Find nine single digit numbers (other than 1, 2, 3, ..., and 9) with a sum of 45 and a product of 9! (362,880).
For instance, 2, 2, 3, 4, 5, 6, 7, 8, 8 add up to 45, but their product is 645,120.
Try finding the answer without the use of a program.
(In reply to
thoughts by Charlie)
I tried, using the thoughts in my previous post, but heck, that was beating my head against the wall for a D2 problem.
CLS
FOR a = 1 TO 5
FOR b = a TO (45  a) / 8
tb = a + b: pb = a * b
FOR c = b TO (45  tb) / 7
tc = tb + c: pc = pb * c
FOR d = c TO (45  tc) / 6
td = tc + d: pd = pc * d
FOR e = d TO (45  td) / 5
te = td + e: pe = pd * e
FOR f = 9  te TO (45  te) / 4
IF f >= e THEN
tf = te + f: pf = pe * f
FOR g = 27  tf TO (45  tf) / 3
IF g >= f THEN
tg = tf + g: pg = pf * g
FOR h = 36  tg TO (45  tg) / 2
IF h >= g THEN
th = tg + h: ph = pg * h
i = 45  th
IF i < 10 AND i >= h THEN
pi = ph * i
IF pi = 362880 THEN
PRINT a; b; c; d; e; f; g; h; i
END IF
END IF
END IF
NEXT h
END IF
NEXT g
END IF
NEXT f
NEXT e
NEXT d
NEXT c
NEXT b
NEXT a
finds
1 2 3 4 5 6 7 8 9
1 2 4 4 4 5 7 9 9
the first being 9! form and the second the desired set.
In terms of my previous thoughts, that involves, relative to the 9 factorial form:
1. splitting apart the 8 into 4 and 2; losing 2 from the total and adding one number
2. splitting 6 into 3 and 2; losing 1 more from the total and adding another number
3. combining the two new 2's into 4; reducing the number of numbers by 1
4. combining the two new 3's into 9; losing all the gained 3 from the total and reducing the number of numbers by 1

Posted by Charlie
on 20050228 16:24:48 