Numbers in a pyramid

The third row of circles would be $ab$ , $bc$ , and $cd$ , the second row of circles would be $ab + bc = b(a + c)$ and $bc + cd = c(b + d)$ , and the top circle would be $72 = bc(a + c)(b + d)$ .

Assume $b \geq 2$ . Since $c > b$ then $c \geq 3$ , and since $d > a$ then $d \geq 2$ , that means $bc(a + c)(b + d) \geq 2 \cdot 3 \cdot (1 + 3) \cdot (2 + 2) = 96$ . However, $72 < 96$ , so $b = 1$ .

Assume $c \geq 4$ . Since $a \neq b = 1$ then $a \geq 2$ , and since $d > a$ then $d \geq 3$ , that means $bc(a + c)(b + d) \geq 1 \cdot 4 \cdot (2 + 4) \cdot (1 + 3) = 96$ . However, $72 < 96$ , so $c \leq 3$ .

Assume $c = 3$ . Since $a \neq b = 1$ then $a \geq 2$ , and since $d > a$ and $d \neq c = 3$ then $d \geq 4$ , that means $bc(a + c)(b + d) \geq 1 \cdot 3 \cdot (2 + 3) \cdot (1 + 4) = 75$ . However, $72 < 75$ , so $c \neq 3$ .

Since $c \leq 3$ and $c \neq 3$ and $c \neq b = 1$ , it must be that $c = 2$ .

Assume $a = 3$ . Then $a + c = 3 + 2 = 5$ , and $bc(a + c)(b + d)$ is divisible by $5$ . However, $72$ is not divisible by $5$ , so $a \neq 3$ .

Assume $a \geq 5$ . Since $d > a$ then $d \geq 6$ , that means $bc(a + c)(b + d) \geq 1 \cdot 2 \cdot (5 + 2) \cdot (1 + 6) = 98$ . However, $72 < 98$ , so $a \leq 4$ .

Since $a \leq 4$ and $a \neq 3$ and $a \neq b = 1$ and $a \neq c = 2$ , it must be that $a = 4$ .

Solving $72 = bc(a + c)(b + d)$ with $a = 4$ , $b = 1$ , and $c = 2$ gives $d = 5$ .

Therefore, the four-digit integer is $\boxed{4125}$ .

Following the rules of the puzzle, the numbers in two circles on the second row can be expressed respectively as $ab + bc$ and $bc + cd$ , so that

$(ab + bc)(bc + cd) = bc(a + c)(b + d) = 72$

Since $a < d$ and $b < c$ , let $c = b + m$ and $d = a + n$ , such that $a, b, m, n$ are all positive integers. Then, let $x_a, x_b, x_m, x_n$ denote possible nonnegative integers, ranging from $0$ to $8$ , such that $a = x_a + 1$ , $b = x_b + 1$ , $c = x_b + x_m + 2$ and $d = x_a + x_n + 2$ . Substituting, we have

$(x_b + 1)(x_b + x_m + 2)(x_a + x_b + x_m + 3)(x_a + x_b + x_n + 3) = 72$

If $b$ is even, then $x_b = 2k - 1$ for positive integer $k$ , which simplifies the expression to

$2k(x_m + 2k + 1)(x_a + x_m + 2k + 2)(x_a + x_n + 2k + 2) = 72$

But if $k = 1$ , then $2k \cdot (2k + 1) \cdot (2k + 2)^2 = 96 > 72$ , which shows that the equation above does not have a solution of nonnegative integers. So since $2k$ is strictly increasing, it suffices to check the possible cases for $b = x_b + 1 = 1$ , where the equation to look at is

$(x_m + 2)(x_a + x_m + 3)(x_a + x_n + 3) = 72$

Either $(x_m + 2)$ is even or odd. As the prime factorization of $72$ is $2^3 \cdot 3^2$ ,

Odd Cases: If $(x_m + 2)$ is odd, the possible value of $c$ is either $3$ ( $x_m = 1$ ) or $9$ ( $x_m = 7$ ). As each of the factors is strictly increasing, $(x_a + x_m + 3)$ is greater than $(x_m + 2)$ , which shows there is no possible solution for $x_m = 7$ . For $x_m = 1$ , we have

$3(x_a + 4)(x_a + x_n + 3) = 72 \Longrightarrow (x_a + 4)(x_a + x_n + 3) = 24$

which implies that the possible factorization of $24$ is $4\cdot 6$ . That simplifies to $x_a = 0$ and $x_n = 3$ . However, $a = b = 1$ , which violates the condition that $a, b, c$ and $d$ are distinct.

Even Cases: Otherwise, if $c$ is even, let $x_m = 2\ell$ , where $\ell$ is a nonnegative integer. Then, the equation to look at is

$(2\ell + 2)(x_a + 2\ell + 3)(x_a + x_n + 3) = 72 \Longrightarrow (\ell + 1)(x_a + 2\ell + 3)(x_a + x_n + 3) = 36$

Following what we learned about the solutions for even values of $b$ , the necessary condition for finding the possible values is $3(\ell + 1)(2\ell + 3) \leq 36$ , which solves to $\dfrac{1}{4} (-5 - \sqrt{97}) \leq \ell \leq \dfrac{1}{4}(\sqrt{97} - 5)$ . As we are interested in nonnegative solutions, the values of $\ell$ to check are $0$ and $1$ as $1 < \dfrac{1}{4}(\sqrt{97} - 5) < 2$ .

If $\ell = 1$ , then $(x_a + 5)(x_a + x_n + 3) = 18$ (from $(\ell + 1) = 2$ ). As no possible factorization of $18$ , involving only two integers, exists, no solution exists.

Otherwise, $\ell = 0$ , where $(x_a + 3)(x_a + x_n + 3) = 36$ . As the possible factorizations of $36$ are $4 \cdot 9$ and $6 \cdot 6$ , the possible solutions of $(x_a, x_n)$ are $(1,5)$ (for $4\cdot 9$ ) and $(3,0)$ (for $6 \cdot 6$ ). Checking, we see that the only existing solution is $(3,0)$ , where $\boxed{a = 4, b = 1, c = 2, d = 5}$ as $(1,5)$ yields $a = c = 2$ .

Ground Level circles : a , b , c , d
First Floor circles (X) : ab , bc , cd
Second Floor circles (+) : b(a + c) , c(b + d)
Top Level circle (X) : 72 = bc(a + c)(b + d)

We were told that a ≠ b ≠ c ≠ d & 0 < a, b, c, d < 10 , so we try factoring 72 into 4 factors, with the smallest 2 being NOT equal.
72
= 1 X 2 X 2 X 18
= 1 X 2 X 3 X 12
= 1 X 2 X 4 X 9
= 1 X 2 X 6 X 6
= 1 X 3 X 3 X 8
= 1 X 3 X 4 X 6
= 2 X 3 X 3 X 4

Since all of the numbers a, b, c & d are positive integers, logically
b < b + d & c < a + c
other than the given
b < c & a < d
Thus, b < c < a + c and also b < b + d simultaneously, making it the littlest of the factors.

72 = 1 X 2 X 2 X 18
18 can only be the sum of two nines if both its summands has to be single digit positive integers, but neither 1 or 2 equal 9

72 = 1 X 2 X 3 X 12
If b = 1, then it can only be that 12 = a + c, and c forced to be 3 and 2 = b + d with b = 1 = d, which is contradictory to the given conditions.

72 = 1 X 2 X 4 X 9
If b = 1, then (c,a,d) = {2, (2,7), (8,3)} for a solution of (b,c,a,d) = {1, 2, 7, 3} , but here a > d and as such is invalid

72 = 1 X 2 X 6 X 6
If b = 1, then we have a solution of (b,c,a,d) = {1, 2, 4, 5} , and here a < d and thus validated

72 = 1 X 3 X 3 X 8
If b = 1, then we have a solution of (b,c,a,d) = {1, 3, 5, 2} , but here a > d and as such is invalid

72 = 1 X 3 X 4 X 6
If b = 1, then (c,a,d) = {(3,4), [(1,3), 2], [(5,3), 2]} for a solution of (b,c,a,d) = {(1, 3, 1, 5), (1, 3, 3, 3), (1, 4, 2, 2)} , but here none of these are completely distinct integers with b = a = 1 or c = a = d = 3 or a = d = 2 respectively and as such are all invalid

72 = 2 X 3 X 3 X 4
If b = 2, then d can only be either 1 or 2, but neither can be right because 0 < a < d and d ≠ b respectively.

Answer : (b,c,a,d) = {1, 2, 4, 5}
Rearranging, we get abcd = 4125

72 factors out as:

1* 72

2* 36

3* 24

4* 18

6* 12

8* 9

Since w=a* b and a≠b≠c≠d (different single digit positive integers) that means w≥2. Same applies to x and y, x=b* c and b≠c and y=c* d and c≠d so x≥2 and y≥2. Now, s=w+x so s≥4 and t=x+y so t≥4.

That means the first three options (1* 72, 2* 36 and 3* 24) don't offer a solution since they all require s<4 or t<4.

To explore the other possibilities we'll simplify the problem a bit by realizing that s and t are interchangeable as are w and y, a and d, and b and c: So we'll only explore for s.

The next option (4* 18) has s=4. If s=4 then either w=1 and x=3 or w=2 and x=2 or w=3 and x=1. Since w≥2, x≥2 and w≠x this doesn't offer a solution.

With the last option (8* 9) we have s=8. If s=8 then w=2 and x=6 or w=3 and x=5.

If w=2 and x=6 then y=3 (t=9=x+y). If y=3 then c=1 and d=3 or c=3 and d=1. But with w=2 that means a=1 and b=2 or a=2 and b=1. So we'd need c=1 or d=1 and a=1 or b=1 and since a≠b≠c≠d that doesn't offer a solution.

If w=3 and x=5 then y=4 and if y=4 then c=1 and d=4 or c=4 and d=1 (c≠d≠2). But with w=3 that means a=1 and b=3 or a=3 and b=1. So we'd need c=1 or d=1 and a=1 or b=1 and since a≠b≠c≠d that doesn't offer a solution either.

So the option (6* 12) has to be the one to go with, it's the only one left.

Now, If s=6 then w=2 and x=4 or w=4 and x=2 (w≠x≠3 and w≥2).

If w=2 and x=4 then either a=1 and b=2 or a=2 and b=1.

If a=1 and b=2 then c=2 (x=4=b* c) and since b≠c that doesn't offer a solution.

If a=2 and b=1 then c=4. With x=4 and t=12 then y=12-4=8. And if y=8 then d=y/c=8/4=2. But a=2 and since a≠d that doesn't offer a solution either.

If w=4 and x=2 then either a=1 and b=4 or a=4 and b=1.

If a=1 and b=4 then c=1 (x=4=b* c) and since b≠c that doesn't offer a solution.

If a=4 and b=1 then c=2. With x=2 and t=12 then y=12-2=10. And if y=10 then d=y/c=10/2=5.

And there we have the solution a=4, b=1, c=2 and d=5.

Of course with the interchangeability of the letters as mentioned, another solution would be a=5, b=2, c=1 and d=4 but since it is stated that a<d and b<c in this case that isn't a valid solution.

SOLUTION:

So abcd=4125.