Question

a number consists of three digits whose sum is 17 the middle one exceeds the sum of the other two by one if the digits be reversed the number is diminished by 396 find the number

Answer 1

Answer:

Step-by-step explanation:

Answer 2

692 is the number.

Given:

Three- digit number whose sum is 17 and the middle one exceeds the sum of the other two by one if the digits be reversed the number is diminished by 396

To find:

The number = ?

Solution:

The number whose digit’s sum is equal to 17 can be taken as XYZ

As the question says the middle digit exceeds the sum of the extreme by 1, which means $X + Z = 1 + Y$

Now if the extremes are interchanged the numbers after subtraction formed are

$Z + Y + X = X + Y + Z -396$

Now let us take the above equation as  

$100Z + 10Y + X = 100X + 10Y + Z - 396$

$396 = 99X - 99Z$

$4 = X - Z$

$X + Z = 8$

$X - Z = 4$

Equation the equations $X + Z = 8$ and$X - Z = 4$, we get the value of X as 6 putting the value of X in $X - Z = 4$, we get the value of Z as 2

Hence, the number is $\bold{XYZ = 692}$.