A certain three-digit number equals fifteen times the sum of its digits. if its digits are reversed, the resulting number exceeds n by 396. the one's digit is one larger than the sum of the other two. give a linear system of three equations whose three unknowns are the digits of n. solve the system and find n.

If the digits are a, b and c, the number n is 100a+10b+c

Here are the equations:

a number equals fifteen times the sum of its digits:

100a+10b+c = 15 (a+b+c)

if its digits are reversed, the resulting number exceeds n by 396:

100a+10b+c+396 = 100c+10b+a

the one’s digit is one larger than the sum of the other two:

a+b+1 = c

Solving it gives you a=1, b=3, c=5, the number is 135.