I can be a sucker for headlines like "Second grade math question has the internet fighting over the answer," and this time I was caught despite knowing it was click-bait. Here's the question:

A mother posted what she said was her child’s second-grade math homework. Here’s how the question goes: “

There are 49 dogs signed up to compete in a dog show. There are 36 more small dogs than large dogs signed up to compete. How many small dogs are signed up to compete?”

I took it seriously, knowing there must be a trick or it wouldn't be controversial. But I quickly dismissed questions like, "How do you draw the line between large and small dogs?" as being uninteresting, and decided to take the problem at face value.

It's one of the easiest, most basic algebra problems. If you are wondering why an algebra problem was given to second graders—well, I've seen children even younger solve such puzzles through logic and an intuitive sense of the world that school hasn't yet had a chance to squeeze out of them.

So. Simple algebra. (It's a lot faster and easier than it looks, written out, but bear with me.)

- Let L = the number of large dogs.
- Let S = the number of small dogs
- We know L + S = 49
- Therefore L = 49 - S
- We know S = L + 36
- Therefore S = (49 - S) + 36 [substituting (49 - S) for L]
- Therefore S = (49+36) - S
- Therefore S = 85 - S
- Therefore 2S = 85 [adding S to both sides]
- Therefore S = 42.5

That's where the Algebra I student stops. The second grader knows better: "Half a dog? Are you crazy? Math makes no sense."

Math makes plenty of sense. All too often, math textbooks and tests do not.

The teacher whose homework assignment caused the kerfuffle admitted that the problem itself was wrong, casting the blame on the school district. She confirmed, however, that 42.5 would be the correct answer, "if done at an age appropriate grade."

No. Just no.

This is why bridges fall down.

Math word problems that purport to be about real-world situations should reasonably conform to reality. If you want to use those numbers, maybe try something like, "Tom and Lisa shared a box of 49 Krispy Kreme donuts. Tom ate 36 more than Lisa. How many did Tom eat?" ("Enough for a massive stomach ache," would be my answer.) Or just change the numbers! (Maybe the error was a misprint.)

The only correct answer to the problem as stated is, "This question is not answerable."

**If we teach our children that it's okay to plug numbers into an algorithm (or a computer model) and make decisions based on the results without considering whether or not the answers actually make real-world sense; if we teach them that giving a wrong answer is better than saying, "we don't know," or even "we can't know"; then bridges are going to collapse, windows are going to blow out of airplanes, economies are going to crash, countries are going to start wars, and people are going to die.**

Although it made Tampa's NBC News affiliate only two days ago, the story is at least five years old, as you can see here (if you don't mind the language).

For what it's worth, this was my thought process once I realized that the problem was the problem.

- The total number of dogs, 49, is an odd number.
- The only way two integers can sum to an odd number is if one of them is odd, and the other is even.
- The number of small dogs and the number of large dogs differs by 36, which is an even number.
- The only way two integers can differ by an even number is if they are either both odd, or both even.
- (2) and (4) are mutually exclusive conditions.
- Since dogs must be counted by positive integers and not fractions, the problem is not solvable.

There's one more difficulty in this puzzling and disturbing report.

The Tampa news team ends the story with their own solution:

The digital team from Nexstar affiliate WTAJ took a crack at it, with their own Olivia Bosar determining the following:

“You first subtract the 36 to group them as small dogs since you need at least 36 small dogs. You then divide the remaining 13 into two categories: large dogs and small dogs. But you have to divide them in a way that would give you the ability to create a class that’s X and a class that’s X+36. It’s not possible in this equation because 13 is a prime number.”

As a prime number, once you try to divide 13, you end up with a .5 in your answer, and, well, obviously, you can’t have half of a dog.

I give them credit for recognizing that you can't have half a dog, at least not if you want him alive enough to compete in a show. But what does the fact that 13 is a prime number have to do with anything? If you make the total number of dogs 51 instead of 49, then after you subtract 36 the number of remaining dogs is 15. Fifteen is not prime, but the problem is jut as insoluble.

Back in elementary school, I loved what was then called the "New Math." Working with sets, and other bases—I ate it up. (Tom Lehrer has a great song about it.) But today's Newer Math? I have my doubts.

I really beat into my students "Make sure your answer makes sense!" They get penalized more for a nonsense answer than a sensible wrong one unless they say "I know my answer is wrong but I can't find my mistake." As soon as I read this problem I thought "parity problem here!"

Which is one reason why you win teaching awards, Kathy. And why your classes in the Gambia were twice the size of other classes. There are still students who realize that more rigor = more learning.

Explain to me why the answer was not a proper second-grade-simplicity 36. Anything else sounds like the gobbledygook of New Math (which my grandchildren suffered through).

Grace, the number of small dogs is 36 MORE THAN the number of large dogs. If the answer were 36, the number of large dogs would have to be 0, which cannot be, since there are 49 total dogs.