The Farmer’s Impossible-Looking Receipt
A farmer buys 10,000 young animals at market: calves, lambs, and piglets. The receipt totals $505,000, the lambs outnumber calves three-to-one, and the piglets arrive in tidy hundreds. Can you reconstruct the herd?
The Farmer’s Impossible-Looking Receipt
At dawn, a farmer walks into the county livestock market with a ledger, a sharp pencil, and a very large truck convoy.
By noon, the auctioneer hands over a receipt for exactly 10,000 animals: calves, lambs, and piglets.
The prices were plain enough:
Calves: $130 each
Lambs: $75 each
Piglets: $25 each
The total bill came to $505,000.
But the farmer’s buying habits add two twists:
1. The farmer bought exactly three times as many lambs as calves.
2. The number of piglets was a multiple of 100.
The challenge:
How many calves, lambs, and piglets did the farmer buy?
Explore this puzzle visually with an interactive diagram — drag sliders, watch the geometry update in real time, and build intuition before you solve.
Solution
Let c be the number of calves, l the number of lambs, and p the number of piglets.
We know:
c + l + p = 10,000
130c + 75l + 25p = 505,000
l = 3c
Step 1 — Use the lamb clue.
Since l = 3c, the total animal count becomes:
c + 3c + p = 10,000
So:
4c + p = 10,000
p = 10,000 − 4c
Step 2 — Put that into the receipt.
The cost equation is:
130c + 75l + 25p = 505,000
Substitute l = 3c and p = 10,000 − 4c:
130c + 75(3c) + 25(10,000 − 4c) = 505,000
That simplifies to:
130c + 225c + 250,000 − 100c = 505,000
So:
255c + 250,000 = 505,000
255c = 255,000
c = 1,000
Step 3 — Finish the herd.
Calves: 1,000
Lambs: 3 × 1,000 = 3,000
Piglets: 10,000 − 4,000 = 6,000
The piglets condition checks out too: 6,000 is a multiple of 100.
Verification:
1,000 calves cost $130,000
3,000 lambs cost $225,000
6,000 piglets cost $150,000
Total: $130,000 + $225,000 + $150,000 = $505,000
Final answer:
The farmer bought 1,000 calves, 3,000 lambs, and 6,000 piglets.
The deeper lesson
This is a small Diophantine puzzle: a problem where whole-number answers matter. The trick is not heavy algebra, but reducing several conditions into one clean equation.
This kind of reasoning appears in budgeting, inventory planning, scheduling, packaging, shipping loads, and any situation where the answer must come in whole units rather than decimals.