Recreational Problems (MM level)

[dcc]

find such that the above system has non-zero solutions for

[Neobeo]

Some number theory while we're at it:

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k=286, the next higher one k=381

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Find (1) k such that 1/(1x2) + 1/(2x3) + 1/(3x4) + 1/(4x5) + ...... + 1/[k(k+1)] = 0.99999 and
(2) k such that 1/(1x2) + 1/(2x3) + 1/(3x4) + 1/(4x5) + ...... + 1/[k(k+1)] = 0.999999 .

[Neobeo]

Yup 286 is right. How did you find it though?

Find (1) k such that 1/(1x2) + 1/(2x3) + 1/(3x4) + 1/(4x5) + ...... + 1/[k(k+1)] = 0.99999 and
(2) k such that 1/(1x2) + 1/(2x3) + 1/(3x4) + 1/(4x5) + ...... + 1/[k(k+1)] = 0.999999 .

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a quick response

same way as you did the last one but additional playing around with factors

[/0]

a quick response

same way as you did the last one but additional playing around with factors

I've gotten to but where do you go from there? 

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k(k+1)(2k+1)=mx6x7x191=mx2x3x7x191=mx2x7x573 (play around)
i.e. k=286=2p, k+1=287=7q where m=pq

[/0]

k(k+1)(2k+1)=mx6x7x191=mx2x3x7x191=mx2x7x573 (play around)
i.e. k=286=2p, k+1=287=7q where m=pq

Thanks, that's pretty neat, especially the m = pq bit.

[Ahmad]

I planned to have them in different sub-forums to somewhat correspond with level of difficulty and/or knowledge requirements, but this is fine I suppose.

[Mao]

I planned to have them in different sub-forums to somewhat correspond with level of difficulty and/or knowledge requirements, but this is fine I suppose.

ooopsies
hope this is slightly better

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Heron's quartet: Find whole numbers a, b, c and A such that

Note: For non-right-angle triangles.

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Examples: Multiples of (a=585, b=1225, c=928, A=261072)

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More examples: Multiples of (a = 13, b = 40, c = 45, A = 252)

[Mao]

do you have an upper limit? e.g. find the sum of whole numbers a,b,c such that A is less than a million?