| Source:
http://www.puzzlet.co.uk/Puzzlets/Puzzlet_002.html
A pandigital is an
integer containing every digit. In this
particular case,
we're dealing with the 9-digit pandigital,
since the zero isn't
used.
Take all the digits
1 through 9 in order, and insert as
many plus and minus
signs as you wish, wherever you want,
to make an arithmetic
sum of 100. For example,
123 + 45 - 67 + 8 - 9 = 100
Are there any more
ways of punctuating a 9-digit pandigital
with plus and minus
signs to make a sum of 100 as in the
example above?
If so, what are they?
.
.
|
 |
Finger pointing down
from darrell94590 on 1/2/2006 |
.
..
THE SOLUTION
Mr. Ellis wrote
a basic program which can be seen at
http://www.puzzlet.co.uk/Solutions/Puzzlet_002_sol.html
which generated
these twelve solutions.
123-45-67+89
= 100
123+45-67+8-9
= 100
123+4-5+67-89
= 100
123-4-5-6-7+8-9
= 100
12+3-4+5+67+8+9
= 100
12-3-4+5-6+7+89
= 100
12+3+4+5-6-7+89
= 100
1+23-4+56+7+8+9
= 100
1+23-4+5+6+78-9
= 100
1+2+34-5+67-8+9
= 100
-1+2-3+4+5+6+78+9
= 100
1+2+3-4+5+6+78+9
= 100
From: lubin100 on
10/25/2009
Dick added
the following solutions:
98+7+6-5-4-3+2-1
= 100
98+7-6+5-4+3-2-1
= 100
98-7+6+5+4-3-2-1
= 100
98-7+6-5+4+3+2-1
= 100
98-7+6+5-4+3-2+1
= 100
98+7-6+5-4-3+2+1
= 100
98-7-6+5+4+3+2+1
= 100
9+8+76+5+4-3+2-1
= 100
9+8+76+5-4+3+2+1
= 100 |
