Fractions sum to 1?

$\large \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} +\dfrac{1}{d} =1$

Since $a\le b\le c \le d$ , then $\frac{1}{d}\le \frac{1}{c}\le \frac{1}{b}\le \frac{1}{a}$

Note: $\frac{1}{a}< \frac{1}{d} +\frac{1}{c} + \frac{1}{b} + \frac{1}{a}\le \frac{1}{a} +\frac{1}{a} + \frac{1}{a} + \frac{1}{a}$

$\frac{1}{a}<1 \le \frac{4}{a}$

$1 < a \le \ 4$

We have $3$ cases: $a=2, a=3$ and $a=4$

Case 1: a=2

$\dfrac{1}{2}+\dfrac{1}{b}+\dfrac{1}{c} +\dfrac{1}{d} =1$

$\dfrac{1}{b}+\dfrac{1}{c} +\dfrac{1}{d} =\dfrac{1}{2}$

Note: $\dfrac{1}{b}< \dfrac{1}{d} + \dfrac{1}{c} + \dfrac{1}{b}\le \dfrac{1}{b} +\dfrac{1}{b} + \dfrac{1}{b}$

$\dfrac{1}{b}< \dfrac{1}{2}\le \dfrac{1}{b} +\dfrac{1}{b} + \dfrac{1}{b}$

$\dfrac{1}{b}< \dfrac{1}{2}\le \dfrac{3}{b}$

$2< b \le 6$

We have $4$ subcases: $b=3, b=4, b=5$ and $b=6$

Case 1.a: a= 2, b=3

$\dfrac{1}{2}+\dfrac{1}{c} +\dfrac{1}{d} =\dfrac{1}{2}$

$\dfrac{1}{c} +\dfrac{1}{d} =\dfrac{1}{6}$

Note: $\dfrac{1}{c}< \dfrac{1}{d} + \dfrac{1}{c}\le \dfrac{1}{c} +\dfrac{1}{c}$

$\dfrac{1}{c}< \dfrac{1}{6}\le \dfrac{2}{c}$

$6< c\le \ 12$

We now have the values for a, b, and c, so we solve for d. We’ll get that the ordered pairs that satisfy $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} +\dfrac{1}{d} =1$ are: $(2,3,7,42), (2,3,8,24), (2,3,9,18), (2,3,10,15), (2,3,12,12)$ . $\boxed{5}$

Case 1.b: a= 2, b=4

$\dfrac{1}{4}+\dfrac{1}{c} +\dfrac{1}{d} =\dfrac{1}{2}$

$\dfrac{1}{c} +\dfrac{1}{d} =\dfrac{1}{4}$

Note: $\dfrac{1}{c}< \dfrac{1}{d} + \dfrac{1}{c}\le \dfrac{1}{c} +\dfrac{1}{c}$

$\dfrac{1}{c}< \dfrac{1}{4}\le \dfrac{2}{c}$

$4< c\le 8$

We now have the values for a, b, and c, so we solve for d. We’ll get that the ordered pairs that satisfy $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} +\dfrac{1}{d} =1$ are: $(2,4, 5, 20), (2,4,6,12). (2,4,8,8)$ . $\boxed{3}$

Case 1.c: a= 2, b=5

$\dfrac{1}{5}+\dfrac{1}{c} +\dfrac{1}{d} =\dfrac{1}{2}$

$\dfrac{1}{c} +\dfrac{1}{d} =\dfrac{3}{10}$

Note: $\dfrac{1}{c}< \dfrac{1}{d} + \dfrac{1}{c}\le \dfrac{1}{c} +\dfrac{1}{c}$

$\dfrac{1}{c}< \dfrac{3}{10}\le \dfrac{2}{c}$

$\dfrac{10}{3}< c\le \dfrac{20}{3}$

Since $5\le c$ , then $5\le c\le 6$

We now have the values for a, b, and c, so we solve for d. We’ll get that the only ordered pair that satisfies $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} +\dfrac{1}{d} =1$ is: $(2,5,5,10)$ . $\boxed{1}$

Case 1.d: a= 2, b=6

$\dfrac{1}{6}+\dfrac{1}{c} +\dfrac{1}{d} =\dfrac{1}{2}$

$\dfrac{1}{c} +\dfrac{1}{d} =\dfrac{1}{3}$

Note: $\dfrac{1}{c}< \dfrac{1}{d} + \dfrac{1}{c}\le \dfrac{1}{c} +\dfrac{1}{c}$

$\dfrac{1}{c}< \dfrac{1}{3}\le \dfrac{2}{c}$

$3< c\le 6$

Since $6\le c$ , then the only possible c is 6.

We now have the values for a, b, and c, so we solve for d. We’ll get that the only ordered pair that satisfies $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} +\dfrac{1}{d} =1$ is: $(2,6,6,6)$ . $\boxed{1}$

Case 2: a=3

$\dfrac{1}{3}+\dfrac{1}{b}+\dfrac{1}{c} +\dfrac{1}{d} =1$

$\dfrac{1}{b}+\dfrac{1}{c} +\dfrac{1}{d} =\dfrac{2}{3}$

$\dfrac{1}{b}< \dfrac{1}{d} + \dfrac{1}{c} + \dfrac{1}{b}\le \dfrac{1}{b} +\dfrac{1}{b} + \dfrac{1}{b}$

$\dfrac{1}{b}< \dfrac{2}{3}\le \dfrac{3}{b}$

$\dfrac{3}{2}< b \le \dfrac{9}{2}$

Since d should be a positive integer,

$2\le b \le 4$

We have $3$ subcases: $b=2, b=3$ and $b=4$ , but we can disregard the first case because we know that $b=2\le a=3$ and it does not satisfy $a\le b\le c \le d$ .

Case 2.a: a= 3, b=3

$\dfrac{1}{3}+\dfrac{1}{c} +\dfrac{1}{d} =\dfrac{2}{3}$

$\dfrac{1}{c} +\dfrac{1}{d} =\dfrac{1}{3}$

Note: $\dfrac{1}{c}< \dfrac{1}{d} + \dfrac{1}{c}\le \dfrac{1}{c} +\dfrac{1}{c}$

$\dfrac{1}{c}< \dfrac{1}{3}\le \dfrac{2}{c}$

$3< c\le 6$

We now have the values for a, b, and c, so we solve for d. We’ll get that the ordered pairs that satisfy $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} +\dfrac{1}{d} =1$ is: $(3,3,4,12), (3,3,6,6)$ . $\boxed{2}$

Case 2.b: a= 3, b=4

$\dfrac{1}{4}+\dfrac{1}{c} +\dfrac{1}{d} =\dfrac{2}{3}$

$\dfrac{1}{c} +\dfrac{1}{d} =\dfrac{5}{12}$

Note: $\dfrac{1}{c}< \dfrac{1}{d} + \dfrac{1}{c}\le \dfrac{1}{c} +\dfrac{1}{c}$

$\dfrac{1}{c}< \dfrac{5}{12}\le \dfrac{2}{c}$

$\dfrac{12}{5}< c\le \dfrac{24}{5}$

Since $4\le c$ , then the only possible c is 4.

We now have the values for a, b, and c, so we solve for d. We’ll get that the only ordered pair that satisfies $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} +\dfrac{1}{d} =1$ is: $(3,4,4,6)$ . $\boxed{1}$

Case 3: a=4

$\dfrac{1}{4}+\dfrac{1}{b}+\dfrac{1}{c} +\dfrac{1}{d} =1$

$\dfrac{1}{b}+\dfrac{1}{c} +\dfrac{1}{d} =\dfrac{3}{4}$

$\dfrac{1}{b}< \dfrac{1}{d} + \dfrac{1}{c} + \dfrac{1}{b}\le \dfrac{1}{b} +\dfrac{1}{b} + \dfrac{1}{b}$

$\dfrac{1}{b}< \dfrac{3}{4}\le \dfrac{3}{b}$

$\dfrac{4}{3}< b \le 4$

Since d should be a positive integer,

$2\le b \le 4$

We have $3$ subcases: $b=2, b=3$ and $b=4$ , but we can disregard the first two cases because we know that $b=2, b=3\le a=4$ and it does not satisfy $a\le b\le c \le d$ .

Case 3.a: a= 4, b=4

$\dfrac{1}{4}+\dfrac{1}{c} +\dfrac{1}{d} =\dfrac{3}{4}$

$\dfrac{1}{c} +\dfrac{1}{d} =\dfrac{1}{2}$

Note: $\dfrac{1}{c}< \dfrac{1}{d} + \dfrac{1}{c}\le \dfrac{1}{c} +\dfrac{1}{c}$

$\dfrac{1}{c}< \dfrac{1}{2}\le \dfrac{2}{c}$

$2< c\le 4$

Since $4\le c$ , then the only possible c is 4.

We now have the values for a, b, and c, so we solve for d. We’ll get that the only ordered pair that satisfies $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} +\dfrac{1}{d} =1$ is: $(4,4,4,4)$ . $\boxed{1}$

$5+3+1+1+2+1+1 = \boxed{14}$