It’s Father’s Day, when America celebrates all our great dads. But one animal dad who gets a shout-out is the gafftopsail catfish. After the mama catfish lays her eggs, the dad carries those eggs around in his mouth for a whole month! He can hold up to 55 of them, each as big as a golf ball. Worse yet, he can’t eat anything that whole time, or he’ll accidentally swallow his kids. When the babies hatch, they’re already 2 inches long, but he carries them for a few weeks until they’re 4 inches long — and he still can’t eat until they move out. That is one patient — and hungry — dad!
Wee ones: When the babies grow from 2 inches to 4 inches, how much longer are they when they finally “move out”?
Little kids: If the dad carries the eggs for 5 weeks, then carries the hatched babies another 4 weeks, how long does he go without eating? Bonus: If you’re counting those 55 eggs, what numbers do you say to count the last 3 eggs?
Big kids: If 1 baby fish finally swims out, then 4 fish in the next batch, then 9 fish in the next…how many swim out in the next batch? Bonus: If the 55 babies hatch and every one of them is 2 inches long, how long an end-to-end chain can they make? (Try feet and inches as an extra bonus!)
The sky’s the limit: If some of the 55 babies swim out, and there are now 7 more babies outside the mouth than inside, how many have swum out?
Wee ones: 2 inches longer.
Little kids: 9 weeks. Bonus: 53, 54, 55.
Big kids: 16 fish…the count jumps by odd numbers (+3, +5, +7) – and it’s always all perfect squares! (2×2, 3×3, 4×4). Bonus: 110 inches, or 9 feet 2 inches of fish.
The sky’s the limit: 31 have swum out, and 24 are still hiding inside. The easiest way to solve is to say, if the bigger number is the smaller number plus 7, then 55 is just two of the smaller number plus an extra 7. Take away 7 and cut in half, and you get half of 48, or 24. In algebra, we’d write it as f fish outside:
f + (f + 7) =55
2f + 7 = 55
2f = 48
f = 24 fish inside, which means 31 outside.