Brain Teasers
2001
Use addition, subtraction, multiplication, division, and parentheses to make the following equation hold.
9 8 7 6 5 4 3 2 1=2001
You can't use two or more of the digits to make multiple-digit numbers (i.e. solutions like (987+6-5+4*3)*2+1=2001 are not allowed.)
9 8 7 6 5 4 3 2 1=2001
You can't use two or more of the digits to make multiple-digit numbers (i.e. solutions like (987+6-5+4*3)*2+1=2001 are not allowed.)
Answer
9*8*(7+6*5)/4*3+2+1=2001.Hide Answer Show Answer
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Comments
This was hard for me, but fun. I didn't get it to work, but then, I have a short attention span.
Brilliant hard for me but easy for you i guess...
WOW, that was a good one, kind of hard for me. Well done
loved it
very challenging
me want more
very challenging
me want more
been too long since I've done this math. fun seeing it again.b
Good teaser. thanks
I think I have found an alternate answer. Please tell me if I am wrong
(9*8*7*4)-6-5-3-2+1
(9*8*7*4)-6-5-3-2+1
Never mind. I now see the digits had to remain in order.
o0
\_/
No emoticon says, "Insanely awesome!"
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Dec 18, 2005
you stumped me!!!
It was pretty difficult, but I eventually got it. However, I found a couple different ones of the 67,436,544 possible combinations of parenthesis and operators that work.
(9 x ( 8 + (7 x 6 x 5) + 4)) + (3 / (2 - 1)) = 2001
(9 x ( 8 + (7 x 6 x 5) + 4)) + (3 x (2 - 1)) = 2001
(There are 1029 ways to group the operations times the 4^8 = 65536 sets of operators; giving 1029 x 65536 = 67,436,544 possible equations.)
(9 x ( 8 + (7 x 6 x 5) + 4)) + (3 / (2 - 1)) = 2001
(9 x ( 8 + (7 x 6 x 5) + 4)) + (3 x (2 - 1)) = 2001
(There are 1029 ways to group the operations times the 4^8 = 65536 sets of operators; giving 1029 x 65536 = 67,436,544 possible equations.)
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