its side-lengths 50 and 64
163 = 100*ln(e)+10*log(1,000,000)+ln(e³)
162 = length of the curve (1/3)x³-x+4 from 3 to 165
161 = cuberoot(4173281)
160 = sum first 11 primes
159 = 2^7+31
158 = ∫dx/x  [e<x<e^159]
157 = Im(z) - Re(z)  [z = (8+9i)(4+9i)]
156 = (10-2i)(15+3i)
155 = gcd(775, 465)
154 = Perimeter of a rectangle with side-lengths 50 and 27
153 = (27/3)/((8/2)/68)
152 = ½(18!/16!)
151 = Smallest palindromic number greater than 145
150 = perimeter of a right triangle with leg lengths 25 and 60

More tomorrow. It's time for bed.

(PS, be sure to check all of these before using them. It's very late, so I might have made some mistakes).

Last edited by bossk171 (2010-06-30 17:24:08)

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There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.

bossk171

Member

Registered: 2007-07-16

Posts: 305

Re: Elementary, yet challenging problem...

Round 2:

149 = ∫|2x|dx  (-7<x<10)
148 = 37*2^(∑2^-i)  (0≤i<∞)
147= 12110 (base 3)
146 = 100 + 50sin(x) + 4sin(3x)  [x=pi/2]
145 = curve length from 0 to arcsec(46) of ln(cos(x))
144 = Only square Fibonacci Number > 1
143 = CXLIII (roman numerals)
142 = ∑(5+n²)²  [0≤n≤2]
141 = ∑(2-|n-1|²)10^n  [0≤n≤3]
140 = x²(x+3)(4x+1)  [x=2]
139 = 2n+1  [n=integral from 0 to 3 of 23 dx]
138 = f'(6)  [f(x) = (2/3)x³ +4x² +18x + 7]
137 = distance between (12,-5) and (100,100)
136 = zz*   [z = 10+6i]  [z* is the conjugate of z]
135 = 5x^x [x=3]
134 = 10n + ((n+1)(n-1)+1)/n -2cos(π)
133 = ∑11^n  (0≤n≤2)
132 = 10^2 + 2^5
131 = (120x +11x -100 -31)(x+1)/(x^2-1)
130 = 104% of 125
129 = distance between (0,0,0) and (4,7,8)
128 = 2^n  [n = lim (n -> 1) (7x-7)/(x-1)
127 = cuberoot(x-2)=5  [solve for x]
126 = exp(ln(2) + 2ln(3) + ln(7))  [exp(x) = e^x]
125 = (n+1)^(n-1)  [n=4]
124 = Surface area of a rectangular prism with side lengths 2, 4 and 9
123 = ∑(3-n)10^n  [0≤n≤3]
122 = |11+i|²
121 = f(6)   [f(n) = 3*f(n-1)+1, f(0)=1]
120 = Cent et vingt
119 = a^b-(a+b)  [a=2, b=7]
118 = 6!/5 - 62
117 = ∑(3n)²  [2≤n≤3]
116 = (-3+3√(3)i)³-100
115 = max(228x-x²)
114 = a^b-(a*b)  [a=2, b=7]
113 = x² = 15² + (x-1)²  [solve for x]
112 = ∑2^n  [4≤n≤6]
111 = 11² - 10
110 = [see below]
109 = |60+91i|
108 = (a^a)(b^b)  [ab = 6 = a+b+1]
107 = 3 + 54 + 31 + 19
106 = zz*  [z = 5+9i]
105 = Product of first three odd primes
104 = exp(3ln(2))*exp(ln(13))
103 = Sum the roots of x² - 103x + 2652]
102 = curve length from 1 to .25(205 + sqrt(42033)) of x²/2 - ln(2x)/2
101 = ∑ n!*(-1)^(n-1)  [1≤n≤5]
100 = ∑ 2n+1  [0≤n≤9]

If any others need to be LaTexed let me know. It's so tedious I tried to do with out it...

What is your target demographic for this? Are you a teacher? If so, what age group are your students? If you are in fact a teacher, I'd suggest having your students do the rest (numbers 1 through 99). There's a lot more to be gained from creating these formulas than there is to solving them.

I think I'm going to pass on the remaining 99, I've been spending too much of my time on this. If you really want me to them, I will, but as I said... they're super educational (and tons of fun).

Last edited by bossk171 (2010-07-01 12:22:25)

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There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.

Ethereal

Guest

Re: Elementary, yet challenging problem...

Thanks a lot!