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Wednesday, July 6, 2022

McNugget numbers

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Henri Picciotto.

Regarding the Frobenius problem uncovered last week, here is Luca Tanganelli’s post:

“With 3 and 5 coins one gets the numbers n ≥ 10 of the form 3k+1 as 3(k–3)+5*2.

The numbers n ≥ 5 of the form 3k+2 are obtained with 3(k–1)+5.

This leaves the Frobenius number at 7.

In general, let a and b, a≤b, be the two currencies. If a and b have a common divisor, then obviously there is no Frobenius number.

If a and b are relatively prime, then a*i are distinct mod b for all 0 ≤ i ≤ b–1 and similarly for b*j, with 0 ≤ j ≤ a–1.

This means that any number greater than or equal to a(b-1) or greater than or equal to b(a-1) can be expressed in the form ax+by. Since a≤b holds, f ≤ ab–b–1, which is not a bad bound”.

For hypothetical 7 and 9 euro coins, the total amounts that we could pay without having to give ourselves change with coins of these denominations alone are: 1, 2, 3, 4, 5, 6, 8, 10, 11, 12, 13, 15, 17, 19, 20, 22, 24, 26, 29, 31, 33, 38, 40 and 47, so the Frobenius number for 7 and 9 is 47 (I’ll leave it to my observant readers) .check this result).

Regarding Bertrand’s paradox, Francisco Montesinos provides an ingenious approach that gives a value of 1/4 for the probability that a randomly drawn chord in a circle is larger than the side of the inscribed equilateral triangle:

If we draw the circle inscribed in the triangle, the center of each chord is either inside or outside that circle, and when inside the chord it is larger than the side of the triangle. Since the radius of the inscribed circle is half that of the circumscribed circle, its area is four times smaller, so the probability that a chord is larger than the side of the triangle is 1/4 (if the chord is larger than the side of the triangle). tangent to the inscribed circle, its length is equal to the side of the triangle).

indigestible numbers

While dining with his son at McDonald’s, mathematician and puzzle expert Henri Picciotto invented a variant of the Frobenius numbers that have become popular as “McNugget numbers” (which I can’t help but mention, as I regret their use) in the sacred realm of mathematics with reference on junk food). It appears that McNuggets were served in boxes of 6, 9, and 20 units, and any number of these indigestible morsels available by purchasing boxes of them were referred to by Picciotto as a McNugget number. And the Frobenius question is: What is the maximum number of McNuggets that cannot be obtained by combining these three types of boxes?

Currently the McNuggets boxes are 4, 6 and 9 pieces (I think they eliminated the 20 by order of the health authorities); what is the Frobenius-McNugget number?

Note that in this case we assume three numbers and not two to find the Frobenius number relative to them, which can greatly complicate the situation; so much so that no general formula is known to directly find the Frobenius numbers corresponding to three or more race numbers.

Carlo Frabetti is a writer and mathematician, a member of the New York Academy of Sciences. He has published more than 50 popular scientific works for adults, children and young people, including “Maldita Physics”, “Malditas Matematicas” and “El Gran Juego”. He was the screenwriter for The Crystal Ball.

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Source elpais.com

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