Monotonize!

Number Theory Level 3

Find the number of triples of positive integers $a \leq b \leq c$ such that

$abc - 2 = a + b + c.$

The answer is 3.

1 solution

Sagnik Dutta
Apr 1, 2014

The equation we must solve is
abc = a+b+c+2.

Monotonize: assume without loss of generality that a ≤ b ≤ c. Then a+b+c ≤ 3c. Hence, we have abc ≤ 2+3c.

If c > 2, then 2 + 3c < 4c, which means that ab < 4. So there are just a few cases: 1. c ≤ 2: This leads to the solution (2,2,2) only.
2. c > 2 and ab = 1: This implies 1 ·c = 1+1+c+2, an impossibility.
3. c > 2 and ab = 2: This leads to (1,2,5).
4. c > 2 and ab = 3: This leads to (1,3,3)

Hence......... the number of solutions=3 Q.E.D............

Should a=1, b=2, c=5 be considered a different solution from a=5, b=2, c=1? If no, please clarify your problem statement.

Edit: the question has been updated. This is no longer an issue.

Calvin Lin Staff - 7 years, 2 months ago

Yeah, that threw me off. :D

Finn Hulse - 7 years, 2 months ago

No. a=5,b=2,c=1 cannot be solution because a is less than or equal to b which is less than or equal to c

Penti Rohit - 7 years, 2 months ago

Note that the comment has been made some time back. The question had been updated to reflect this, and has been corrected. Let me edit my original comment for clarity.

Calvin Lin Staff - 7 years, 2 months ago

sorry... Sir...! I forgot to mention that condition.....!

Sagnik Dutta - 7 years, 2 months ago

Yes, Calvin. This is another solution. And because of this mistake, my rating had decreased

Dang Anh Tu - 7 years, 2 months ago

Your answer is so simply. I solved it in a more complex way. The first, I must prove a and b not equal 1, simultaneously. Then, based on monotone function to find all solutions.

Ân Thiện - 7 years, 2 months ago

good....Q: Find the equation whose roots are 1-a/1+a, 1-b/1+b, 1-c/1+c if a,b,c are the roots of x^3 - x - 1 =0

vinod trivedi - 7 years, 1 month ago

Monotonize!

The answer is 3.

1 solution

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