| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Here is the problem:
"A farmer sells 7 cows at a certain price. Later he sells
6 more for 4 dollars per head less. The sum totals in the
two cases contained the same two digits in reverse order.
Find the cost of the cows."
Enjoy,
Kostas G.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 628.1 | Piece of cake | SMURF::DIKE | Thu Dec 11 1986 15:39 | 22 | |
He sells 7 cows at $x apiece.
He sells 6 cows at $x-4 apiece.
7x = (a)(b) (the digit a followed by the digit b)
6x-24 = (b)(a) Subtracting...
x + 24 = (a)(b) - (b)(a), which is divisible by 9.
x is in the set { 3, 12, 21, ...), since x + 3 = 27 = 3*9
x + 12 = 36 = 4*9, etc.
If x=3, then he paid the recipient of the 6 cows $1 apiece to take
them, so I'll throw it out.
If x=12, then he got 7*12 = $84 for the first lot, and
6*8 = $48 for the second lot.
If x=21, then he got 7*21 = $147, which is not a 2-digit number,
so we stop.
Jeff
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