Question

A five-digit number.
The sum of all numbers is 21.
The 1st number minus the 3rd number is equal to 1.
The 2nd number is smaller than the 4th number by 1, which is smaller than the 5th number by 1. There are no repeating digits.
What is this number?

Answer 1

Answer:

Step-by-step explanation:

i dont know sorry

Answer 2

Answer:

The numbers are 81723 and 25167.

Solution:

Let the digits are a, b, c, d, e from the 10000th place to unit place respectively.

By the given conditions,

a + b + c + d + e = 21 ..... (1)

a - c = 1 ..... (2)

b = d - 1 ..... (3)

d = e - 1 ..... (4)

From (3) and (4), we get

b = (e - 1) - 1

or, b = e - 2 ..... (5)

From (1), (2) and (4), we get

c + 1 + e - 2 + c + e - 1 + e = 21

or, 2c + 3e = 23

or, c = (23 - 3e)/2 ..... (6)

Then a = (23 - 3e)/2 + 1, by (2)

or, a = (25 - 3e)/2 ..... (7)

So all the digits a, b, c, d, e can be expressed in terms of e only,

a = (25 - 3e)/2

b = e - 2

c = (23 - 3e)/2

d = e - 1

e = e

We recall the fact that a, b, c, d, e are positive integers lying in [0, 9]. We put values e = 0, 1, 2, 3, ..., 9 and try to find out suitable five digit numbers.

• If e = 0, b and d become negative, so e ≠ 0
• If e = 1, b becomes negative, so e ≠ 1
• If e = 2, a and c become fractions, so e ≠ 2
• If e = 3, a = 8, b = 1, c = 7, d = 2

∴ the five digit number be 81723

• If e = 4, a and c become fractions, so e ≠ 4
• If e = 5, a = 5, b = 3, c = 4, d = 4

∴ the five digit number be 53445

• If e = 6, a and c become fracture, so e ≠ 6
• If e = 7, a = 2, b = 5, c = 1, d = 6

∴ the five digit number be 25167

• For any higher values of e, like e = 8 and 9, we will see that the values of a and c become fractions and negative respectively, so e ≠ 8, 9

It is given that the five digit number will have no repeating digits, then we can't take 53445 as our solution.

∴ the required five digit numbers are 81723 and 25167