Easy math puzzles
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43 members have voted
March 27th, 2020 at 7:35:48 PM
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It isn't easy finding math problems that I feel fit in the sweet spot of being hard enough to be beer worthy but not too hard that I couldn't solve it.
That said, this thread is for problems I feel are too easy for a beer, but might be a good challenge for the members who are not at the elite level here.
All are welcome to pose problems. Please put answers and solutions in spoiler tags until I've declared a winner.
That said, here is the first problem.
x = sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(....))))))))))))))))
What is x?
That said, this thread is for problems I feel are too easy for a beer, but might be a good challenge for the members who are not at the elite level here.
All are welcome to pose problems. Please put answers and solutions in spoiler tags until I've declared a winner.
That said, here is the first problem.
x = sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(5*sqrt(....))))))))))))))))
What is x?
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
March 27th, 2020 at 8:04:21 PM
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No beer club rule?
Let x = sqrt(5*sqrt...
x^2 = 5*sqrt....
x^2/5 = x
x/5 = 1
x = 5
Let x = sqrt(5*sqrt...
x^2 = 5*sqrt....
x^2/5 = x
x/5 = 1
x = 5
March 27th, 2020 at 8:05:06 PM
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x=5
The infinite (or semi-infinite) expression in the problem is equivalent to 5 raised to the power of an infinite series of terms, starting with 1/2 and each step involving adding 1 and dividing by 2:
(((...((((((((1/2)+1)/2)+1)/2)+1)/2+ ...)
The sum approaches 1. For various finite sub-series of the terms, the exponents of 5 would be 1/2, 3/4, 7/8, 15/16, 31/32, 63/64, etc.
x = 5^1 =5.
The infinite (or semi-infinite) expression in the problem is equivalent to 5 raised to the power of an infinite series of terms, starting with 1/2 and each step involving adding 1 and dividing by 2:
(((...((((((((1/2)+1)/2)+1)/2)+1)/2+ ...)
The sum approaches 1. For various finite sub-series of the terms, the exponents of 5 would be 1/2, 3/4, 7/8, 15/16, 31/32, 63/64, etc.
x = 5^1 =5.
March 27th, 2020 at 8:24:42 PM
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I agree, the answer is as quoted above.
I'll post a new problem shortly, but anybody else is welcome to do so before me.
I'll post a new problem shortly, but anybody else is welcome to do so before me.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
March 27th, 2020 at 8:49:07 PM
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Anaswer = 5^0.5 * 5^0.25 * 5^0.125 * 5^0.0625 *........
= 5^(0.5+0.25+0.125+0.0625+ .....)
= 5^1
= 5
0.5+0.25+0.125+0.0625+ .... is a geometric series, r=0.5, Sn = 0.5( 1- 0.5^infinity)/(1-0.5) = 2 * 0.5 = 1
March 27th, 2020 at 9:46:00 PM
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Quote: djtehch34t
Let x = sqrt(5*sqrt...
x^2 = 5*sqrt....
x^2/5 = x
...
Instead of next dividing both sides by x, try subtracting x from both sides and factoring.
March 28th, 2020 at 4:56:46 AM
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Quote: ChesterDog
Instead of next dividing both sides by x, try subtracting x from both sides and factoring.
Dividing by x should be legit because we can clearly tell that x > 0?
March 28th, 2020 at 5:41:40 AM
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Quote: djtehch34tDividing by x should be legit because we can clearly tell that x > 0?
It took me a while, but now it's clear to me that x must be greater than zero.
Thanks.
March 28th, 2020 at 4:07:01 PM
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If x = SQRT(5*SQRT(5*....)) = SQRT(5*x). It then seems obvious x=5, but mathematically squaring each side to get x^2 = x*5, so x=5.
March 29th, 2020 at 8:09:01 PM
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Here is your next puzzle that is too simple to be beer-worthy, but is nevertheless pretty challenging.
A 3-4-5 triangle is inscribed in a square of length x.
Find x.
A 3-4-5 triangle is inscribed in a square of length x.
Find x.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
