Need help on a math puzzle...

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2 total messages Started by James Micha Brabson Sat, 07 Sep 1996 00:00

#36

Author: James Micha Brabson
Date: Sat, 07 Sep 1996 00:00

42 lines
1969 bytes

Hi there guys, we were looking for some help on something.
My sister is a senior in high school, and was given a homework assignment
that has all of us stumped.  She's taking math analysis, and we were
hoping someone would have enough of a clue to figure this out.

You have to use exactly 4 "2"'s, along with a variety of math symbols
and get every number from 1 through 20.  You cannot use any number BUT 2.
You have to use exactly four 2's.  You cannot combine 2 "2"'s to make 22.
(Below are the ones I've gotten so far.)
Examples:
(2+2)/(2+2)	= 1
(2/2)+(2/2)	= 2
(2/2)+(sqrt(2))^2 = 3
2*2+2-2		= 4
2*2+(2/2)	= 5
2*2+(sqrt(2))^2 = 6
???????????????	= 7
2^2 + 2^2	= 8
((2/2)+2)^2	= 9
2*2*2+2		= 10
((2^2)!-2)/2	= 11
(2^2)!/(sqrt(2)^2) = 12
((2^2)!+2)/2	= 13
((2^2)^2)-2	= 14
???????????????	= 15
2^(2+(sqrt(2)^2)= 16
??????????????? = 17
((2^2)^2)+2	= 18
??????????????? = 19
((2^2)!)-(2^2)	= 20

Fill in the "????"'s.  We think factorial should be allowed, as well as
sqrt's, since we're just using that little symbol we learned in high
school.  No using calculus or above, HS algebra I, HS trig, HS geometry, 
and HS algebra II.  Remember, this is a public school, so don't expect
them to have learned much.  We can get the last 4 using "i", but we're
pretty darned sure that's cheating.  (eg. (2^2)*2 + i^2 = 7)
Any takers?  We want 7, 15, 17, & 19.  No summations allowed.
If any of you find answers, (Or even some questions) please e-mail me!

-James

Re: Need help on a math puzzle...

#35

Author: Janusz Murakowski
Date: Thu, 12 Sep 1996 00:00

24 lines
1168 bytes

In article <50t5sa$b6j@chopin.udel.edu>, synic@chopin.udel.edu (James Micha Brabson) says:
>Examples:
>(2+2)/(2+2)     = 1
>(2/2)+(2/2)     = 2
>(2/2)+(sqrt(2))^2 = 3
>2*2+2-2         = 4
>2*2+(2/2)       = 5
>2*2+(sqrt(2))^2 = 6
>??????????????? = 7 = (2+2)!! - 2/2 = 2*2*2 - sgn(2)
>2^2 + 2^2       = 8
>((2/2)+2)^2     = 9
>2*2*2+2         = 10
>((2^2)!-2)/2    = 11
>(2^2)!/(sqrt(2)^2) = 12
>((2^2)!+2)/2    = 13
>((2^2)^2)-2     = 14
>??????????????? = 15 = (2+2+2/2)!! = (2+2)^2 - sgn(2)
>2^(2+(sqrt(2)^2)= 16
>??????????????? = 17 = (2+2+sgn(2))!! + 2 = (2+2)^2 + sgn(2)
>((2^2)^2)+2     = 18
>??????????????? = 19
>((2^2)!)-(2^2)  = 20
>

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