Quadruple Reciprocal Sum

WLOG, let $a < b < c < d$ . We will multiply our result by $4!$ at the end. If $a\geq 3$ then $\frac 1 a+\frac 1 b+\frac 1 c+\frac 1 d \leq \frac 1 3+ \frac 1 4 + \frac 1 5+ \frac 1 6 <1$ . Clearly $a=1$ doesn't work. Hence $a=2$ is the only possible $a$ . Hence $\frac 1 b + \frac 1 c + \frac 1 d = \frac 1 2$ , so $\frac 1 b > \frac 1 6$ , i.e. $3\leq b\leq5$ .

Case 1. $b=3$ : In a similar way $\frac 1 c+\frac 1 d=1-\frac 1 2-\frac 1 3=\frac 1 6$ , so $\frac 1 c > \frac 1 {12}$ , i.e. $7\leq c\leq11$ . We test these possible $c$ and find that $c=7,8,9,10$ gives integer $d=42,24,18,15$ respectively.

Case 2. $b=4$ : $\frac 1 c + \frac 1 d = 1-\frac 1 2 - \frac 1 4 = \frac 1 4$ , so $\frac1c>\frac18$ , i.e. $5\leq c\leq7$ . We test these possible $c$ and find that $c=5,6$ gives integer $d=20,12$ respectively.

Case 3. $b=5$ : $\frac 1 c + \frac 1 d = 1-\frac 1 2 - \frac 1 5 =\frac 3 {10}$ , so $\frac 1 c > \frac 3 {20}$ , i.e. $c=6$ . ( $c<6$ will not satisfy $b<c$ ). $c=6$ does not give an integer value for $d$ .

Hence there are $6$ solutions: $(a,b,c,d)=(2,3,7,42),(2,3,8,24),(2,3,9,18)$ , $(2,3,10,15),(2,4,5,20),(2,4,6,12)$ Then we multiply $4!=24$ as this question wants ordered quadruples (each of the above $6$ solutions can be reordered in $24$ different ways). Hence $6\times 24=144$ .

As the numbers a,b,c and d are distinct positive integers, they will follow some order.

Let us take , for once, a > b > c > d. Then, 1/a < 1/b < 1/c < 1/d

So the given expression 1/a+ 1/b + 1/c+ 1/d (=1) < 4/d i.e., d<4 So d can attain values 1,2 ,3

d=1 is not possible since 1/a + 1/b+ 1/c cannot be 0 Now we have two possibilities, d=2 and d =3

Case 1: d=2

 1/a +1/b+ 1/c = 1-1/2 = 1/2

As we have taken a >b>c>d , 1/a + 1/b + 1/c < 3/c i.e. 1/2 < 3/c i.e., c < 6 Since c is to be an integer, c can assume values 1,2,3,4,5 c = 1 and c=2 are clearly not possible.

(1) c= 3 1/a +1/b= 1/2 - 1/3 = 1/6

By the same assumption, 1/6 < 2/b i.e., b< 12 b can have values 1, 2,3,4,5,6,7,8,9,10,11 out of which b=1 or 2 or 3 are clearly not possible.

b=4 gives a = -2/24 not possible b=5 gives a = -1/30 not possible b=6 gives 1/a = 0 not possible b=7 gives a= 42 b=8 gives a= 24
b=9 gives a= 18 b=10 gives a= 15 b=11 gives a= 5/66 not possible

(2) c = 4 1/a + 1/b = 1/2 - 1/4 = 1/4 By the similar argument, 1/4 < 2/b i.e. b< 8
Then, b=1,2, 3,4 can be rejected . b =5 gives a= 20 b =6 gives a= 12 b=7 gives a=3/28 not possible

(3) for c=5 , 1/a+ 1/b = 3/10

b< 20/3 i.e, b can have values 1,2,3,4,5,6 of which all values are ineligible. Hence no possible solution for c=5.

Case 2: d=3 By similar calculations, we can check that there is no possible solution for d= 3 also.

Hence the final quadruples are- ( 42,7,3,2 ) , ( 24,8,3,2) , ( 18,9,3,2) , ( 15,10,3,2), (20,5,4,2) , and (12,6,4,2). Now that we had assumed in the beginning that a>b>c>d and worked for it, total number of quadruples would be 6* 4! i.e. total 144 possible quadruples {since we can arrange the integers a,b,c and d in 4! ways }

Without lost of the generality, assume $a < b < c< d$ . It implies $\frac{1}{a}>\frac{1}{b}>\frac{1}{c}>\frac{1}{d}$ . Then, $1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}<\frac{4}{a}\Longleftrightarrow a<4$ but $a\neq 1$ . So there are only two possible value for $a$ i.e. $a=2$ or $a=3$ .

Case 1 : $a=2$

Hence, $\frac{1}{2}=\frac{1}{b}+\frac{1}{c}+\frac{1}{d}<\frac{3}{b}\Longleftrightarrow b<6$ and also $b>a=2$ .

@ If $b=3$ then $\frac{1}{6}=\frac{1}{c}+\frac{1}{d}<\frac{2}{c}\Longleftrightarrow c<12$ and $c>6$ .Therefore the possible value for $c$ is $7,8,9,10,11$ . After checking one by one we get solution $(2,3,7,42),(2,3,8,24),(2,3,9,18)$ and $(2,3,10,15)$ .

@ If $b=4$ then $\frac{1}{4}=\frac{1}{c}+\frac{1}{d}<\frac{2}{c}\Longleftrightarrow c<8$ and $c>4$ .Therefore the possible value for $c$ is $5,6,7$ . After checking one by one we get solution $(2,4,5,20)$ and $(2,4,6,12)$ .

@ If $b=5$ then $\frac{3}{10}=\frac{1}{c}+\frac{1}{d}<\frac{2}{c}\Longleftrightarrow c<\frac{20}{3}$ and $c>5$ implies $c=6$ but there is no solution for $d$ .

Case 2 : $a=3$

Hence $\frac{2}{3}=\frac{1}{b}+\frac{1}{c}+\frac{1}{d}<\frac{3}{b}\Longleftrightarrow b<\frac{9}{2}$ and since $b>a=3$ , implies $b=4$ . Then we get $\frac{5}{12}=\frac{1}{c}+\frac{1}{d}<\frac{2}{c}\Longleftrightarrow c<\frac{24}{5}$ . Since $c>b=4$ there is no solution for $c$ .

Therefore we get six solution, $(2,3,7,42),(2,3,8,24),(2,3,9,18),(2,3,10,15),(2,4,5,20),(2,4,6,12)$ . Since all of its permutation is also a solution we get total $6\times 4!=144$ solution.

Let us consider the lower bound for any of a, b, c or d.

Considering that each of the integers are distinct, consider the highest possible reciprocal sum of 4 numbers that do not include 2.

This is:

(1/3)+(1/4)+(1/5)+(1/6) = 19/20 < 1

Therefore, in order to get a sum total of 1, we need to have one of the integers as 2.

For now, assume that d = 2. Note that no other integers can be 2, as that would give us a sum of 1 already. So, we are left with:

1/a +1/b + 1/c = 1/2

Now, as 1/6 + 1/6 + 1/6 = 1/2, we notice that the average of the 3 fractions is 1/6, but, as they are all distinct, we can infer that atleast one of the fractions must be greater than 1/6, let us assume this is 1/c.

So,1/2> 1/c > 1/6

\Rightarrow 2< c <6.

So, c must be either 3, 4 or 5.

Let us consider these individual possibilities.

If c= 3,

then, 1/a + 1/b = 1/6

\Rightarrow (a-6)(b-6) = 36

As a-6 and b-6 are integers, let us consider the distinct integer pairs that multiply into 36,

(a-6)(b-6) = 36 x 1 = 18 x 2 = 12 x 3= 9 x 4

Pairing these individually, for example, a-6 = 36 and b-6 = 1 or a-6 = 18 and b-6= 2. we get 4 pairs of distinct integers, namely,

(42,7), (24,8), (18, 9), (15, 10).

Now let us consider the case that c = 4.

Following the same process, we get integer pairs, (20,5) and (12, 6).

Considering when c = 5, we find that there are no pair of unit fractions that satisfy the equation.

So, we have 4+2= 6 integer quadruples in total. Accounting for all the possible permutations,

we have 6 x 4!= 144 possible distinct integer quadruples.

First of all, let $(w, x, y, z)$ be the ordered form of $(a, b, c, d)$ , in other words, it is a permutation of $(a, b, c, d)$ such that $w<x<y<z$ .

We notice that $w$ must be smaller than $3$ or $\frac1w+\frac1x+\frac1y+\frac1z \leq \frac13+\frac14+\frac15+\frac16=\frac{19}{20}<1$ . Also $w$ cannot be equal to $1$ or $\frac1x+\frac1y+\frac1z=0$ contradicts that they are positive. So $w=2$ .

So $\frac1x+\frac1y+\frac1z=\frac12$

Now since $\frac16+\frac17+\frac18=\frac{73}{168}<\frac12$ , using the same logic above we know that $x<6$ . So $x=3, 4$ or $5$ .

If $x=3$ , we obtain $\begin{aligned} \frac1y+\frac1z&=\frac16\\ 6(y+z)&=yz\\ (y-6)(z-6)&=36\\ &=1\cdot36=2\cdot18=3\cdot12\\ &=4\cdot9=6\cdot6\\ \end{aligned}$ since $y<z$ , we could have $(y,z)=(7,42), (8,24), (9,18)$ or $(10, 15)$ .

If $x=4$ , similar to above we obtain $(y-4)(z-4)=16=1\cdot16=2\cdot8=4\cdot4$ so $(y,z)=(5,20)$ or $(6,12)$ .

So $(w,x,y,z)=(2,3,7,42),(2,3,8,24),(2,3,9,18),(2,3,10,15),$ $(2,4,5,20)$ or $(2,4,6,12)$ . For each $(w,x,y,z)$ there are $4!=24$ solutions to $(a,b,c,d)$ . Thus there are $6\cdot24=\boxed{144}$ quadruples in total.

Let $a<b<c<d$ .

Suppose $a\geq3$ .

Then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d} \leq \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}<1$

Therefore $a=2$ , since $a \neq 1$ , else $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}>1$

Thus $\frac{1}{b}+\frac{1}{c}+\frac{1}{d} = \frac{1}{2}$ .

Suppose $b>5$ . Then $\frac{1}{b}+\frac{1}{c}+\frac{1}{d} \leq \frac{1}{6}+\frac{1}{7}+\frac{1}{8} < \frac{1}{2}$ . Thus $b \leq 5$

Suppose $b=5$ . Then $\frac{1}{c}+\frac{1}{d} = \frac{3}{10}$ .

If $c=6$ , it is easy to see $d$ is not an integer.

If $c\geq7$ , $\frac{1}{c}+\frac{1}{d} \leq \frac{1}{7}+\frac{1}{8} < \frac{3}{10}$ .

Thus we have:

Case 1: $b=3$ and Case 2: $b=4$

Case 1:

We see $\frac{1}{c}+\frac{1}{d} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$

Multiplying by $6cd$ , we get

$6c+6d = cd$

By rearranging, we get

$cd-6c-6d+36=36$

$(c-6)(d-6)=36$

By factoring 36, we see $c-6=1, 2, 3,$ or $4$ , since $c-6<d-6$

That makes 4 solutions in case 1.

Case 2:

We see $\frac{1}{c}+\frac{1}{d} = \frac{1}{2} - \frac{1}{4} = \frac{1}{64}$

Multiplying by $4cd$ , we get

$4c+4d = cd$

By rearranging, we get

$cd-4c-4d+16=16$

$(c-4)(d-4)=16$

By factoring 16, we see $c-4=1$ or $2$ , since $c-4<d-4$

That makes 2 solutions in case 2.

Since the 6 solutions we found are ordered, we have to multiply by $4!$

Total quadruples $=6 \times 4! = 144$

WLOG, let $a>b>c>d$

Now if $d=1$ it would not be possible as it implies $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$

If $d \geq 3$ then;

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d} \leq \frac{1}{6}+\frac{1}{5}+\frac{1}{4}+\frac{1}{3}=\frac{19}{20}<1$

Which is not what we want, so we have $d<3$ or $d=2$

Hence; $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$

Now if $c \geq 6$ , then;

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq \frac{1}{8}+\frac{1}{7}+\frac{1}{6}=\frac{73}{168}<\frac{1}{2}$

Which is not what we want, so we have $c<6$

So we need to consider the cases where $c=3,4,5$

Case 1: $c=3$

Hence $\frac{1}{a}+\frac{1}{b}=\frac{1}{6} \Rightarrow (a-6)(b-6)=36$

Since $a$ and $b$ are integers, we would then only need to consider the factors of 36. And as;

$36=36 \times 1=18 \times 2=12 \times 3=9 \times 4=6 \times 6$

We will get solutions $(a,b)$ :

$(42,7), (24,8), (18,9), (15,10)$ respectively. Note we reject the possibility of $6 \times 6$ as $a$ and $b$ are distinct.

Case 2: $c=4$

Hence $\frac{1}{a}+\frac{1}{b}=\frac{1}{4} \Rightarrow (a-4)(b-4)=16$

By a similar argument as above and that; $16=16 \times 1=8 \times 2=4 \times 4$

We will get solutions $(a,b)$ :

$(20,5), (12,6)$ respectively. Note we reject the possibility of $4 \times 4$ as $a$ and $b$ are distinct.

Case 3: $c=5$

If so, $\frac{1}{a}+\frac{1}{b}=\frac{2}{5}$

If $b \geq 7$ ,

$\frac{1}{a}+\frac{1}{b}\leq \frac{1}{8}+\frac{1}{7}=\frac{15}{56}<\frac{2}{5}$

So $b=6$ , but this means that $a=\frac{30}{7}$ , where it is not an integer. Hence there is no possible solutions for this case.

Summing up, the solutions possible for $(a,b,c,d)$ are $(42,7,3,2), (24,8,3,2), (18,9,3,2), (15,10,3,2), (20,5,4,2), (12,6,4,2)$ .

Hence we need to find the permutations of the 6 possible combinations of quadruples found, which is $6 \times 4!= 144$

First, we assume that $1 < a < b < c < d$ and count the number of solutions. Since $\frac {1}{3} + \frac {1}{4} + \frac {1}{5} + \frac {1}{6} = \frac {19}{20} < 1$ , thus we must have $a < 3$ . Since $a > 1$ , thus $a = 2$ . Now we want to solve $\frac {1}{b} + \frac {1}{c} + \frac {1}{d} = \frac {1}{2}$ . Since $\frac {1}{6} + \frac {1}{7} + \frac {1}{8} < \frac {3}{6} = \frac {1}{2}$ , hence we must have $2 < b < 6$ . We have the following cases:

Case 1: $b=3$ , $\frac {1}{c} + \frac {1}{d} = \frac {1}{6}$ . We have $\frac {1} {6} > \frac {1} {c}$ and $\frac {1}{6} = \frac {1}{c} + \frac {1}{d} < \frac {1}{c} + \frac {1}{c} = \frac{2}{c}$ which gives $6 < c < 12$ . Thus, the solutions are $(c, d) = (7, 42), (8, 24), (9, 18), (10, 15)$ , and $c=11$ does not yield an integer $d$ .

Case 2: $b=4$ , $\frac {1}{c} + \frac {1}{d} = \frac {1}{4}$ . Similar to the argument above, we have $\frac {1}{8} < \frac {1}{c} < \frac {1}{4}$ which gives $4<c<8$ . Thus the solutions are $(c,d) = (5, 20), (6, 12)$ , and $c=7$ does not yield an integer $d$ .

Case 3: $b = 5$ , $\frac {1}{c} + \frac {1}{d} = \frac {3}{10}$ . Similar to the argument in Case 1, we have $\frac {3}{20} < \frac {1}{c} < \frac {3}{10}$ , which gives $4 \leq c \leq 6$ . Since $c > b=5$ , we have $c=6$ , but this doesn’t yield an integer $d$ .

Hence, with the assumption $a < b < c < d$ , there are $6$ solutions. Thus, there are $6 \times 4! = 144$ solutions in all.

Note: The standard way to solve $\frac {1}{c}+\frac {1}{d} = \frac {1}{6}$ is to cross multiply to obtain $cd = 6c+6c \Rightarrow (c-6)(d-6)=36$ , and so $c-6$ and $d-6$ are (possibly negative) factors of 36. However, we cannot use the same procedure for $\frac {1}{c}+\frac {1}{d} = \frac {3}{10}$ .