| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 222.1 | | METOO::TOOLSHED | | Tue Feb 12 1985 14:56 | 10 |
| Assuming that all prices are given in pennies, the four prices must be
divisors of 711000000 = 2^6*3^2*5^6*79, and <711. The factor 79 can only
appear as 79, 158, 237, 316, 395, or 474, which further reduces the maximum of
the other factors to 711-79=642. Furthermore, 711^(1/4) ~1.6, so the prices
must be toward the low end of the scale. There are only a few possible
factors ending in something besides 0 and 5, which further narrows the
search. A few minutes of work yields the solution
{3.16, 1.50, 1.25, 1.20}.
Lynn Yarbrough
|
| 222.2 | | HARE::STAN | | Thu Feb 28 1985 16:41 | 12 |
| A similar problem, by J. A. H. Hunter. Crux Mathematicorum 8(1982)246,
problem 776.
Ann watched in amazement as Sam made out the check. "I said two
mugs and three plates," she reminded him.
Sam nodded. "That's right. There's no quantity discount, so it's
$4.05 for the five pieces."
"But you multiplied the two amounts instead of adding," Ann
protested.
"Sure I did, lady," replied the old man. "But it made no
difference to the total."
He was right! So what were the two prices?
|
| 222.3 | | METOO::YARBROUGH | | Fri Mar 01 1985 09:40 | 2 |
| IN this case the two prices are $2.25 and $1.80. So the mugs were $.90
and the plates $.75. - Lynn
|
| 222.4 | | HARE::STAN | | Fri Mar 01 1985 11:37 | 4 |
| Okay, now can you generalize to other sets of numbers?
For example, can you find a price such that 3 item's costs (uniquely)
both sum and multiply to this value?
|
| 222.5 | | METOO::YARBROUGH | | Fri Mar 01 1985 12:55 | 4 |
| I wrote a quick and dirty BASIC program to look for them. It found
1.00+1.25+9.00 = 11.25 and
1.00+2.25+2.60 5.85
and there are probably more.
|
| 222.6 | | METOO::YARBROUGH | | Fri Mar 01 1985 13:19 | 12 |
| After further investigation I found the following:
1.50+1.75+2.00 = 5.25 (Palindrome; arithmetic progression; smallest product)
1.20+1.80+2.50 = 5.55 (Repdigit)
1.36+1.50+2.75 = 5.61
1.25+1.60+2.85 = 5.70
1.25+1.30+4.08 = 6.63
1.21+1.25+4.80 = 7.26
1.20+1.25+4.90 = 7.35
1.05+1.25+7.36 = 9.66
.75+1.75+8.00 = 10.50
There may yet be more.
|
| 222.7 | | METOO::YARBROUGH | | Fri Mar 01 1985 13:20 | 1 |
| Ooops - I mistyped 1.20+1.85+2.50 = 5.55.[29~
|
| 222.8 | | HARE::STAN | | Fri Mar 01 1985 14:40 | 1 |
| Is there a set of n such numbers for each n?
|
| 222.9 | | HARE::STAN | | Sat Mar 02 1985 00:20 | 22 |
| Here are some solutions for 5 prices:
0.30 2.20 2.40 2.50 2.50
0.40 2.00 2.00 2.30 2.50
0.50 1.30 2.00 2.50 2.80
0.50 1.50 2.00 2.00 3.00
0.50 1.60 2.00 2.20 2.50
0.60 1.00 2.40 2.50 2.50
0.70 0.80 2.50 2.50 2.60
0.70 1.20 2.00 2.00 2.50
0.80 0.80 2.20 2.50 2.50
0.80 1.00 1.50 2.50 2.90
0.80 1.00 1.70 2.50 2.50
0.80 1.00 2.00 2.10 2.50
0.80 1.20 1.40 2.50 2.50
0.80 1.50 1.50 1.80 2.50
0.90 1.20 1.50 2.00 2.50
1.00 1.00 1.00 3.00 3.00
1.00 1.00 1.40 2.00 3.00
1.00 1.00 1.80 1.80 2.50
1.00 1.00 2.00 2.00 2.00
1.00 1.30 1.50 2.00 2.00
|
| 222.10 | | HARE::STAN | | Sat Mar 02 1985 00:26 | 34 |
| Here are some solutions for 6 prices:
0.10 2.40 2.50 2.50 3.00 3.00
0.20 1.50 2.40 2.50 2.50 2.60
0.20 2.00 2.00 2.00 2.50 2.90
0.20 2.00 2.00 2.30 2.50 2.50
0.30 0.80 2.50 2.50 2.60 3.00
0.30 1.50 1.60 2.40 2.50 2.50
0.30 1.50 2.00 2.00 2.00 3.00
0.40 0.90 2.00 2.00 2.50 3.00
0.40 1.00 1.50 2.40 2.50 3.00
0.40 1.00 2.00 2.10 2.50 2.50
0.40 1.20 1.20 2.50 2.50 3.00
0.40 1.20 1.40 2.50 2.50 2.50
0.40 1.50 1.60 1.70 2.50 2.50
0.50 0.80 1.50 2.00 3.00 3.00
0.50 0.80 2.00 2.00 2.50 2.60
0.50 1.00 1.20 2.50 2.50 2.80
0.50 1.00 1.70 2.00 2.00 3.00
0.50 1.00 2.00 2.00 2.00 2.50
0.50 1.10 1.80 2.00 2.00 2.50
0.50 1.20 1.50 2.00 2.20 2.50
0.50 1.40 1.40 2.00 2.00 2.50
0.50 1.50 1.60 2.00 2.00 2.00
0.60 0.60 1.60 2.50 2.50 3.00
0.60 1.00 2.00 2.00 2.00 2.00
0.60 1.10 1.20 2.00 2.50 2.50
0.70 0.80 1.00 2.50 2.50 3.00
0.80 0.80 1.50 2.00 2.00 2.50
0.80 0.90 1.00 2.20 2.50 2.50
0.80 1.00 1.10 1.50 2.50 3.00
0.80 1.00 1.20 1.60 2.50 2.50
1.00 1.00 1.00 1.60 2.00 3.00
1.00 1.00 1.50 1.50 2.00 2.00
|