Question

In a three-digit number, the difference between
hundreds digit and the tens digit is equal to the
difference between the tens digit and the units digit.
If the sum of the digits is 9, how many numbers
satisfy the given condition?​

Answer 1

unit digit = x , tens digit= y , hundreds digit = z

the number = (100z+10y+x)

according to question

z - y = y - x ............... (1) equation

x + y + z = 9................(2) equation

from (1) -

x+z = 2y put this value in equation (2)

2y+y= 9 ; or 3y = 9 or y = 3

now, because both the equations have same variable

so, put y = 3 for all digits

i.e 100×3 + 10×3 + 3 = 300+30+3 = 333.

Answer 2

Answer:

9

Step-by-step explanation:

Hundred's digit = x, Ten's digit = y, Unit's digit = z

x + y + z = 9

Two conditions are possible from the given conditions:

1. x - y = y - z therefore, x + z = 2y

2. y - x = y - z therefore, x = z

possible numbers when x+z = 2y

135, 234, 333, 432, 531, 630

possible numbers when x = z

171, 252, 333, 414

Total number of numbers possible are 10, but 333 is common in both cases, hence we will count it only once.

So possible numbers are 9.