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Posts posted by BMAD
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That is true. For this problem it doesn't matter whether it is the corresponding side on the left or right. Just be consistent.
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Suppose we have a quadrilateral with Angles A,B,C,D, corresponding sides a,b,c,d, and the following fact:
CosA/a=CosB/b=CosC/c=CosD/d
What can be claimed about this quadrilateral?
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On 8/25/2023 at 8:58 PM, EventHorizon said:
Proof 1: Modulus 10
The last digit is just the number mod 10.
(x * y) mod z = ((x mod z) * (y mod z)) mod z, so we just need to check x^5 mod 10 for each of 0-9.
x,^2,^3,^4,^5
0,0,0,0,0
1,1,1,1,1
2,4,8,6,2
3,9,7,1,3
4,6,4,6,4
5,5,5,5,5
6,6,6,6,6
7,9,3,1,7
8,4,2,6,8
9,1,9,1,9Done.
Proof 2: Modulus 5
Let n be a natural number. n can be represented as 5x + a, where x is an integer and a is one of {-2,-1,0,1,2} .
Case 1: a = 0
Any power of an odd multiple of 5 remains an odd multiple of 5, and must have a 5 in the ones digit.
Any power of an even multiple of 5 remains an even multiple of 5, and must have a 0 in the evens digit.
So n^5 has the same ones digit as n.Case 2: a = 1 or -1
(5x +/- 1) * (5x +/- 1)
= 25x^2 +/- 10x + 1
= 1 mod 5
Squaring again it remains 1 mod 5.Case 3: a = 2 or -2
(5x +/- 2) * (5x +/- 2)
= 25x^2 +/- 20x + 4
= 4 mod 5
From case 2, we see if we square something 4 mod 5 (a=-1), it becomes 1 mod 5.Finishing cases 2 and 3:
n^4 = 1 mod 5, so it can be represented as 5y+1 for an integer y.
(5y+1)*(5x+a)
= 25xy + 5x + 5ay + a
= a mod 5.
So n^5 mod 5 = a mod 5 = n mod 5.
n^5 mod 2 = n mod 2 since any power of an even number remains even and any power of an odd number remains odd.
n^5 mod 10 = n mod 10, because 5 and 2 are relatively prime.
Same mod 10 = same last digit.Done.
It's also interesting from this proof... n^4 ends in 1 for odd non multiples of 5, n^4 ends in 6 for even non multiples of 5.
Also, no fourth power of any integer ends in 2,3,4,7,8, or 9.I have a third proof idea... I'll see if I can flush it out later.
Are you a math professor/researcher? If not, where do you find / get inspiration for these?
I am a math professor.
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Assume that n is a natural number, prove that n and n5 will always have the same one's digit. e.g. 13 and 135=371,293 both end in 3.
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For any Convex Quadrilateral, show that the ratio of the Area to its Perimeter^2 is always <1/16, bonus points if you can show that it holds for concave quadrilaterals (not squares).
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Let F(t)=f(t)/g(t) be a rational function with integer coefficients, assume g(0)=1, then the Taylor expansion of F(t) at 0 has integer coefficients, and more over, these coefficients satisfy a recursion relation of the form c_n+k=a_{k-1}c_{n+k-1}+ ... + a_0c_n (k and all a_i are all fixed integers) for all but finitely many n?
(for example try computing a MacLauren series for (1+2x)/(1-x^3)
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(f(x+y)-f(xy))/(3x) = f(y/(3x))-11-y
Find f(x) where f(x) is a polynomial.
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I get two possible solutions: 1100 or 76461. Though if we want only positive values for each emoji then my answer of 1100 is the correct one.
Though I am treating the fact that like how one row has two emojis of alligators and is different than the other rows, then
the eagles being doubled is significant.
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2f(1/x)-f(x)+2f(2/x)-f(x/2) = x, x is defined on the reals except where x =0
find f(x) =
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h(f(x)) + g(h(x)) + f(g(x)) = 2x^2 + 11x + 14
f(h(x)) + h(g(x)) + g(f(x)) = 2x^2 - 15x + 66
f(g(x)) = g(f(x))
h(g(x)) - g(h(x)) = -16x + 72
h(f(x)) + f(h(x)) = 2x^2 + 10x + 30
f(x) * g(x) = h(x) - 3x - 40
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f(x) =
g(x) =
h(x) = -
How many different paths can I make up a flight of 20 stairs if I can take the steps either one at a time or two at a time (in any order)?
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I have a negative value for x as my min and a different x max
36 minutes ago, Nightfox__ said:Max of x is 1/2+(1/√2) and min 0??
This is not the answer but as an example of this possibility:
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Suppose we have the following system
x^2+y^2=r; x+y=r, such that the line crosses the circle at exactly two places.
Obviously with two equations and three variables, we have a solution set of answers that can satisfy the given conditions. What I want to know is of the given solutions that satisfies this problem, what is the smallest and largest values x can possibly be?
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I think you are on the right track rocdocmac
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I have in mind a number which, when you remove the units digit and place it at the front, gives the same result as multiplying the original number by 2. Am I telling the truth?
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On 4/17/2019 at 6:02 AM, rocdocmac said:
116 seconds (or possibly even 106 seconds) presuming first two pieces are already in place.
Toasting alone takes 90 seconds for slices numbered 1, 2 and 3 with sides A and B.
Stages: (1) 1A and 2A under grill (30 sec), followed by (2) 1B and 3A under grill (30 sec); finally (3) 2B and 3B under grill (30 seconds)
Normal additional time for single movements:
End of 1st stage: Slice 1 flipped from 1A to 1B (3 sec), 2A removed (5 sec), 3A inserted (5 sec)
End of 2nd stage: Slice 1 removed (5 sec.), slice 3 flipped from 3A to 3B (3 sec), slice 2 (2B) re-inserted with flip side up (5 sec)
End of 3rd stage: Done
Normal additional time = 26 seconds
Total time = 116 seconds.(1:56)
Exchange in one go:
If one is allowed to do simultaneous movements (exchanging in one switch of 5 sec), 10 seconds can be deducted, i.e. a total time of 106 seconds (1:46)
14 hours ago, Pirate8 said:Left
side
Right
side
1-5
A1 in
6-10
B1 in
6-35
A1 toast
11-40
B1 toast
36-38
A flip
41-45
B out
39-68
A2 toast
46-50
C in
69-74
A out
51-80
C1 Toast
75-79
B2 in
81-83
C flip
80-109
B2 toast
84-113
C2 toast
110-114
B out
114-116
C out
Manual operations must not overlap.
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Three slices of bread are to be toasted under a grill. The grill can hold two slices at once but only one side is toasted at a time. It takes 30 seconds to toast one side of a piece of bread, 5 seconds to put a piece in or take a piece out and 3 seconds to turn a piece over. What is the shortest time in which the three slices can be toasted?
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I wish to share 30 identical individual sausages equally amongst 18 people. What is the minimum number of cuts I need to make? What is the minimum number of pieces I need to create?
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Assume that only four buttons work on your calculator: 5, 7, enter, and plus. What whole numbers can you not use your calculator to make?
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20 hours ago, EventHorizon said:
I'm sure I know what BMAD was saying. Yes, ignore the units / squared units difference. The problem doesn't involve a circle though. You need to find a function f(x) such that the area under the curve between two points is the same as the arc length between those two points along the function. Here the arc length is like the length of a string representing the function cut at the two points.
How to find the arc length of a function:
If the function was a straight line, you could use the Pythagorean theorem to find it.
sqrt( (y2-y1)^2 + (x2-x1)^2 )
If you assume the arc length was a straight line, and slowly bring the points on the function closer together, eventually you get the change in y approaching the slope of the line times the difference in x's.
The arc length for that infinitesimally small section is then sqrt( ( f'(x)dx )^2 + dx^2 ) = sqrt( dx^2 * (f'(x)^2 + 1 ) = sqrt( f'(x)^2 + 1 ) dx. Where f'(x) is the derivative of f(x).
Integrate that between two points to get the arc length between those two points.
Example: f(x)=2x
f'(x) = 2
integrate( sqrt( 2^2 + 1 ) dx ) = integrate( sqrt(5) dx) = x*sqrt(5) + c
If you evaluate the integral between 0 and 1 you end up with arc length = sqrt(5). Which is the same as if you just used the Pythagorean theorem. Yeah, I'm too lazy right now to find a function that ends up with a nice arc length function that isn't just a straight line.
Restatement of problem using previous spoiler:
One answer is simple, the other is less so.
yes, this is what I mean.
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James and Mike are running in a race. They both walked and ran for part of the rate. They each walked and ran at the same speed. James ran for half the distance and walked for half the distance. Mike ran for half of his time and walked for half of his time. Who finished first?
A simple Coffee Problem
in New Logic/Math Puzzles
Posted · Edited by BMAD
Every morning, I brew 3 cups of coffee in my French press. I prepare a large mason jar with ice, sweetener, and cream which fills the container half-way. I add enough coffee to fill to the top of the jar. Throughout the morning I drink the jar down halfway just to top it up once again. I am able to refill it twice fully. For the third refill, I am only able to add 2 FL oz. of coffee. How big is my Mason Jar?
Bonus question: what % of coffee is in my last cups mixture after adding the 2 oz?
For added clarity, when I top it up, I mean that I am only adding my coffee to the mixture.