Aliens.

mfgoode · 2006-05-25, 16:21

100 aliens attended the Intergalactic meeting on Earth.

73 had 2 heads
28 had 3 eyes.
21 had four arms.
12 had 2 heads and 3 eyes.
9 had 3 eyes and 4 arms.
8 had 2 heads and 4 arms.
3 had all 3 unusual features.
How many had none of these unusual features.?
Mally

alpertron · 2006-05-25, 17:04

My solution is:

A = 1 head, 2 eyes, 2 arms
B = 1 head, 2 eyes, 4 arms
C = 1 head, 3 eyes, 2 arms
D = 1 head, 3 eyes, 4 arms
E = 2 heads, 2 eyes, 2 arms
F = 2 heads, 2 eyes, 4 arms
G = 2 heads, 3 eyes, 2 arms
H = 2 heads, 3 eyes, 4 arms

100 = A+B+C+D+E+F+G+H
73 = E+F+G+H
28 = C+D+G+H
21 = B+D+F+H
12 = G+H
9 = D+H
8 = F+H
3 = H

A = ?

D = 9-3 = 6
F = 8-3 = 5
G = 12-3 = 9
C = 28-6-9-3 = 10
B = 21-6-5-3 = 7
E = 73-5-9-3 = 56
A = 100-7-10-6-56-5-9-3 = 4

So the requested number of aliens is four.

Citrix · 2006-05-25, 18:39

This is a simple question of sets. Add all the 3 sets, subtract the overlaping.
My answer is 4.

alpertron · 2006-05-25, 19:10

I found the drawing features of Tex very interesting:

$\Large\hspace{5}\unitlength{1}\picture(150,150){~(50,50){\circle(100)}~(100,50){\circle(100)}~(75,100){\circle(100)}(c75,120){3$7}(c75,75){3$3}(c50,86){3$5}(c25,50){3$56}(c100,86){3$6}(c75,30){3$9}(c125,50){3$10}(c10,135){3$4}(c130,140){3$4a}(c140,7){3$3e}(c10,7){3$2h}}$

Sorry but spoiler and tex tags are incompatible.

victor · 2006-05-25, 22:17

If you really don't want to spoil, give us a link : answer ;)

Xyzzy · 2006-05-25, 22:28

Quote:

Originally Posted by alpertron

$\Large\hspace{5}\unitlength{1}\picture(150,150){~(50,50){\circle(100)}~(100,50){\circle(100)}~(75,100){\circle(100)}(c75,120){3$7}(c75,75){3$3}(c50,86){3$5}(c25,50){3$56}(c100,86){3$6}(c75,30){3$9}(c125,50){3$10}(c10,135){3$4}(c130,140){3$4a}(c140,7){3$3e}(c10,7){3$2h}}$

Very cool!

davar55 · 2006-05-26, 01:22

Using the Inclusion-Exclusion Principle, the answer can be
immediately computed as:

100 - (73+28+21-12-9-8+3) == 100 - 96 == 4

-- davar55

mfgoode · 2006-05-26, 07:10

All the answers given above are correct.

The Venn diagram given by Alpetron was what I was in search of but do not know how to use Tex. These were followed up by Mike and Vctor.

Citrix and Davar gave the short cuts.

Thanks to you Davar Im happy that you are now flexing your muscles!

Inclusion - Exclusion principle ? One for my note book!

Mally

2006-05-25, 16:21	#1
mfgoode Bronze Medalist Jan 2004 Mumbai,India 2052₁₀ Posts	Aliens. 100 aliens attended the Intergalactic meeting on Earth. 73 had 2 heads 28 had 3 eyes. 21 had four arms. 12 had 2 heads and 3 eyes. 9 had 3 eyes and 4 arms. 8 had 2 heads and 4 arms. 3 had all 3 unusual features. How many had none of these unusual features.? Mally

2006-05-25, 18:39	#3
Citrix Jun 2003 2²×397 Posts	This is a simple question of sets. Add all the 3 sets, subtract the overlaping. My answer is 4. Last fiddled with by Citrix on 2006-05-25 at 18:40

2006-05-26, 07:10	#8
mfgoode Bronze Medalist Jan 2004 Mumbai,India 4004₈ Posts	Aliens. All the answers given above are correct. The Venn diagram given by Alpetron was what I was in search of but do not know how to use Tex. These were followed up by Mike and Vctor. Citrix and Davar gave the short cuts. Thanks to you Davar Im happy that you are now flexing your muscles! Inclusion - Exclusion principle ? One for my note book! Mally

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2006-05-25, 17:04	#2
alpertron Aug 2002 Buenos Aires, Argentina 10110101110₂ Posts	My solution is: A = 1 head, 2 eyes, 2 arms B = 1 head, 2 eyes, 4 arms C = 1 head, 3 eyes, 2 arms D = 1 head, 3 eyes, 4 arms E = 2 heads, 2 eyes, 2 arms F = 2 heads, 2 eyes, 4 arms G = 2 heads, 3 eyes, 2 arms H = 2 heads, 3 eyes, 4 arms 100 = A+B+C+D+E+F+G+H 73 = E+F+G+H 28 = C+D+G+H 21 = B+D+F+H 12 = G+H 9 = D+H 8 = F+H 3 = H A = ? D = 9-3 = 6 F = 8-3 = 5 G = 12-3 = 9 C = 28-6-9-3 = 10 B = 21-6-5-3 = 7 E = 73-5-9-3 = 56 A = 100-7-10-6-56-5-9-3 = 4 So the requested number of aliens is four.

2006-05-25, 19:10	#4
alpertron Aug 2002 Buenos Aires, Argentina 2·727 Posts	I found the drawing features of Tex very interesting: $\Large\hspace{5}\unitlength{1}\picture(150,150){~(50,50){\circle(100)}~(100,50){\circle(100)}~(75,100){\circle(100)}(c75,120){3$7}(c75,75){3$3}(c50,86){3$5}(c25,50){3$56}(c100,86){3$6}(c75,30){3$9}(c125,50){3$10}(c10,135){3$4}(c130,140){3$4a}(c140,7){3$3e}(c10,7){3$2h}}$ Sorry but spoiler and tex tags are incompatible.

2006-05-25, 22:17	#5
victor Oct 2005 Fribourg, Switzerlan 2²·3²·7 Posts	If you really don't want to spoil, give us a link : answer ;)

2006-05-26, 01:22	#7
davar55 May 2004 New York City 5×7×11² Posts	Using the Inclusion-Exclusion Principle, the answer can be immediately computed as: 100 - (73+28+21-12-9-8+3) == 100 - 96 == 4 -- davar55