prime factors

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motorhero

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Hello all, I was asked by my daughter below question.
A number N has 3 and 7 as the only prime factors. If we multiply it by 63 the number of factors increases by 10, but if we multiply it by 49, the number of factors increases by 4. Find the number N. Justify your answer.
Can someone help me out? thank you.
 
motorhero said:
Hello all, I was asked by my daughter below question.
A number N has 3 and 7 as the only prime factors. If we multiply it by 63 the number of factors increases by 10, but if we multiply it by 49, the number of factors increases by 4. Find the number N. Justify your answer.
Can someone help me out? thank you.
Click to expand...
Hi motorhero,
Can you clarify what it means by "the number of factors increases by 10" ?
:)
 
You have not said what your daughter is studying. If she knows algebra, this is a tedious but easy problem.

[math]n = 3^p * 7^q, \text { where } p \text { and } q \text { are integers } \ge 0.[/math]
Therefore n has [imath]pq + p + q + 1[/imath] distinct divisors. Do you see why?

Therefore 63n has how many distinct divisors? How many distinct divisors does 49n have?

Now what?
 
JeffM said:
You have not said what your daughter is studying. If she knows algebra, this is a tedious but easy problem.

[math]n = 3^p * 7^q, \text { where } p \text { and } q \text { are integers } \ge 0.[/math]
Therefore n has [imath]pq + p + q + 1[/imath] distinct divisors. Do you see why?

Therefore 63n has how many distinct divisors? How many distinct divisors does 49n have?

Now what?
Click to expand...
hi Jeff
she is in year 7, (13 years old). I am surprised her math question is so hard. After I read your post, the question became easier now :)
thanks
 
JeffM said:
You have not said what your daughter is studying. If she knows algebra, this is a tedious but easy problem.

[math]n = 3^p * 7^q, \text { where } p \text { and } q \text { are integers } \ge 0.[/math]
Therefore n has [imath]pq + p + q + 1[/imath] distinct divisors. Do you see why?

Therefore 63n has how many distinct divisors? How many distinct divisors does 49n have?

Now what?
Click to expand...
I don't see it. Can you explain?
 
AvgStudent said:
I don't see it. Can you explain?
Click to expand...
Given [imath]n=3^p7^q[/imath], any factor (divisor) of n will be [imath]3^x7^y[/imath] where [imath]0\le x\le p[/imath] and [imath]0\le y\le q[/imath]. There are [imath](p+1)[/imath] ways to choose x, and [imath](q+1)[/imath] ways to choose y, for a total of [imath](p+1)(q+1)=pq+p+q+1[/imath] possible divisors.
 
AvgStudent said:
I don't see it. Can you explain?
Click to expand...
Did Dr. Peterson address your question?

We end up with a system of two linear equations in two unknowns that is quite easy to solve. Getting to those two equations, however, may require a lot of thought about counting divisors. I am not sure that the problem is fair to a 13 year old, who is extremely unlikely to have even heard about number theory. I am not saying that an intelligent 13 year old is unable to think it through, but I suspect it would be a painful process. Number theory and discrete mathematics are usually taught in college.
 
This post is under probability. This means to me that the student should know how to count. There are p+1 possibilities for the power of 3 and q+1 possibilities for the power of 7. The student should be able to count all the different numbers of powers.

I too am surprised that this problem is for a 7th grader. However, if she is learning probability, then she should have learned how to handle such a problem.

This problem can be taught/come from a number theory or probability theory course.
 
Steven G said:
This post is under probability. This means to me that the student should know how to count. There are p+1 possibilities for the power of 3 and q+1 possibilities for the power of 7. The student should be able to count all the different numbers of powers.

I too am surprised that this problem is for a 7th grader. However, if she is learning probability, then she should have learned how to handle such a problem.

This problem can be taught/come from a number theory or probability theory course.
Click to expand...
the is question is not from school my daughter is in, it is from enhanced coaching class but it is for year 7. I have master degree and I did not how to do it until I found this wonderful website and really helpful people. thank you all.
 
Steven G said:
This post is under probability. This means to me that the student should know how to count. There are p+1 possibilities for the power of 3 and q+1 possibilities for the power of 7. The student should be able to count all the different numbers of powers.

I too am surprised that this problem is for a 7th grader. However, if she is learning probability, then she should have learned how to handle such a problem.

This problem can be taught/come from a number theory or probability theory course.
Click to expand...
Steven, I must respectfully disagree. At least in the U.S., a child in seventh grade has no background in number theory or discrete mathematics, and has at best an introduction to combinatorics. In general, they are just starting to learn algebra. The application of combinatorics to this problem is possible only if you know enough algebra and number theory to formulate

[math]n = 3^p7^q, \text { where } p \text { and } q \text { are integers} \ge 0.[/math]
Teaching students about exponents (including zero and one, which do not fit into the repeated multiplication definition frequently used as an introduction to exponents) usually comes late in the first year of high school algebra. Teaching basic combinatorics is usually in second year algebra.

This strikes me as a problem that someone who already understands the rudiments of combinatorics and the Fundamental Theorem of Arithmetic sees as a sort of neat little puzzle that leads to a system of two linear equations in two unknowns. No deep knowledge of any branch of mathematics is required. What is required is bits of knowledge drawn from several disparate areas of mathematics. AvgStudent is far more than an average student, and I suspect that he studied first-year algebra at least two or three years ago. But his admission that he did not understand my first post seems to prove that attacking this problem requires a breadth of mathematical knowledge that is very rare among students just being introduced to algebra.
 
Jeff, I didn't exactly disagree with you. At the end I said that I too am surprised that this problem is for a 7th grader. My attitude is that if a student is given a problem, 7th grade or not, then they should have learned the material to work out the problem. The only exceptions that I can think of is if the problem is a competition problem or a challenge problem.
 
Steven G said:
Jeff, I didn't exactly disagree with you. At the end I said that I too am surprised that this problem is for a 7th grader. My attitude is that if a student is given a problem, 7th grade or not, then they should have learned the material to work out the problem. The only exceptions that I can think of is if the problem is a competition problem or a challenge problem.
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Sorry. I missed your meaning. Yes, I fully agree: a student should not be given a problem that requires tools and knowledge that they have not been at least exposed to.

But it is easy to forget just how much a child of 13 has not been exposed to.
 
JeffM said:
Did Dr. Peterson address your question?

We end up with a system of two linear equations in two unknowns that is quite easy to solve. Getting to those two equations, however, may require a lot of thought about counting divisors. I am not sure that the problem is fair to a 13 year old, who is extremely unlikely to have even heard about number theory. I am not saying that an intelligent 13 year old is unable to think it through, but I suspect it would be a painful process. Number theory and discrete mathematics are usually taught in college.
Click to expand...
Yes. At first, I didn't understand why (p+1) instead of just p ways. But then I realized we're counting from 0 and not 1. Similarly for (q+1).
 
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