Restricted Triples

First, setting $a = b = c$ would show that $a \leq 18$ . [Note, this is not a correct argument. How would you show that $a \leq 18$ ?]

Then change the equation from $ab+ac+bc=1001$ to $(a+b)(a+c)=1001+a^2$ .

For $a = 1$ , we get that $(1+b)(1+c)=1002=2 \times 3 \times 167$ , using the requirement that $a < b < c$ , we get that $(1+b) = 3$ or $6$ . Continue this with $a = 2, \ldots, 18$ .

You will find that for:
$a = 1$ , there are 2 solutions
$a = 2$ , there are 2 solutions
$a = 3$ , there are 1 solutions
$a = 4$ , there are 1 solutions
$a = 5$ , there are 3 solutions
$a = 6$ , there are 1 solutions
$a = 7$ , there are 4 solutions
$a = 8$ , there are 0 solutions
$a = 9$ , there are 0 solutions
$a = 10$ , there are 0 solutions
$a = 11$ , there are 1 solutions
$a = 12$ , there are 0 solutions
$a = 13$ , there are 1 solutions
$a = 14$ , there are 0 solutions
$a = 15$ , there are 0 solutions
$a = 16$ , there are 0 solutions
$a = 17$ , there are 0 solutions
$a = 18$ , there are 0 solutions

Totaling to 16 solutions.

From rearrangement inequality we know that ab+bc+ca will have the greatest value when a>b>c and least value when a<b<c now we can find the upper and lower bound from this condition. 3 a^2<ab+bc+ca<3 c^2 . Now from the given value of ab+bc+ca we can find the upper bound on a . 3 a^2<1001 , this reduces to a<19 which tells us, a can be from 1 to 18. Now if we put the value of a in the equation given we will have ( lets put 18 first ) 18b+18c+bc=1001 adding 18^2 at both sides we get (b+18)(c+18)=1001+18^2 , now 1001 +18^2 = 5^2 53 now in no way this number (1001+18^2) can be written as a product of two numbers which will yield such a value of b and c satisfying the given condition (a<b<c) . Similarly we can put a=17,16,15, up to 1 and can look for the values of b and c which satisfy the given condition. The solution triplets will be ( 1,2,333), (1,5,166), (2,13,65),(2,3,199),(3,7,98), (4,5,109),(5,14,49),(5,13,52),(5,22,23) ,(6,11,55), (7,23,28), (7,18,35),(7,8,63),(7,14,43), (11,22,23) ,(13,17,26 )

Consider the case a=1. Then bc+b+c=1001, so (b+1)(c+1)=1002=2 3 167. b>a=1, so b+1>2. Then b+1 can be 3 or 167. There are 2 solutions here.

If a=2, bc+2b+2c=1001, so (b+2)(c+2) = 1005=3 5 67. Since b>2, b+2>4, so b+2 can be 5 or 15. There are 2 solutions here.

Continuing on, we see that a=3 gives 1 solution, a=4 gives 1 solution, a=5 gives 3 solutions, a=6 gives 1 solution, a=7 gives 4 solutions, a=11 gives 1 solution, and a=13 gives 1 solution for 16 solutions. [When I originally did this I found 2 solutions for a=6, but somehow added wrong, giving the correct answer of 16 through an incorrect method. I reviewed my work and found the error in a=6.]

(1,2,333) (1,5,166)

(2,3,199) (2,13,65)

(3,7,98)

(4,5,109)

(5,13,52) (5,14,49) (5,22,33)

(6,11,55)

(7,8,63) (7,14,43) (7,18,35) (7,23,28)

(11,22,23)

(13,17,26) 16 triples of positive integers (a,b,c)

I use $(b+a)(c+a)=1001+a^{2}$ And take a=1,2,3,4,5,6,...,17 because a<b<c.

you have given ab+ac+bc=1001 and a<b<c.so, c=[(1001-ab)/(a+b)] so, 0<a<b<sqrt[(a^2)+1001]-a so, 0<a<sqrt(1001/3)=18.266...... so, the integer solutions for a is =1,2,3,.....16,17,18 similarly replacing a with b we get the same result for c ...so, b can also take any value from 1 to 18....but, experiment shows neither a nor b can take the value 18. hence calculating we get 16 results....for example take a=1 then b+c+bc=1001 and find how many integer pairs there can hold for the relation...the solutions i got are ::::

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a = 1, b = 2, c = 333

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a = 1, b = 5, c = 166

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a = 2, b = 3, c = 199

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a = 2, b = 13, c = 65

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a = 3, b = 7, c = 98

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a = 4, b = 5, c = 109

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a = 5, b = 13, c = 52

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a = 5, b = 14, c = 49

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a = 5, b = 22, c = 33

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a = 6, b = 11, c = 55

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a = 7, b = 8, c = 63

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a = 7, b = 14, c = 43

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a = 7, b = 18, c = 35

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a = 7, b = 23, c = 28

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a = 11, b = 22, c = 23

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a = 13, b = 17, c = 26

First, we have ab + bc + ac = 1001 and a < b < c so a < 18. With each value of a, we have 16 triples of positive integers (a, b, c) that a < b < c: (1, 2, 333), (1, 5, 166), (2, 3, 199), (2, 13, 65), (2, 3, 199), (3, 7, 98), (4, 5, 109), (5, 13, 52), (5, 14, 49), (5, 22, 33), (6, 11, 55), (7, 8, 63), (7, 14, 43), (7, 18, 35), (7, 23, 28), (11, 22, 33) and (13, 17, 26). Hence, we have 16 trples of positive integers (a, b, c) that satisfy the problem.

I was unsure as to how to go about solving this mathematically. But I am also a computer scientist and I knew that the numbers involved were small enough that I could brute force the solution easily with a simple python program.

My program: total = 0
for i in range(1,500):
for j in range(i+1,500):
for k in range(j+1,500):
if i * j + i * k + j * k==1001:
print "("+str(i)+","+str(j)+","+str(k)+")"
total+=1
print total

I knew I didn't have to test beyond 500 because 1 500+1 1+1*500=1001 (which is not a valid solution, but still provides a simple upper limit).

The program's output is below. It lists the triples and finally the solution to the problem. (The solution is 16.)

(1,2,333)
(1,5,166)
(2,3,199)
(2,13,65)
(3,7,98)
(4,5,109)
(5,13,52)
(5,14,49)
(5,22,33)
(6,11,55)
(7,8,63)
(7,14,43)
(7,18,35)
(7,23,28)
(11,22,23)
(13,17,26)
16

In order to produce this solution, I'll be using the following results:

I) If a number n is factored like this: $n = p_{1}^{k_{1}}*p_{2}^{k_{2}}*...*p_{n}^{k_{n}}$ , where $p_{i}$ is the i-th prime factor of n, then the number of divisors it has is given by the product:

$d = (k_{1} + 1)*(k_{2} + 1)*...*(k_{n} + 1)$

And if the number of divisors $d$ is even, then the number of ways we can write n as a product of two distinct postive integers, without taking into account the order of the numbers in the multiplication, is $d/2$ .

II) Given that $ab + ac + bc = 1001$ , take the numbers ab, ac and bc. By the AM-GM inequality, $cbrt(ab*ac*bc) <= (ab + ac + bc)/3$ .

Thus, $cbrt(a^{2}b^{2}c^{2}) <= 1001/3$ , which leads us to $(abc)^{2} <= (1001/3)^{3}$

We now have $abc <= sqrt((1001/3)^{3})$ . Also, since $a < b < c$ , we can safely say that $a^{3} < abc$ , so $a^{3} < sqrt((1001/3)^{3})$ , or $a < sqrt(1001/3)$

Since $1001/3 = 333 + 2/3$ , and the greatest square number that is smaller than this value is $324 = 18^{2}$ , we can write that $a <= 18$ .

III) $(a + b)(a + c) = a^{2} + ab + ac + bc = 1001 + a^{2}$ . We'll be using the identity $(a + b)(a + c) = 1001 + a^{2}$ from now on. We must find $b$ and $c$ such that $b < c$ , and $b > 0$ . Also, $b > a$ .

Now, onto the brute-force math, where we see the values b and c can assume when $1 <= a <= 18$ .

Case 1) $a = 1$ .

$(b + 1)(c + 1) = 1002 = 2*3*167$ . We have 8 divisors, thus 4 ways to divide 1002 as a product of distinct numbers. Ignore cases where the smallest number is 1 ( $b$ is not a positive number) or 2 ( $b <= a$ )

1) $b + 1 = 3$ , $c + 1 = 334$ , which gives us $b = 2$ and $c = 333$ . This solution is valid.

2) $b + 1 = 6$ , $c + 1 = 167$ , which gives us $b = 5$ and $c = 166$ . This solution is valid.

Case 2) $a = 2$

$(b + 2)(c + 2) = 1005 = 3*5*67$ We have 8 divisors, thus 4 ways to divide 1005 as a product of distinct numbers. Ignore cases where the smallest number is 1 ( $b$ is not a positive number) or 3 ( $b <= a$ )

1) $b + 2 = 5$ , $c + 2 = 201$ , which gives us $b = 3$ and $c = 199$ . This solution is valid.

2) $b + 2 = 15$ , $c + 2 = 67$ , which gives us $b = 13$ and $c = 65$ . This solution is valid.

Case 3) $a = 3$

$(b + 3)(c + 3) = 1010 = 2*5*101$ We have 8 divisors, thus 4 ways to divide 1005 as a product of distinct numbers. Ignore cases where the smallest number is 1, 2 ( $b$ is not a positive number) or 5 ( $b <= a$ )

1) $b + 3 = 10$ , $c + 3 = 101$ , which gives us $b = 7$ and $c = 98$ . This solution is valid.

Case 4) $a = 4$

$(b + 4)(c + 4) = 1017 = 3^{2}*113$ We have 6 divisors, thus 3 ways to divide 1005 as a product of distinct numbers. Ignore cases where the smallest number is 1 or 3 ( $b$ is not a positive number).

1) $b + 4 = 9$ , $c + 4 = 113$ , which gives us $b = 5$ and $c = 109$ . This solution is valid.

Case 5) $a = 5$

$(b + 5)(c + 5) = 1026 = 2*3^{3}*19$ We have 16 divisors, thus 8 ways to divide 1026 as a product of distinct numbers. Ignore cases where the smallest number is 1, 2, 3 ( $b$ is not a positive number), 6 or 9 ( $b <= a$ )

1) $b + 5 = 18$ , $c + 5 = 57$ , which gives us $b = 13$ and $c = 52$ . This solution is valid.

2) $b + 5 = 19$ , $c + 5 = 54$ , which gives us $b = 14$ and $c = 49$ . This solution is valid.

3) $b + 5 = 27$ , $c + 5 = 38$ , which gives us $b = 22$ and $c = 33$ . This solution is valid.

Case 6) a = 6. $(b + 6)(c + 6) = 1037 = 17*61$ . This number has 4 divisors, thus there are 2 ways to divide 1037 as a product of 2 distinct numbers without taking into account the order of the factors. Ignore the case where the smallest number is 1 ( $b$ is not a positive number). We have the single case:

1) $b + 6 = 17$ , $c + 6 = 61$ , which gives us $b = 11$ and $c = 55$ . This solution is valid.

Case 7: $a = 7$ $(b + 7)(c + 7) = 1050 = 2*3*5^{2}*7$ . We have then 24 divisors - 12 ways to divide in a product of 2 different numbers. Ignore cases where the smallest number is 1, 2, 3, 5, 7 ( $b$ is not positive), 10 or 14 ( $b <= a$ ) Cases remaining:

1) $b + 7 = 15$ , $c + 7 = 70$ . $b = 8$ and $c = 63$ . Valid solution.

2) $b + 7 = 21$ , $c + 7 = 50$ . $b = 14$ and $c = 43$ . Valid solution.

3) $b + 7 = 25$ , $c + 7 = 42$ . $b = 18$ and $c = 35$ . Valid solution.

4) $b + 7 = 30$ , $c + 7 = 35$ . $b = 23$ and $c = 28$ . Valid solution.

Case 8: $a = 8$ . $(b + 8)(c + 8) = 1065 = 3*5*71$ . We have 8 divisors, 4 ways to divide in a product of 2 different numbers. Ignore cases where the smallest number is 1, 3, 5 ( $b <= 0$ ) or 15 ( $b <= a$ ). No cases remain.

Case 9: $a = 9$ . $(b + 9)(a + 9) = 1082 = 2*541$ . 4 divisors, 2 ways to divide in a product of 2 different numbers. Ignore cases where smallest number is 1, 2 ( $b <= 0$ ). No cases remain.

Case 10: $a = 10$ . $(b + 10)(a + 10) = 1101 = 3*367$ . 4 divisors, 2 ways to divide in a product of 2 different numbers. Ignore cases where smallest number is 1 or 3 ( $b <= 0$ ). No cases remain.

Case 11: $a = 11$ . $(b + 11)(a + 11) = 1122 = 2*3*11*17$ . 16 divisors, 8 ways to divide in a product of 2 different numbers. Ignore cases where smallest number is 1, 2, 3, 6, 11 ( $b <= 0$ ), 17 or 22 ( $b <= a$ ) Only case remaining:

1) $b + 11 = 33$ , $c + 11 = 34$ . $b = 22$ and $c = 23$ . Valid solution.

Case 12: $a = 12$ . $(b + 12)(c + 12) = 1145 = 5*229$ . 4 divisors, 2 ways to divide in a product of 2 different numbers. Ignore cases where smallest number is 1, 5 ( $b <= 0$ ).. No cases remain.

Case 13: $a = 13$ $(b + 13)(c + 13) = 1170 = 2*3^{2}*5*13$ . 24 divisors, 12 ways to divide in a product of 2 different numbers. Ignore cases where smallest number is 1, 2, 3, 5, 6, 9, 10, 13 ( $b <= 0$ ), 15, 18 or 26 ( $b <= a$ ). Only case remaining:

1) $b + 13 = 30$ , $c + 13 = 39$ . $b = 17$ and $c = 26$ . Valid solution.

Case 14: $a = 14$ $(b + 14)(c + 14) = 1197 = 3^{2}*7*19$ . 12 divisors, 6 ways to divide in a product of 2 different numbers. Ignore cases where smallest number is 1, 3, 7, 9 ( $b <= 0$ ), 19 or 21 ( $b <= a$ ). No cases remain.

Case 15: $a = 15$ . $(b + 15)(c + 15) = 1226 = 2*613$ . 4 divisors, 2 ways to divide in a product of 2 different numbers. Ignore cases where smallest number is 1 or 2 ( $b <= 0$ ). No cases remain.

Case 16: $a = 16$ . $(b + 16)(c + 16) = 1257 = 3*419$ . 4 divisors, 2 ways to divide in a product of 2 different numbers. Ignore cases where smallest number is 1 or 3 ( $b <= 0$ ). No cases remain.

Case 17: $a = 17$ . $(b + 17)(c + 17) = 1290 = 2*3*5*43$ . 16 divisors, 8 ways to divide in a product of 2 different numbers. Ignore cases where smallest number is 1, 2, 3, 5, 6, 10, 15 ( $b <= 0$ ) or 30 ( $b <= a$ ) No cases remain.

Case 18: $a = 18$ . $(b + 18)(c + 18) = 1325 = 5^{2}*53$ . 6 divisors, 3 ways to divide in a product of 2 different numbers. Ignore cases where smallest number is 1, 5 ( $b <= 0$ ) or 25 ( $b <= a$ ) No cases remain.

The valid triplets are: $(1, 2, 333), (1, 5, 166), (2, 3, 199), (2, 13, 65), (3, 7, 98), (4, 5, 109), (5, 13, 52), (5, 14, 49), (5, 22, 33), (6, 11, 55),(7, 8, 63), (7, 14, 43), (7, 18, 35), (7, 23, 28), (11, 22, 23), (13, 17, 26)$ , which give us a total of 16 triplets which satisfy the problem's conditions.