A certain three-digit number equals fifteen times the sum of its digits. if its digits are reversed, the resulting number exceed s 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.
1 answer:
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.
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