Math Problem for 2017
January 3, 2017 at 10:11 pm 7 comments
Happy New Year! Welcome to 2017.
Here are some interesting facts about the number 2017:
- It’s prime. (Okay, so that fact isn’t very interesting. But just you wait…)
- Insert a 7 between any two digits of 2017, and the result is still a prime number. That is, 27,017, 20,717, and 20,177 are all prime. (See? Told you it was gonna get better.)
- The cube root of 2017 is approximately 12.63480759, which uses all ten digits 0‑9, and 2017 is the least positive integer that has this property. (Mind blown yet?)
- The decimal expansion of 20172017 has 6,666 digits.
- 2017 = 442 + 92
- 2017 = 123 + 63 + 43 + 23 + 13 = 103 + 93 + 63 + 43 + 23
If you need some more, check out Matt Parker’s video.
Sorry, no video from me. But in honor of our newly minted prime year, I have created a problem for you to solve.
In the area model below (not to scale), the area of the five blue regions is indicated by the number inside the rectangle. What is the area of the yellow region with the question mark inside?
Sorry, I don’t give answers. Feel free to have at it in the comments.
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Entry filed under: Uncategorized. Tags: 2017, facts, Matt Parker, prime, problem.
Math Jokes 4 Mathy Folks 
1.
xander | January 4, 2017 at 4:19 pm
I’m super-lazy. The five rectangles give five equations in six variables: a*x = 20; a*y=17; b*z=20; c*x=2017; c*z = 17 (the heights of the rectangles are a,b,c from top to bottom, and the widths are x,y,z from left to right). Plug those into Maple to get solutions, then have Maple multiply b*y to find that the mystery area is 2017.
One could solve this system by hand, but that seems rather tedious, and therefore can’t possibly be the intended method of attack.
2.
venneblock | January 5, 2017 at 7:51 am
You can’t really “solve” the system by hand, since there are six unknowns but only five equations. I didn’t intend for you to use Maple, but I’m curious… what values did it return for each variable?
I had an algebraic trick in mind to find the solution, though working with my sons last night, they were far more keen on trying to find values for a, b, c, x, y, and z and then using those values to determine b*y.
3.
xander | January 5, 2017 at 10:36 am
You can “solve” it by hand in the sense that the system is underdetermined, hence one of the variables is free. Pick your independent variable, then solve everything else in terms of that variable.
By the way, by hand, I get:
b*y
= (20/z)*y (since b*z = 20)
= (20/17)*c*y (since c*z = 17)
= (20/17)*(2017/x)*y (since c*x = 2017)
= (2017/17)*a*y (since a*x = 20)
= 2017 (since a*y = 17).
Maple, by the way, does a more complete job of describing all possible solutions, and returns
{a = 17/y, b = 2017/y, c = 34289/(20*y), x = (20/17)*y, y = y, z = (20/2017)*y}
4.
puntomaupunto | January 5, 2017 at 9:08 am
well, 2017 is a solution, but this would mean that the image was taken directly above the model and that what we perceive as “squares” are wiggled surfaces (very wiggled when the value is 2017)
5. Carnevale della matematica #105 | backup del Post | January 13, 2017 at 10:10 pm
[…] un quizzino di Patrick Vennebush, da cui ho rubato anche le proprietà di cui sopra. Nel disegno qui a destra (non in scala), l’area delle […]
6.
margit | February 8, 2017 at 4:49 pm
(the heights of the rectangles are a,b,c from top to bottom, and the widths are x,y,z from left to right)
Since
ax = bz (=20) ===> b = ax / z
and
ay = cz (=17) ===> y = cz / a
well we are looking for by:
by = (axcz) / (az) ===> by = xz (and this is 2017)
7.
margit | February 8, 2017 at 5:20 pm
Sorry
little mistake in the end
must be by = xc ( and this is 2017)