Finding a Product

$(ab-1)(bc-1)(ca-1) = a^2b^2c^2 - abc(a+b+c) + (ab+bc+ca) - 1$ . So, the problem is equivalent with $abc|ab+bc+ca-1$ . Here, $x|y$ means $x$ divides $y$ . Since $ab+bc+ca-1 \not= 0$ , $abc \le ab+bc+ca-1$ must hold. The last inequality implies $abc < ab+bc+ca \Leftrightarrow 1 < \frac{1}{a} + \frac{1}{b} + \frac{1}{c} (1)$ .

Without loss of generality, assume that $a < b < c$ . If $a \ge 3$ , then $b \ge 4$ and $c \ge 5$ . Consequently, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} < 1$ , contradicting (1). So $a = 2$ , and $\frac{1}{2} < \frac{1}{b}+\frac{1}{c} (2)$ . Similarly, if $b \ge 4$ , then $c \ge 5$ and $\frac{1}{b}+\frac{1}{c} < \frac{1}{2}$ , a contradiction. So $b = 3$ . Finally, $\frac{1}{6} < \frac{1}{c}$ , which means the possible values for $c$ are 4 and 5. Checking into the original problem, we will see that $c = 5$ is the only possible solution. So, $abc = 2\times 3 \times 5 = 30$ .

Clearly $a,b,c$ are pairwise relatively prime since $ab\equiv 1 \mod c$ , so $(a,c)=(b,c)=1$ and $(a,b)=1$ similarly. Now since $(ab-1)(bc-1)(ca-1)=(abc)^2-(a+b+c)(abc)+ab+bc+ca-1$ , we know $abc|ab+bc+ca-1$ Since $ab+bc+ca-1$ is clearly positive, we have $abc\le ab+bc+ca-1$ If $a,b,c\ge 3$ , then WLOG $a<b<c$ , so $3bc\le abc\le ab+bc+ca-1<3bc-2$ Which is a contradiction. So WLOG $a=2$ . Then $b,c$ are odd, and $bc-2b-2c+4\le 5$ so $(b-2)(c-2)\le 5$ Now assume $b,c\ge 4$ . Since they are distinct, this implies $6=(5-2)(4-2)\le (b-2)(c-2)\le 5$ Which is a contradiction, so WLOG $b=3$ . Finally, since $c|ab-1=5$ and $c\neq 1$ , we know $c=5$ (which does give a solution), so $abc=2\cdot 3\cdot 5=\boxed{30}$ .

As given in the task statement, $abc \mid (ab-1)(bc-1)(ac-1)$ .

By expanding the expression on the right, we get: $(ab-1)(bc-1)(ac-1) = a^2b^2c^2 - abc(a + b + c) + ab + bc + ac - 1$ . The first two elements of the expanded expression are clearly divisible by $abc$ , hence we can conclude that $abc \mid ab + bc + ac - 1$ , which can also be written as

$ab + bc + ac - 1 = kabc$

for some integer $k \ge 1$ . Now let's try and determine $k$ . First of all notice that because $a, b, c > 1$ , $\underline{ab + bc + ac - 1 > 0}$ . Also, $ab \le \frac{abc}{2}$ , $bc \le \frac{abc}{2}$ and $ac \le \frac{abc}{2}$ .

We can sum the previous three inequalities together to get $ab + bc + ac \le \frac{3}{2}abc$ , hence $\underline{ab + bc + ac - 1 < \frac{3}{2}abc}$ .

From the underlined inequalities we can deduce that $0 < k < \frac{3}{2}$ , and hence the only possible integral solution to this is $k = 1$ . We have successfully derived that $ab + bc + ac - 1 = abc$ .

Now, assume $a, b, c \ge 3$ . From this we could derive, similarly to the way we did before, that $ab \le \frac{abc}{3}, bc \le \frac{abc}{3}, ac \le \frac{abc}{3}$ . Adding the inequalities together yields $ab + bc + ac - 1 \le abc - 1$ , which is contradictory to the relation we have proven before, implying our assumption was wrong; hence, at least one of the integers $a,b,c$ must be equal to 2.

Let $a=2$ . Substituting the value in the obtained relation, we have $2b + bc + 2c - 1 = 2bc$ , i.e.

$(b-2)(c-2) = 3$ .

Since $b, c \ge 2$ , the only possible factorization of 3 we could use in this case is $1 * 3$ , hence we can let $b = 3, c = 5$ . Finally, we may calculate the desired value: $\boxed{abc = 2 * 3 * 5 = 30}$

Expanding $(ab-1)(bc-1)(ca-1)$ gives $a^2b^2c^2-a^2bc-ab^2c-abc^2+ab+bc+ac-1$ . Since $a^2b^2c^2\equiv0\pmod{abc}$ and $-a^2bc-abc^2-abc^2=abc(-a-b-c)\equiv0\pmod{abc}$ , the question is reduced to finding distinct positive integers $a$ , $b$ , and $c$ such that $\frac{ab+bc+ac-1}{abc}$ is an integer. WLOG, let $a<b<c$ . We have $\begin{aligned} \frac{ab+bc+ac-1}{abc}&=&\frac1a+\frac1b+\frac1c-\frac1{abc} \end{aligned}$ Note that the maximum of $\frac1a+\frac1b+\frac1c$ is when $a=2$ , $b=3$ , and $c=4$ . This gives $\frac12+\frac13+\frac14=\frac{13}{12}$ , which is greater than 1 but less than 2. Therefore, we need $\frac1a+\frac1b+\frac1c-\frac1{abc}=1$ .

We will prove that the only solution is $a=2$ , $b=3$ , and $c=5$ . If $a\not=2$ , then the maximum value of $\frac1a+\frac1b+\frac1c$ occurs when $a=3$ , $b=4$ , $c=5$ . This gives $\frac13+\frac14+\frac15=\frac{47}{60}<1$ , so there are no solutions. Thus, $a=2$ . If $b\not=3$ , then the maximum value of $\frac12+\frac1b+\frac1c$ occurs when $b=4$ and $c=5$ . This gives $\frac12+\frac14+\frac15=\frac{19}{20}<1$ , so $b=3$ .

We can now solve for $c$ from the equation $\frac12+\frac13+\frac1c-\frac1{6c}\;\;=\;\;1$ Solving gives $c=5$ . Therefore, $abc=\boxed{30}$ .

Suppose that $abc$ divides $(ab-1)(bc-1)(ca-1)$ , then exist an integer $t$ such that $abct = (ab-1)(bc-1)(ca-1)$ . If two or more of $a$ , $b$ or $c$ is even, then the product $(ab-1)(bc-1)(ca-1)$ will be odd, hence at most one of $a$ , $b$ or $c$ is even.

As $abct = (ab-1)(bc-1)(ca-1) = a^2b^2c^2 - a^2bc - ab^2c - abc^2 + ab + ac +bc -1$

then $abc$ divide $ab +ac + bc -1$ , so that $abc \leq ab + ac + bc - 1 < ab + ac + bc$ Hence $1 < \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$

Then, the sets of integers strictly greater than 1 which satisfy the above inequality are $\{2,3,4\}$ and $\{2,3,5\}$ , but the first set may not be the solution because it has two even numbers. Therefore $abc = 2\cdot3\cdot5 = 30$

Intuitively, I first expanded $(ab-1)(bc-1)(ca-1)$ ------------ (1) into

$a^{2}b^{2}c^{2} - a^{2}bc - ab^{2}c - abc^{2} + ab +bc +ca -1$ .
Then I observed that this expansion is simply $\equiv ab + bc +ca -1 \pmod{abc}$ .

This tells us that for (1) to be divisible by $abc$ , $abc | ab +bc +ca -1$ . Thus, $abc \leq ab +bc +ca -1$ .

Since, the question asks for $a,b$ and $c$ to be distinct positive integers strictly greater than $1$ , note that the smallest $abc = 2*3*4 = 24$ . Using trial and error, when $(a,b,c) = (2,3,4)$ , $abc \nmid ab +bc +ca -1$ . The next smallest $(a,b,c) = (2,3,5)$ works. Hence, $abc = 2*3*5 = \boxed {30}$ .

Note that any $abc > 30$ will lead to $abc > ab +bc +ca -1$ , hence not fulfilling the above conditions. Thus, no other solutions are possible. [I am not sure why so though. Hopefully someone can post a more insightful solution.]

1/a+1/b+1/c-1/abc is integer, less than 1/2+1/3+1/4<2 so must be 1, and a,b,c must be 2,3,5

Since $abc$ divides $(ab-1)(bc-1)(ca-1)$ , it should divides $ab+bc+ca-1$ , and thus $abc \leq ab+bc+ca-1$ . Without loss of generality, we can assume $a <b<c$ . Now for $a \geq 3$ , we have $3(ab+bc+ca-1) < 3abc-3 <3abc$ which lead to contradiction. Thus we must have $a=2$ . For $b \geq 4$ , we have $2(ab+bc+ca-1) <4bc-2 <2abc$ which lead to contradiction. Thus we must have $b=3$ and we have $6c$ divides $5c+5$ , which give $c=5$ as the only positive integer solution. Thus $abc=30$ .

If $abc$ divides $(ab-1)(bc-1)(ca-1)$ , then $abc$ divides $ab+ac+bc - 1$ , because:

$(ab-1)(bc-1)(ca-1) = a^2b^2c^2-a^2bc-ab^2c-abc^2$

$+ab+ac+bc-1 = abc(abc-a-b-c) + ab + ac + bc - 1$ .

Therefore, we only need to see if $ab + ac + bc - 1$ is divisible by $abc$ . However, consider that $a=b=c=3$ (which is clearly not a possible set to this problem). Using this set, we may see that:

$\frac{ab + ac + bc - 1}{abc} = \frac{26}{27} < 1$ ,

that is, if $min\{a,b,c\} = 3$ , then $abc$ will already be higher than $ab+ac+bc-1$ . That's easy to see, because:

$\frac{ab+ac+bc-1}{abc} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} - \frac{1}{abc} < \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq \frac{3}{min\{a,b,c\} } \leq 1$ if $min\{a,b,c\} \geq 3$ .

Knowing this, it's easy to see that one of our values is $2$ . Let's say that $a=2$ . Using the same process used above

$\frac{2(b+c) + bc - 1}{2bc} = \frac{1}{b} + \frac{1}{c} + \frac{1}{2} - \frac{1}{2bc} < \frac{1}{b} + \frac{1}{c} + \frac{1}{2} \leq \frac{1}{2} + \frac{2}{min\{b,c\} } \leq 1$ , which is only valid if $min\{b,c\} \geq 4$ .

Therefore, we may conclude that one of our numbers is $3$ . Let's say that $b=3$ . Using the previous process:

$\frac{ab + ac + bc - 1}{abc} = \frac{5 + 5c}{6c} = \frac{5}{6c} + \frac{5}{6} \leq 1$ if $c \geq 5$ .

Therefore, our only possibilities are $c=4$ or $c=5$ . It's easy to see that $c=4$ is not a solution, because:

$\frac{ab + ac + bc - 1}{abc} = \frac{25}{24}$ .

This shows us that $c=5$ (it's easy to see that this is a solution to the problem). Therefore, $abc=30$ .

Let $n= (ab-1)(bc-1)(ca-1)$ . We have $n = a^2b^2c^2-a^2bc-ab^2c-abc^2+ab+bc+ca-1$ So $abc\mid n\iff abc\mid ab+bc+ca-1$ but if $abc\mid ab+bc+ca-1$ then it must be $abc\le ab+bc+ca-1 \ (1)$ since $a,b,c$ are positive integers. Now let WLOG $a\ge b\ge c$ (we can do it because the inequality is symmetric) so we must have $abc\le ab+bc+ca-1\le 3ab-1$ i.e. $ab(3-c)\ge1$ If $c\ge3$ we would have $0\ge ab(3-c)\ge1$ , clearly impossible, so we must have $c=2$ Now, back to the text: we must have $c\mid abc\mid (ab-1)(bc-1)(ca-1)$ but $(c,bc-1)=1$ and $(c,ca-1)=1$ so it must be $c\mid ab-1$ i.e $2\mid ab-1$ so $a$ and $b$ must be odd (if at least one of the two is even, then $ab-1$ is odd). Back to $(1)$ an substitute $c=2$ : we must have $2ab\le 2a+2b+ab-1$ that is $ab-2a-2b+1\le0$ and adding 3 and factoring $(a-2)(b-2)\le 3 \ (2)$ If $b\ge4$ we would have $4\le(b-2)(b-2)\le(a-2)(b-2)\le3$ , impossible. So, $b$ can be only $2$ or $3$ , but since it must be odd $b=3$ Now, putting $b=3$ in $(2)$ we have $a\le5$ , i.e. $a=3,5$ Checking, we have $18=3\cdot3\cdot2\mid(3\cdot3-1)(3\cdot2-1)(2\cdot3-1)=8\cdot5\cdot5$ clearly false, but $30=5\cdot3\cdot2\mid(5\cdot3-1)(3\cdot2-1)(2\cdot5-1)=14\cdot5\cdot9$ is true, so $(a,b,c)=(5,3,2)$ satisfies the condition and is the only triple that works. Hence we finally have $abc=5\cdot3\cdot2=\boxed{30}$

$(ab-1)(bc-1)(ca-1) = a^2b^2c^2-a^2bc-ab^2c-abc^2+ab$

$+ac+bc-1$

$\frac{a^2b^2c^2-a^2bc-ab^2c-abc^2+ab+ac+bc-1}{abc} = abc-a-b-c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}$

Now $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}$ has to be an integer.

Clearly this must equal one, since it cannot be $0$ or $2$ . This also means that the least of $a,b,c$ cannot be greater than $3$ since $\frac{1}{4}+\frac{1}{4}+\frac{1}{4}-\frac{1}{64} < 1$ .

Setting $a$ as $2$ , $\frac{1}{2} + \frac{1}{b}+\frac{1}{c}-\frac{1}{2bc} = 1$

$\frac{2b+2c-1}{2bc} = \frac{1}{2} \Rightarrow 2bc = 4b+4c-2 \Rightarrow (b-2)(c-2) = 4$ which yields $(2,3,5)$ as a solution.

$2*3*5 = 60$ is then the answer

Assuming $a<b<c$ , since $abc|ab+bc+ac-1$ , $abc<3bc-1$ , thus $a<3$ . $a$ can only be 2. We have then $2bc|2(b+c) + bc - 1$ . Again, because $2(b+c)<2bc$ , we have $(2(b+c) + bc -1) / 2bc < 3/2$ , therefore $2(b+c) + bc - 1 = 2bc$ , or $bc = 2(b+c) -1 < 4c - 1$ . So $b\leq3$ , which means $b=3$ . Solving for $c$ gives $c=5$ .

Notice that $(a,b,c) = (2,3,5)$ works, so the answer is $\boxed {30}$ .

Just take (ab - 1), ignore the other, because what control divisible only : $\frac{ab - 1}{abc} = k$ Then simplify : $abk - k = abc$ $ab(k - c) = k$ $1 - \frac{c}{k} = \frac{1}{ab}$ $1 = \frac{c}{k} + \frac{1}{ab}$ Minimum a.b is 2 x 3 = 6, then c = 5 and k = 6. Finally $abc = 2(3)(5) = 30$

Seems like missing "minimum" word in problem because it's valid for infinity pair of (a, b, c). Try (2, 7, 13).

First, WLOG, $a < b < c$ . Now, note that $(ab-1)(ac-1)(bc-1) = abc(abc-a-b-c)+ab+ac+bc-1$ . Therefore, our problem is reduced to $abc \vert ab+ac+bc-1$ , which implies $ab+ac+bc>abc$ .

Before proceeding, let's notice that $a \nmid ab-1$ and $a \nmid ac-1$ , which means that $a \mid bc-1$ . Similarly, $b \mid ac-1$ and $c \mid ab-1$ .

Now, as we assumed $a < b < c$ , $bc=max\{ab,ac,bc\}$ , so $ab+ac+bc>a \cdot bc$ if $a<3$ , i.e. $a=2$ .

Then, let's plug $a=2$ into our inequality. We get $2(b+c)>bc$ . Let's prove that this implies $b<4$ . Let $c=b+k, k \in \mathbb{N}$ (we can do this because we assumed $a < b < c$ ). Our inequality then produces a quadratic equation $b^2+(k-2)b-k<0$ , whose roots are $\frac{2-k \pm \sqrt{k^2+4}}{2}$ . If we prove that the larger one is lesser than 4, we prove $b<4$ .

So, $\frac{2-k + \sqrt{k^2+4}}{2}\ < 4$ is equivalent to $\sqrt{k^2+4} < 6+k$ , which, when we square (the term under the square root is positive, and so is $6+k$ , since $k \in \mathbb{N}$ ), we get $-32<12k$ , which is true for all $k \in \mathbb{N}$ . Therefore, $b=3$ (since $a \neq b$ and $a<b$ and $a=2$ ).

Finally, plugging $a=2, b=3$ into $c \mid ab-1$ gives us $c=5$ .

By experience, this typical problem is likely to have the solution $(2,3,5)$ , since that solution works, and by logic since this problem must have a unique solution, therefore the solution must be $(2,3,5)$ , hence $abc=30$

Restate as $(ab-1)(bc-1)(ca-1) \equiv 0 \pmod {abc}$

Expand this, leaving out all terms that are a multiples of abc to get

$ab+bc+ca-1 = kabc$ for some integer k. Note that k has to be positive because a, b and c are greater than 1. Bring multiples of a to the left side:

$a(b+c-kbc)=1-bc$

$\Rightarrow a=\frac{bc-1}{kbc-b-c}$ with positive numerator and denominator.

Since it was given that $a>=2$ we have

$bc-1>=2(kbc-b-c)$ which we rewrite as

$2(b+c)-1>=(2k-1)bc$ .

$k <= \frac{b+c-\frac{1}{2}}{bc} + \frac{1}{2}$

For both $b,c >= 2$ we have $bc >= 2c$ and also $bc >=2b$ so that $bc >= b+c$ . Using this we can write

$k <= \frac{b+c-\frac{1}{2}}{b+c}+\frac{1}{2}$

$k <= 1 -\frac{1}{2b+2c}+\frac{1}{2} <=\frac{3}{2}$

Conclusion: $k=1$ , the only positive number smaller than $3/2$ .

We observe

• In $ab+ac+bc=abc+1$ , if a,b,c are all odd we 'd get: odd=even; if two of them, or all, are even we get: even=odd. So one is even and the other two are odd.

• Because a,b,c are different, any product ab is at least 6. If the other factor (c) gets large, the product abc soon outruns the sum ab+bc+ac+1.

So first try the lowest possible case (a=2,b=3,c=5): 6+10+15=31=30+1. This works.

abc=30. It is easily checked that indeed 5×9×14=(5×3×2)×(3×7) is a multiple of 30.

Given : abc | (ab-1)(bc-1)(ca-1) where x | y means x divides y. i.e. abc | (abc)^2 - abc(a+b+c) + ab + bc + ca - 1 Clearly abc | (abc)^2 - abc(a+b+c) So we have abc | ab + bc + ca - 1 .................(1) 4 possibilities arise in this case: i) all of a,b,c are Odd ii) all of a,b,c are Even iii) 1 of a,b,c is Even & the other 2 are Odd iv) 1 of a,b,c is Odd & 2 the other are Even Clearly (i) , (ii) & (iv) do not satisfy (1). So there is only 1 possibility - 1 of a,b,c is Even & the other 2 are Odd. Since a, b & c are distinct positive integers strictly greater than 1, considering the first & most basic case: a=2 , b=3 & c=5 we get abc=30

here abc divides (ab-1)(bc-1)(ca-1) therefore we can conclude that c divides (ab-1) because since abc divides (ab-1)(bc-1)(ca-1) therefore c divides (ab-1)(bc-1)(ca-1). but (bc-1)(ca-1) leaves remainder 1 when it is divided by c .therefore c must divides (ab-1).similarli we can say that a divides (bc-1) and b divides (ac-1). let us consider that the numbers are a, a+1,2a+1 .then we see that (a+1)(2a+1)=1(mod a) again, a(2a+1)=2a^2+a=a^2+a+a^2-1+1=a(a+1)+(a+1)(a-1)+1=(a+1)(2a-1) +1=1(mod a+1). but a(a+1)-1 not divisible by (2a+1) for all a therefore consider that , a(a+1)-1=2a+1 or, a^2+a-1=2a+1 or, a^2-a-2=0 or, (a-2)(a+1)=0 hence a=2 or a=-1 since a>1 so a=2,therefore b= a+1=3 and c=2a+1=5. hence (2,3,5)are the solution and abc = 2 3 5=30

Claim 1: $a|bc-1$ , $b|ac-1$ , $c|ab-1$ .

Proof: ac & ab are multiples of a( $\neq 1$ ),so a doesn't divide $(ac-1)$ or $(ab-1)$ ,so a must divide $(bc-1)$ . Similar case follows for b & c.

Claim 2: If any,then exactly one even lies among a,b & c.

Proof: All can be odd,but our claim is "If any,then exactly one even". So we prove any 2 can't be even. If a & b are even,then (ac-1) is odd,but $b|ac-1$ . Contradiction! Since no two can be even,clearly all 3 can't be likewisely even. Let us test when any one,say a is even. Then (bc-1) is even,as b & c are odd,(ac-1) & (ab-1) are also odd. So no parity error arises.

Without loss of generality let us take $a<b<c$ & take 'a' even. Since we are asked to find one such pair, assume that a is atleast 2,then it divides wide range of (bc-1). Then $b|2c-1$ & $c|2b-1$ . If 2c-1=mb, where m is an odd positive integer then $c=\frac {mb+1}{2}$ . Since $c|2b-1$ $\Rightarrow \frac {mb+1}{2} | 2b-1$ . Then $mb+1 | 4b-2$ . So m can be 1 or 3 (<4). If m=1, then $b+1|4b-2$ i.e., $b+1|4(b+1)-6$ i.e., $b+1|6$ . So $b \neq 0$ , $b \neq 1$ , $b \neq 2$ , b=5. Then c=(5+1)/2=3, inadmissible as b<c.

If m=3,then $3b+1|4b-2$ . Now at most $3b+1=4b-2$ ,as 4b-2 can't be a greater multiple of (3b+1) (then b would become negative)...So b=3, $c=\frac{3 \times 3 + 1}{2}=5$ . So a=2,b=3,c=5. So abc=30,as required.

P.S. a,b,c can be all odd as well, & can produce such a triplet e.g. 3,5,7.

let abc =x then x is a factor of (x-a)(x-b)(x-c)/x then we get a cubic equation and on solving with condition as x is a factor we get (1/a)+(1/b)+)(1/c)+(1/abc)=1 hence we get abc=30

By expanding the product of the 3 brackets, we get a number of terms. Some of which 'contains' the factor of abc. For eg, ab(c^2) is a multiple of abc. In other words, they are a multiple of abc. The terms which do not contain the factor abc are as follows. bc+ac+ab-1=(abc)[(1/a)+(1/b)+(1/c)- (1/abc)]

Thus, we require (1/a)+(1/b)+(1/c)-(1/abc) to be an integer. Since a, b, c are distinct positive integers greater than 1. We know that (1/a)+(1/b)+(1/c)-(1/abc) <(1/a)+(1/b)+(1/c) <(1/2)+(1/3)+(1/4) <2 This implies that (1/a)+(1/b)+(1/c)-(1/abc)=0 or 1 First consider the case of it being equal to 0. From the first equality, this is equivalent to having bc+ac+ab=1 which is impossible. Second case, (1/a)+(1/b)+(1/c)-(1/abc)=1. => bc+ac+ab-1=abc By tryin a few numbers, we know that 2, 3, 5 satisfies the eqn. Thus, abc=30.

Cheers. Mystery solved..

( ab - 1): that "-1" makes even things odd and odd things even. so we should take care. (a,b,c) cannot be all even. as we will end up with odd term in numerator and odd term in denominator which will not divide. (a,b,c) cannot have 2 even numbers and one odd number, as once again we will get odd over even. so one should be even and 2 should be odd to yield a result. smalest possible combination is (a,b,c) = (2,3,5) 2 x 3 x 5 = 30