Homework problem

This was in the 8yr old’s homework.
This is the question in its entirety.
“A number has eight factors. Two of the factors are 10 and 14. What is the number and what are all of the factors?”

What keeps the factors (and the subsequent multiple) from being any arbitrary numbers?

Am I missing something?

This is stated strangely, but here’s what it means: A (positive) number is (evenly) divisible by (exactly) eight (positive) numbers, including 1 and itself. Two of those numbers are 10 and 14. It must be true then that the number is divisible by 1, 2, 5, 7, and 14. If that is true, it must also be divisible by 35 and 70. 70 is divisible by 1, 2, 5, 7, 10, 14, 35, and 70, so 70 must be the answer.

Nm - hit send too early.

And then Wendall Wagner jumped in while I was still doing sums on my fingers!

Oh to answer the question in your last line: if it has only 8 factors, that is the limiting fact. You have to find a number that is evenly divisible by 10 and 14, with no more than 8 factors, so it can’t be purely arbitrary or infinite.

Factors in Grade 2 or 3? That seems early, unless I’m misremembering.

Well, we know that the factors of 10 are 1, 2, 5 and 10. The factors of 14 are 1, 2, 7, and 14.

The answer, therefore will have 1, 2, 5, 7, 10, 14 and itself as factors. 5 times 7 = 35, so that’s another factor. and 2 * 35 = 70. Which is the answer.

So the 8 factors are 1, 2, 5, 7, 10, 14, 35 and 70.

qed or something.

I agree with your solution, but that seems like a really tough problem for an 8-year-old. What is that, third grade? I was just learning what multiplication was in third grade. This looks more like an 8th grade problem at least, and I wouldn’t be surprised to find it in a high school text.

I’m curious about something. (And in case this sounds censorious or snarky, I apologize in advance; that’s not my intent. I’m just trying to get an idea of how you approached the problem).

Before you posted your question, did you try to discover at least two numbers that fit all the conditions? In other words, your concern was that there were an infinitely large set of solutions that fit the conditions – did you stop with that general sense or did you actually generate a couple of solutions?

That was my thought as well. I don’t exactly remember when I did factors, but I think it was maybe around 5th or 6th grade. I would have a tough time doing the problem today (I just skipped to the answer, so I’m not sure how long it would have taken me to work it out).

Do they teach the idea of unique prime factorization in third grade now? One way to think about this is that any number can only be written in one way as the product of prime numbers. If a number is divisible by 10, then 2 and 5 must be two of those primes. If a number is divisible by 14, then 2 and 7 must be two of those primes. So the primes in its unique factorization must include 2, 5, and 7. 2 times 5 times 7 equals 70.

To know how hard this is for the students, I’d have to know much more about what is being taught to the students.

The first time factors appear in the Common Core is Grade 6, although there it’s in the context of finding Greatest Common Factors and Least Common Multiples.

It seems like a tough problem for Grade 3, although I can see it as somewhat approachable with guidance or just as an advanced “challenge” problem. Seems tough for the regular curriculum.

I did not generate any attempts at a solution.
:smack:
I read it, posted then went to bed.

the way I had been thinking of it, a set of 6 prime numbers > 7 would have been what was needed [though I hadn’t gotten as far as articulating of the prime part yet]

e.g.
10, 14, 11, 13, 17, 19, 23, 29

but that doesn’t count that the 10 and fourteen each have factors which are necessarily included

When I think “factors,” I think “prime factors.” So, seeing 10 and 14 in the list was confusing.

I had the same problem. I at first assumed that “factors” meant “prime factors.” I would call 10 and 14 divisors of 70, not factors of it. I had to think for a bit before realizing that by the term “factors” they meant “divisors.” PatriotX, I would really like to know what the book your child uses for arithmetic says about this subject and what the teacher says about the subject.

I ask this because the following situation happens too often: The parent, who hardly ever talks to their child about the arithmetic they are studying and hardly ever looks through their arithmetic book, one day glances at a homework sheet. There they see a problem that they can’t immediately solve. Often this is not because the problem is particularly hard for a child of that age but because it uses a term that the parent isn’t familiar with, although the term refers to an idea that the parent knows well but under another name. Or the problem will be of a sort that isn’t particularly hard for a child of that age, although it is different than the problems that the parent did at that age but is about the same difficulty. The parent will then explode in anger, complaining that the child is being forced to do hopelessly difficult arithmetic problems. I would honestly like to know if your child is being asked to do really hard problems or just different problems from what you know or just different terms.

No one has quite explained why there cannot be two solutions. I was thinking about a complicated explanation involving multiplicative functions when my non-mathematician wife came up with a simple explanation. A number divisible by 10 and 14 must be divisible by their least common multiple 70. 70 does have exactly 8 divisors, as does any product of three distinct prime. Any other solution would have to divisible by 70 and hence would have more divisors, itself and the 8 divisors of 70.

But really, 8 year olds? This is crazy. I assume they had seen similar problems solved, but even so.

Idk the name of the book, but here’s the shot I was sent.
I looked at it last night when I noticed that I had gotten a text from the teacher earlier in the day.
https://www.remind.com/messages/f842b91e-6b6c-4206-9fc4-50024c53226f
(hopefully that will work for you guys)

Not sure what the teacher has said. But it was not an issue for the kid–I asked him about it this afternoon.

I didn’t get angry or anything.
It was just the one problem on the page whose answer didn’t “just come to me.”
I figured that I was missing something but hoped that I had found an error.

When I asked the kid this afternoon, he found the problem unremarkable.

When I read “factors”, I was thrown that the numbers weren’t prime.
I think I thought, “If the factors aren’t going to be prime, then they could be anything.”
And from there to any other random six numbers. Multiply 10 * 14 * X * Y * Z etc and then you have a number with 8 factors, 2 of which are 10 and 14.

I failed to sufficiently appreciate how the problem was limited.

I do rarely see the kid’s homework.
The kid usually completes his homework at school in the time alotted.

Thanks, PatriotX, you’ve answered my question.

That happened to me at the worst possible time — when I took the SAT. I was a math whiz, and I could factor quadratic equations at a glance, but the math portion had several questions asking for the roots of an equation, and I’d simply never heard that term before, except in the context of square and cube roots. Probably lost 200 points on it, between missing the questions themselves, and spending way too much time trying to figure out what they were asking (this was long before test prep was a big industry).