Question

Find an odd number with 3 digits such that all the digits are different and add up to 15. The difference between the first two digits equals the difference between the last 2 digits. The hundred digit is greater than the sum of the tens and ones digits

Answer 1

Answer:

The three digit number = 951

Step-by-step explanation:

Let suppose the numbers are:

= abc

According to given condition:

a+ b + c = 15  -------------eq1

Also, given the difference between the first two digit = the difference between the last two digits:  

==> l a-b l = l b-cl  

==> (a-b) = (b-c)  

==> (a+c) = 2b

Now we will substitue in eq1

==> a+ b + c = 15

==> 2b + b = 15

==> 3b = 15

Dividing both sides by 3 we get:

b =5

a + c = 2b

a+ c = 10

a = 10 -c ..........(2)

We know that"  

(a-b) = (b-c)

==> a > b+c

==> a > 5 + c

==> 10 -c > 5 +c  

==>5 > 2c

==> 2.5 > c

As c is an odd number so c will be equal to 1

c = 1

a = 10 -1

a = 9

The three digit number = 951

The hundred digit is greater than the sum of the tens and ones digits

i hope it will help you!