General Question

nailpolishfanatic's avatar

How do I solve this problem?

Asked by nailpolishfanatic (6617points) October 1st, 2010

This is the question, and I really need help on it.

Problem 5.
The numbers 4 and 12 have a special property.
and 16 is a factor of 48.

Find some other examples of pairs of numbers such that their sum is a factor of their product.

What condition must exist for pairs of numbers to be related on this way?

Thank you in advance!

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62 Answers

Kayak8's avatar

Sounds like homework to me . . .

talljasperman's avatar

@Kayak8 yes… when he said Problem 5… it gave it away as homework

crisw's avatar

Think about squares.

bobbinhood's avatar

A couple examples to get you started:

12 is a factor of 36

25 is a factor of 100

As you find other examples that work, look for a pattern. From the pattern, try to figure out a rule. To test the rule, use it to choose other examples and see if they work (I’m assuming your class is not so advanced as to require you to actually prove the rule).

Seek's avatar


@Thesexier is specifically asking how to find the answer, not looking for the answer itself. Thus, the question is acceptable.

That said, I really can’t help you, as numbers are static in my brain. Looks like @bobbinhood is on to something though.

Kayak8's avatar

@Seek_Kolinahr so he can ask homework questions in the context of what it means, but not looking for the specific answer?

Dog's avatar

Yup! So long as the asker is not seeking the answer directly and wishes help in learning how to solve the problem it is allowed according the guidelines.

nailpolishfanatic's avatar

@bobbinhood , I came up with this:

is it fine?
But I don’t understand the last sentences :/

bobbinhood's avatar

You don’t understand the part that says “16 is a factor of 48”? That basically means that 16 times some number equals 48. In other words, if you divide 48 by 16, you get a whole number (no fractions or decimals). Does that answer your question?

bobbinhood's avatar

Oh, and yes, that example works because 16 times 4 is 64. That means that 16 is a factor of 64.

Kayak8's avatar

welcome to the number “3” as they say on Sesame Street

nailpolishfanatic's avatar

@bobbinhood , I understand now.

But then now I have:

Q: what condition must exist for pair of numbers to be related in this way?

what is the teacher talking about here? :O

Kayak8's avatar


Somehow, the number “3” is looking promising, isn’t it?

Kayak8's avatar

Many teachers care about the property of numbers, and it seems that your teacher is stuck on the number 3.

bobbinhood's avatar

@Thesexier That is the rule I was talking about. You need to come up with at least a few more examples and see what they all have in common. Write a rule that will tell you whether any two numbers will work.

For example:
Every even number with itself has this property.

That rule is true, but it is too specific. You know it is not the right rule because you have examples of numbers that work (4 and 12, 5 and 20) that do not fit in this rule. So you need to find more examples and figure out what they have in common so you can write the correct rule. The correct rule will tell you what must be true for any two numbers to have the property.

Kayak8's avatar


Kayak8's avatar

Or maybe he or she is stuck on the number 4.

bobbinhood's avatar

@Kayak8 We already have examples that have nothing to do with 3. Also, I don’t understand what your examples have to do with the property in question. Perhaps you could explain?

Kayak8's avatar

@bobbinhood I changed my answer to the number 4 instead of the number 3 as clearly, 3 is not applicable in all situations encountered thus far. I am sticking with 4, the square root of 64.

bobbinhood's avatar


Another example:
9 is a factor of 18

Even if 4 was the magic answer, I still don’t see how your list of multiplication facts helps with this…

Just so you are aware, 4 is the cube root of 64; 8 is the square root of 64.

Kayak8's avatar

OK, I give up. 3 is not the answer and 4 is not the answer. Maybe 5 is the answer.

Kayak8's avatar

@bobbinhood and you are right about the cube root/square root thing. My bad.

bobbinhood's avatar

@Kayak8 “The answer” is a general property, not a specific number.

Seek's avatar

@Kayak8 I have a feeling it’s not a number that’s the answer, but a rule. Like (this is not the answer) “All numbers are a multiple of pi” or something.

Kayak8's avatar

@bobbinhood I guess I am not much help with homework then . . .

Response moderated (Off-Topic)
bobbinhood's avatar

@Kayak8 Don’t feel bad. I’ve devoted my life to math (and the teaching of it). I’m sure there are other topics in which you could run circles around me.

Response moderated (Off-Topic)
Kayak8's avatar

@bobbinhood Homework isn’t my thing. I do OK with math in general. But thanks for the props.

Response moderated (Off-Topic)
nailpolishfanatic's avatar

so I’ve done more examples:



The thing I notice here is that all the numbers are composite. So I guess that’s the MUST condition?

@Kayak8 , yes I was reading about it on Perez Hilton and it was all over Youtube. How sad:/
I guess five isn’t a magical number it’s a “friday the 13th number“xD

bobbinhood's avatar

@Thesexier Can you find a counter example? That means, can you find two composite numbers for which this doesn’t work? If you can, then this is not the rule.

nailpolishfanatic's avatar

@bobbinhood , I found:


so the rule isn’t:
The thing I notice here is that all the numbers are composite

bobbinhood's avatar

@Thesexier Correct. Keep looking at examples and thinking. I haven’t really thought it out (I have plenty of my own work that needs doing), but I think @crisw was onto something with the suggestion to think about squares.

BonusQuestion's avatar

@Thesexier, they do not necessarily need to be both composite. I can find a general formula for all pairs of numbers satisfying that condition but a bit of work needs to be done and I am not sure how much number theory you know.

These pairs of examples have the condition and in each pair there is one prime number:

(3, 6); (5, 20); (7, 42), (11, 110)

bobbinhood's avatar

@BonusQuestion I don’t think she knows any number theory. Based on her question history, I’m fairly certain she’s in high school.

BonusQuestion's avatar

@bobbinhood I know I had some courses in number theory in high school and could fully solve this problem.

@Thesexier, Look at the problem this way: Lets say (a, b) is a pair of integers satisfying the condition. It means ab is a multiple of a+b, which means there is another integer k such that ab = k(a+b). Take everything to one side and you get ab-ka-kb =0. Can you factorize ab-ka-kb? If not, what is this expression missing in order to be factorized?

Let me know if you are following this logic and I will continue walking you through the rest of the solution.

bobbinhood's avatar

@BonusQuestion Huh. I never heard of number theory being done in high school. If I had known that was a possibility, I would have approached this question differently. Hopefully she can unterstand your approach since it makes much more sense than simply tilting your head until you can see a pattern.

nailpolishfanatic's avatar

@BonusQuestion , I still don’t understand.
But the teacher also gave us hint sheet and for Problem 5 it says:

You are looking for a formula which gives pairs of numbers a and b such that a+b is a factor of ab

@bobbinhood , I am from Iceland.

crisw's avatar

Look at the numbers that you get that fit the equation,for example:



How are 5 and 25 related? How are 6 and 36 related?

nailpolishfanatic's avatar

@crisw , you can devide 5 into 25 and 6 into 36, I dont know what they call those kind of numbers.

crisw's avatar

What do these numbers have in common?

2 and 4
3 and 9
4 and 16
5 and 25

and so on?

bobbinhood's avatar

@Thesexier Not quite. Keep thinking.

LostInParadise's avatar

@BonusQuestion indicated the general solution, but I don’t think it is reasonable to expect high school students to be able to come up with this. I do think that it is possible to come up with special cases of the general solution.

We have xy is divisible by x+y. We see in the example using 4 and 12 that 12 is divisible by 4. In this case 12=3*4. So we consider cases where x is divisible by y. Suppose, to be as concrete as possible, that x=5y. Then we have y*5y is divisible by (y+5y). That is 5y^2/6y is an integer. We can cancel y from numerator and denominator, giving 5y/6. This will be an integer if y is divisible by 6. For example, if y=6 then x = 5*6=30 and 6*30 is divisible by 6+30.

How did we get from 5 to 6? Do you see how we can generalize for numbers other than 5?

BonusQuestion's avatar

FWIW, here is the solution, but you need to fill in some details. If you can do that, you deserve to hand it in as a solution to your teacher.

We can rewrite the equality as (a-k)(b-k) = k^2. Assume c^2 and d^2 are the largest squares that divide a-k and b-k, for two non-negative integers c and d . One can look at the prime factorization of a-k and b-k and notice that:

a-k = e c^2 & b-k = e d^2

Therefore k = +/- ecd and hence:

a = ec(c+d)
b = ed(c+d)


a = ec(c-d)
b = ed(d-c)

Replace c, d ≥ 0 and e with integers and you get all integer solutions.

nailpolishfanatic's avatar

@all, shit this is so difficult.
i think am gonna give up this one…

chocolatechip's avatar

@BonusQuestion That is needlessly complicated.

Here’s an easy test to figure out the rule.

We have 2 numbers, a and b, and a sum and product, c and d.

a + b = c
a * b = d

a * X = b
c * Y = d

Try a bunch of different sets of numbers. For example, a=2, b=2, a=5, b=20, a=3, b=5. Fill in the equations above, and figure out what X and Y are.

For the sets of numbers that fulfill the special property, what is the relationship between X and Y?

nailpolishfanatic's avatar

I don’t know:/

LostInParadise's avatar

@BonusQuestion , Well done and easy to follow, but I don’t think there are many of us here, let alone a typical high school student, who could have come up with the solution.

gasman's avatar

The original question:
Find…pairs of numbers such that their sum is a factor of their product.

@BonusQuestion‘s solution is elegantly stated!

We can rewrite the equality as (a-k)(b-k) = k^2. Assume c^2 and d^2 are the largest squares that divide a-k and b-k, for two non-negative integers c and d . One can look at the prime factorization of a-k and b-k and notice that: a-k = e c^2 & b-k = e d^2 Therefore k = +/- ecd and hence:

a = ec(c+d) [and] b = ed(c+d)
a = c(c-d) [and] b = ed(d-c)

Replace c, d ≥ 0 and e with integers and you get all integer solutions.

Sorry but I’m a little fuzzy on why the same factor e must be shared by (a-k) and (b-k). I agree with the basic equation. Do both c^2 and d^2 each divide both factors a-k and b-k? Then I think I get it. I’m having trouble reconciling the solution pairs (a,b) which you exemplified earlier by ”(3, 6); (5, 20); (7, 42), (11, 110)” in an earlier post. If the given pair is (n, n*(n-1)) then what are the corresponding values c, d, and e?

I, too, took a course in number theory in high school, which included proving the fundamental theorem of arithmetic, aka unique factorization theorem. Although I generally excelled in math I found number theory (‘queen of mathematics’) to be quite difficult. I guess my brain doesn’t visualize modular congruences that well… ~

When I saw this problem yesterday I soon realized that a and b can’t both be odd integers, reasoning by parity—an even sum won’t divide an odd product. Thus, at least one of the numbers must be even.

This didn’t seem to lead anywhere, however, so I dismissed it as a useless and irrelevant fact. Apparently this was a wise choice…~

LostInParadise's avatar

e is the greatest common divisor (gcd) of (a-k) and (b-k). If (a-k) and (b-k) do not have any prime factors in common then e=1. The significant thing is that if you divide (a-k) and (b-k) by e then you must get squares c^2 and d^2 in order for (a-k)(b-k) to be a perfect square.

The solution a=n, b=n(n-1) corresponds to c=1, d= n-1, e=1

gasman's avatar

@LostInParadise OK, that helps. And it’s easy to show that b is always an even number.

LostInParadise's avatar

One final comment on this problem and @BonusQuestion’ s solution.
It should be apparent from the problem statement that if a and b form a solution then so do multiples ma and mb. We can simplify things if we look only for those solutions that are not multiples of another solution. To do this, we eliminate e from
giving a=c(c+d)
We also have to require that c and d do not have a common factor.

The same thing applies to the second set of equations, which will always have one negative value.

LostInParadise's avatar

Correction: It is okay for c and d to have a common factor

gasman's avatar

This problem—so simply stated—turned out to be a real gem. Hey, @Thesexier, what kind of math(s) class are you taking? If that’s Problem 5, I’d like to see Problem 6 ! Do you still live in Iceland?

BonusQuestion's avatar

@LostInParadise If you are only looking for the “smallest” solutions you need to require c and d to be co-prime. Otherwise when you divide c and d by their g.c.d you get a new solution. Here is an alternative solution that also gives us those “smallest” solutions.

Assume g is the g.c.d. of a and b and say a = mg and b = ng. We have (m+n)g divides mng^2. Therefore m+n divides mng. But since m and n are co-prime, so are mn and m+n. Hence m+n divides g. Therefore we should have, g = (m+n)k and:

a = k (m+n)m and b = k (m+n) n

where m and n are co-prime.

Since we are looking for the “smallest” solutions we need to set k=1.

LostInParadise's avatar

You are correct. We don’t need e for any of the solutions. We can simply multiply c and d by a common factor m. This will increase a and b by a factor of m^2, In order to get all values of a and b, we would then have to allow multiplication by square root of m, which may give non-integer values for c and d.

nailpolishfanatic's avatar

@gasman , A gem? – Do you mean that it’s hard? – if so I agree!

The math book am using is Mathematics for the international student pre-diploma SL and HL (MYP 5 PLUS)
Second Edition.

But the problems am doing are just hand outs. It’s called :Algebra Challenge 2.
I got 7 on the first Algebra Chanllenge.

Problem 6:
An AP rectangle is one whose area is numerically equal to its perimeter. For example the rectangle 10 units by 2.5 units has area 25 square units and petimeter 25 units. If you are given the length of a side can you always fine an AP rectangle with one side the given length? Explain your answer.

I don’t understand ANYTHING!

LostInParadise's avatar

This one I think you might be able to do. First understand the example. The rectangle is 10 by 2.5 so its area is 10*2.5=25. The perimeter = 10 + 10 + 2.5 + 2.5 = 25, which is the same as the area. They want to know if you are given one side of a rectangle, can you always find the length of the other side so that its area and perimeter are equal.

I suggest doing the following steps:

1. Choose a number, say 7, to be one of the lengths of a rectangle. Call the other side x. Write the formulas for area and perimeter in terms of x. Set the two equations equal to one another. Solve for x. This gives the rectangle given a specific starting point. We want to generalize from a specific starting point like 7 to any starting point.

2. Replace 7 in the previous step with a variable k. This is the known side of the rectangle. Again let x stand for the unknown side. Write the equations for area and perimeter in terms of k and x and, as in the previous step, set the equations equal to one another, and solve for x in terms of k.

3. Look at the equation from the last step and see if there are any values of k that make it impossible to solve for x. If such k values exist that would mean that there are starting lengths for which it is impossible to find a second length which will make the area and perimeter equal.

nailpolishfanatic's avatar

@LostInParadise , thank you very much I am now finally finished with my work.
I understand it now after translating perimeter and area into Icelandic.
Also my mom read your response and she was doing it with me.

@all, thank you everyone else for the help! I appreciate it very well.

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