### Brain Teasers

# 1000 Point Star

Logic
Logic puzzles require you to think. You will have to be logical in your reasoning.

Typical "stars" are drawn in connected, but not repeated, line segments. For example, a 5-point star is drawn as such - line segments AC, CE, EB, BD, DA. The segments must always alternate a constant number of points (in the above case, skipping 1 point in between).

Given the information that there is only 1 way to draw a 5-point star, and that there is NO way to draw a 6-point star (in continuous lines, that is), and there are 2 ways to draw a 7-point star, how many different ways are there to draw a 1000-point star?

Given the information that there is only 1 way to draw a 5-point star, and that there is NO way to draw a 6-point star (in continuous lines, that is), and there are 2 ways to draw a 7-point star, how many different ways are there to draw a 1000-point star?

### Hint

It's not guess work. Think of factors and mirror images. Try drawing out examples.### Answer

199.First, let's examine the 5-point star and why there is only that one, described, way to draw it. Starting at A, you cannot go straight to point B, otherwise you will have a pentagon, not a star. Same with point E. AC and AD (and their progressions) yield identical pictures (notice that the last segment in the AC one is DA, literally reverse all the letters to get AD ending in CA).

Now, for a 6-point star, there are points A, B, C, D, E, and F. Starting at A, we cannot go to B or F, because then we have a hexagon. If we go AC, then we must go CE, then CA... making a triangle, not a 6-point star. Going backwards, AE, EC, EA; mirror image. If we skip 2 points, we go AD, then DA, a straight line, hardly a 6-point star.

For a 7-point star, A-G, AB and AG are not possible, as that's a heptagon. AC, CE, EG, GB, BD, DF, FA is one way to draw the star, skipping one point in between. Also, AD, DG, GC, CF, FB, BE, EA, skipping two points in between each, is the other way. The last two line possibilities, AF and AE are mirror images of the two working scenarios.

Examining the above information a little more deeply, we see that we must divide all possible drawings by two, to get rid of the mirror images. Also, we cannot use the point immediately to the left, that is AB as a segment. This explains why there is only 1 way for a 5-point, and 2 for a 7-point, but does not explain the 6-point. Of those remaining points (dealing only with the first half, again, mirror images), those that are factors of the number of total points do not work, because you will end up back at the starting point before hitting every number. AC is a factor in six because it passes point B, so it's 2. AD is a factor of 3 (B, C and D). Not only factors, but multiples of factors, as well (for example, skipping 4 points in the 6-point still does not work).

Using the knowledge of mirrors, first point impossibility (AB) and factors, we see that the number 1000 is made up of 10^3, or 2^3*5^3. Without going through every factor, it's quickly and easily discernible that of the numbers 1-1000, 2 or multiples of it encompass 500 numbers, all evens. Any multiple of 5 includes 200 numbers, half of which end in 0 and are already included with the 2's, so there are 600 unique factors and multiples between 1 and 1000. That leaves us with 400 that are *not*, which should explain how many unique stars we can make. Remember to cut it in half because of mirroring, leaving us with 200, and to discount the AB segment (factor of 1), leaving us with 199. (If you discount AB, and THEN halve it, remember you needed to remove A[Z], the mirror image of that point).

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## Comments

Wow, very nice. It looks liek you invested a lot of time into that one!!! Very nice

Someone has wayyy too much free time. xD

Interesting, though...That was cool.

Interesting, though...That was cool.

how ling did it take u to do this?

My goodness!

Waaaaay too complicated or me.

Great job. Must've been very time-consuming.

Great job. Must've been very time-consuming.

Got me on this one! I've been interested in multi-pointed stars ever since a 9-pointed star showed up in a math contest I once entered in middle school. So this one was a lot of fun for me, even though I came up with a wrong answer.

Good puzzle, not so diffucult of one know about phi

Ugh! Math is my best subject, but at the end I'm like "WHAT?!?!?!? Is this English???"

In the course of my job I usually have to do alot of modulus arithmetics, so this took me less than 10 seconds to figure out.

Cool one though, this is the first time I think of stars as ranges.

Keep up the good work

Cool one though, this is the first time I think of stars as ranges.

Keep up the good work

Nice problem!

Confusing use of "star" and "repeated line segment."

Hope to effect correction.

Hope to effect correction.

...what?...

wow...that was hard

the answer was long. very long. it scared me

good teaser

scary teaser

the answer was long. very long. it scared me

good teaser

scary teaser

i dunno, a million? yet i think that defeats the purpose of brain logic. still just grab a pencil and that ones pretty easy. for some people

please help me out.my answer is 407 or 408.

my working

all points lie on a circle .

there will be n points and gaps between points.

let us fix a point from where we start .

now for moment we move clockwise .

the number of gaps between starting point and first joined point must not be factor of n .if we move around joining points with strategy given the nearest point in the direction opposite to witch we start will have let us say r number of gaps with starting point(in one rotation).let us say number of gaps we are skipping is p.then r is remainder if we divide n by p.

from easy visualization we can observe that must not be a factor of p

if r is a factor of p then r is factor of n buy euclid rules .n=p(some value let us say q)+r .

we want to exclude all p such that r is factor of it .so we must exclude all divisors of n-r where r is factor of n.since in the above case we have to skip no of gaps less than 500 to avoid mirror images .so p have a limit .now using counting of all the number to be excluded .i got my answer which is no close to the given one .there might be a counting mistake but not so much please help

my working

all points lie on a circle .

there will be n points and gaps between points.

let us fix a point from where we start .

now for moment we move clockwise .

the number of gaps between starting point and first joined point must not be factor of n .if we move around joining points with strategy given the nearest point in the direction opposite to witch we start will have let us say r number of gaps with starting point(in one rotation).let us say number of gaps we are skipping is p.then r is remainder if we divide n by p.

from easy visualization we can observe that must not be a factor of p

if r is a factor of p then r is factor of n buy euclid rules .n=p(some value let us say q)+r .

we want to exclude all p such that r is factor of it .so we must exclude all divisors of n-r where r is factor of n.since in the above case we have to skip no of gaps less than 500 to avoid mirror images .so p have a limit .now using counting of all the number to be excluded .i got my answer which is no close to the given one .there might be a counting mistake but not so much please help

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