#### Otis

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- Thread starter Otis
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26178

I have the actual work in the spoiler to support the answer, including that it shows

that there is a unique solution.

Let the listing of digits of the number from the left to the right be:

a, b, c, d, e.

a + b = e

d = e - 1

b = a + 4

a + b + c + d + e = 24

----------------------------------

d = e - 1 ---> e = d + 1

So, a + b = d + 1 --->

d = a + b - 1

Recall e = a + b.

Substitute what d & e equal into the fourth equation at the top:

a + b + c + (a + b - 1) + (a + b) = 24 --->

3a + 3b + c = 25 *

b = a + 4 ---> b - a = 4 ---> 3b - 3a = 12 **

Add equations * and ** together:

6b + c = 37 ---> c = 37 - 6b.

Potential possible choices for b (so c can be legit) are: 5 or 6. Then start

back-substituting for the other possible digits.

b = 5, c = 7, a = 1, d = 7 . . . . Stop. There is a repeated digit in this route.

b = 6, c = 1, a = 2, d = 7, e = 8 . . . . . There are no repeated digits here.

This work shows that there are no other possible answers.

The 5-digit number is \(\displaystyle \ \)26,178.

that there is a unique solution.

a, b, c, d, e.

a + b = e

d = e - 1

b = a + 4

a + b + c + d + e = 24

----------------------------------

d = e - 1 ---> e = d + 1

So, a + b = d + 1 --->

d = a + b - 1

Recall e = a + b.

Substitute what d & e equal into the fourth equation at the top:

a + b + c + (a + b - 1) + (a + b) = 24 --->

3a + 3b + c = 25 *

b = a + 4 ---> b - a = 4 ---> 3b - 3a = 12 **

Add equations * and ** together:

6b + c = 37 ---> c = 37 - 6b.

Potential possible choices for b (so c can be legit) are: 5 or 6. Then start

back-substituting for the other possible digits.

b = 5, c = 7, a = 1, d = 7 . . . . Stop. There is a repeated digit in this route.

b = 6, c = 1, a = 2, d = 7, e = 8 . . . . . There are no repeated digits here.

This work shows that there are no other possible answers.

The 5-digit number is \(\displaystyle \ \)26,178.

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