SOLUTION: A 3 digit number such that it is equal to 19 times the sum of its digit. What is the largest possible value? example 114 = 19 * (1+1+4) or 133 = 19 * (1+3+3)

Click here to see ALL problems on Numeric Fractions

Question 1132364: A 3 digit number such that it is equal to 19 times the sum of its digit. What is the largest possible value? example 114 = 19 * (1+1+4) or 133 = 19 * (1+3+3)
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website!
.

ANSWER. The maximal such a number is 399.

Solution

Let "a" be the "hundredth" digit; "b" be the "tens" digit and "c" be the "ones" digit. Then the number is 100a + 10b + c, and the condition says 100a + 10b + c = 19*(a+b+c), or 81a - 9b - 18c = 0, 9a - b - 2c = 0. Then a= 3, b= 9 and c= 9 is a) the solution to the last equation, and b) represents/provides the maximal such a number. It is obvious, taking into account that 0 <= a <= 9, 0 <= b<= 9, 0 <= c <= 9.

Solved.

Answer by greenestamps(13200) (Show Source):
You can put this solution on YOUR website!

Given the 3-digit number "abc", the value of the number is 100a+10b+c; the sum of the digits is a+b+c. So the statement of the problem says

We want to find the largest 3-digit number that satisfies the condition of the problem.

To get the largest such number, we want the value of a to be as large as possible.

All three of a, b, and c have to be single digit integers.

In particular, the largest possible value for both b and c is 9.

So the largest possible value for a is found from

And so the largest 3-digit number that satisfies the conditions is with a = 3 and b = c = 9: 399.