4

The product of two numbers is 10. One of them is $a$. Express their sum in terms of $a$.

The factors of 10 are 1,2,5,10. Thus $a \in {1,2,5,10}$ Thus the sum can be either 7 or 11.

Now how does one express the sum in terms of $a$?

How do I express the non-$a$ factor?

I can only think of $a + 10/a$. But that seems too simple, and I feel like I am not understanding the question correctly.

yiyi
  • 7,352
  • 2
    That seems like the correct answer to me. (Also note that $a$ is never assumed to be an integer, so your factor analysis, while correct, doesn't seem relevant to me.) – Potato Aug 23 '13 at 02:11
  • @Potato Thanks, didn't realize that $a$ could be a non-integer. Mostly worried because English is not my first language, so I felt like I wasn't understanding the question correctly. – yiyi Aug 23 '13 at 02:14

3 Answers3

3

Your expression is alright

Let the two numbers be $a$ and $b$ then $a*b=10$ or $b=\frac{10}a$

The sum of the two numbers would be $a+b=a+\frac{10}a$.

Shobhit
  • 6,902
3

$a$ is not assumed to be an integer, so $a$ does not have to be one of $1, 2, 5, 10$. But you are correct that the sum is $a + \frac{10}{a}$.

3

While this is unlikely the question that is really being asked, but if we do assume the "numbers" are positive integers, we can give a purely polynomial expression for $a+10/a$ in terms of $a$. Consider

$$\frac{(a-5)(a-2)(a-10)11}{-36}+\frac{(a-5)(a-2)(a-1)11}{360}+\frac{(a-10)(a-2)(a-1)7}{-60}+\frac{(a-5)(a-10)(a-1)7}{24}$$

Note that three of the terms vanish for $a=1,2,5$ or $10$, so it is enough to choose coeffiecnts of each term such that the non-vanishing term returns the correct answer.

PVAL-inactive
  • 8,551
  • it would be nice if you explained how you came up with that answer. – yiyi Aug 23 '13 at 02:56
  • Pval, I really like your solution, its neat; however, I don't understand how to make a general solution for this style of problem, could you explain the process used? – yiyi Aug 23 '13 at 03:09
  • I allowed a term for each value of the polynomial I wanted to force. Since I wanted $f(5)=7,f(2)=7,f(10)=11,f(1)=11$, I looked at polynomials of the form $r_1{(a-5)(a-2)(a-10)}+r_2(a-5)(a-2)(a-1)+r_3(a-10)(a-2)(a-1)+r_4(a-5)(a-10)(a-1)$. Since all except one of these terms vanish at each divisor of ten, for each divisor of ten I can just look at one of the $r_i$ to force the value I want. – PVAL-inactive Aug 23 '13 at 03:18