Prime and Composite Possibilities

Number Theory Level 3

Let $x$ be a positive integer which can be written as $a \times b$ , where $a$ and $b$ are primes (positive). We define $x_\$=ab\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)$ . Find the sum of all possible values of $x$ such that $x_\$=600$

The answer is 3335.

1 solution

Trevor Arashiro
Sep 1, 2014

First we multiply and factor our equation from $x_\$=ab(1-\frac{1}{a}-\frac{1}{b}+\frac{1}{ab})\Rightarrow ab-a-b-1$

$(a-1)(b-1)=600$ .

Next, because a,b must be prime a-1,b-1 must be even (with the exception of a or b=2). So we hvae to find the pairs of even factors of 600= (t,v)= (1, 600) (2, 300) (4, 150) (6, 100) (8, 75) (10, 60) (12, 50) (20, 30).

Next, we must check which pairs of t,v are prime when 1 is added to both numbers. The only ones that work are (2,601), (5, 151), (7,101),and (11, 61).

Thus $2\times601+5\times151+7\times101+11\times61=1202+755+707+671=3335$

(minor note) I disagree with "because a,b must be prime, a-1, b-1 must be even". Note that you gave a solution with 1, which is an odd number.

Calvin Lin Staff - 6 years, 9 months ago

Ah, thank you. Didn't catch that.

Trevor Arashiro - 6 years, 9 months ago

because 2 is only even prime

Bhuwan Vyas - 6 years, 7 months ago

Nice problem @Trevor Arashiro

Krishna Ar - 6 years, 8 months ago

Rather than multiply out $(1-\frac{1}{a})(1-\frac{1}{b})$ , it is easier to just re-arrange, and distribute: $a(1-\frac{1}{a})b(1-\frac{1}{b})=(a-1)(b-1)$

I like the problem, though.

Doug Boyd - 6 years ago

Prime and Composite Possibilities

The answer is 3335.

1 solution

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