Problema Solution
the digits of 3 digit natural number are in A.P and their sum is 18.The number obtained by reversing digits is 594 less than original number,find the original number.
Answer provided by our tutors
The digits of 3 digit natural number are in A.P = Arithmetic Progression that is we can write them as
x, x + k, x + 2k
Since they are digits they are elements of the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Arithmetic progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant - in our case the difference is 'k'.
Their sum is 18:
x + x + k + x + 2*k = 18
3*x + 3*k = 18
x + k = 6
The number that we are looking for A = x + 10*(x+k) + (10^2)(x+2k)
Let B the number with reversed digits that is B = (x+2k) + 10*(x+k) + (10^2)*x
A - B = 594 that is
[x + 10*(x+k) + (10^2)(x+2k)] - [(x+2k) + 10*(x+k) + (10^2)*x] = 594
Lets find x and k as a solutions of the systems equations
x + k = 6
x + 10*(x+k) + (10^2)(x+2*k) - (x+2*k) - 10*(x+k) - (10^2)*x = 594
x = 3
k = 3
The digits are 3, 6, 9 hence the number is 963.
Indeed 963 - 369 = 594.