perplexus.info :: Just Math : Zero to 100 with 12345

Zero to 100 with 12345 (Posted on 2005-10-29)

Find an expression for each integer zero to 100 which satisfies the following conditions:

- Each of the numbers 1, 2, 3, 4, and 5 appears exactly once in each expression and no other numbers. There is no concatenation or decimal points.
- Each of the four elementary operations (+ - * /) appears exactly once in each expression.
- The only other items which may be in an expression are factorial (!) and parentheses. Factorials and parentheses may occur as many times as needed or not be used at all.

Example: 0=5+3-4*2/1 or 0=(3!+4)/5-2*1

Submitted by Brian Smith

Rating: 2.0000 (2 votes)

Solution: (Hide)

0=5+3-4*2/1 1=(4+5)/3-2*1 2=(3+5-4)/2*1 3=4+5-3*2/1 4=(3+5)/(4-2*1) 5=4*3/(5-1)+2 6=4*3/(5-(1+2)) 7=3*2/1+5-4 8=4*2/(5-(3+1)) 9=4*3/1-5+2 10=5*(3-4/2+1) 11=5*2+4/2-1 12=5*2+4/(3-1) 13=5*3/1-4+2 14=4*(5+3-1)/2 15=5*(4-3+2)/1 16=(3+5)*(4-2)/1 17=5*3+4-2/1 18=3*(5+4/2-1) 19=5*4-3/(2+1) 20=(3+4/2-1)*5 21=(4+5-2)*3/1 22=(3!+5)*(4-2/1) 23=(5*4+3)/(2-1) 24=(3+5-2/1)*4 25=(4+3-2/1)*5 26=4!+2/(3!-5*1) 27=(4+2)*5-3/1 28=4!/(5-2)!*(3!+1) 29=5*3!-4/2+1 30=(5+3)*4-2/1 31=5*3!+4/2-1 32=5*3!+4/1-2 33=5*(4+3)-2/1 34=5*3!+4/(2-1) 35=5*(4!/3-2+1) 36=3!*(5+4/2-1) 37=5*(4!/3-1)+2 38=5!/3-4+2*1 39=2*(5!-4+1)/3! 40=5!/(4-3+2)*1 41=2*(5!+4-1)/3! 42=5!/3+4-2*1 43=(4!-2*5)*3+1 44=4*(3!+5/(2-1)) 45=5*(4!/3+2-1) 46=5*(4!/2-3)+1 47=5!/3+4*2-1 48=(1+(5-3)/2)*4! 49=(3!+1)*(4!/2-5) 50=(5!/4-3!+1)/2 51=1*(5!+3!-4!)/2 52=4!*(2+1)-5!/3! 53=5!*2/4-(3!+1) 54=1*(5!-4!)/2+3! 55=5!*2/4-3!+1 56=(5+1)!/(2*3!)-4 57=(5!/(2+4)-1)*3 58=(5+1)!/(4*3)-2 59=5!/2+3-4*1 60=5!*(3-2+1)/4 61=5!*(4-3)/2+1 62=5!/3+4!-2*1 63=5!/(4-2)+3*1 64=1*(5!-4)/2+3! 65=2*(5!-4!)/3+1 66=4!*(5+1)/2-3! 67=3*4!+5/(1-2) 68=5!/(3-1)+4*2 69=5!/2+(4-1)*3 70=3*4!+(1-5)/2 71=5!/2+4*3-1 72=(2+1)*(5!/4-3!) 73=3*(4!+2)-5/1 74=4*5!/3!-(2+1)! 75=2*5!/3-(4+1) 76=4!*3!/2+5-1 77=3*4!/(2-1)+5 78=3*(4!+(5-1)/2) 79=4*5!/3!-2+1 80=5!*(3!-4)/(1+2) 81=4*5!/3!+2-1 82=4*(5!/3!+1)-2 83=2*5!/3+4-1 84=3*5!/4-(2+1)! 85=3*(4!+5)-2/1 86=3!*(4!+5)/2-1 87=4*(5!/3!+2)-1 88=(5+3)*(4!/2-1) 89=(5!*3)/4-2+1 90=(4+1)*(5!/3!-2) 91=(5!*3)/4+2-1 92=5*(4!-3!)+2/1 93=5!-(4!+1*3!/2) 94=5!-(4*3!+2/1) 95=(5!*4)/(2+3)-1 96=(5!-4!)*3/(2+1) 97=(3!)!/5+1-2*4! 98=(5!*4)/(3!-1)+2 99=5!+3!/2-4!*1 100=5*(4!-3!+2)/1

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
re(4): computer solution -- minimum use of factorial	Cory Taylor	2005-11-02 10:32:22
re(3): computer solution -- minimum use of factorial	Charlie	2005-11-02 08:40:11
re(2): computer solution -- minimum use of factorial	SilverKnight	2005-11-01 19:08:37
re: computer solution -- minimum use of factorial	Cory Taylor	2005-11-01 18:14:02
re: computer solution -- minimum use of factorial	Alex Korn	2005-10-30 21:08:13
computer solution -- minimum use of factorial	Charlie	2005-10-29 14:02:18
Solution	goFish	2005-10-29 12:56:06

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