Suppose that a, b, and c are digits for which the number
708,16b,8c9 is a multiple of 99. Find a+b+c.
To determine whether the number 708,16b,8c9 is a multiple of 99, we can use the divisibility rule for 99.
The divisibility rule for 99 states that a number is divisible by 99 if and only if the sum of its digits is divisible by 9 and the number formed by the tens and units digit is divisible by 11.
Let's first find the sum of the digits:
7 + 0 + 8 + 1 + 6 + b + 8 + c + 9
Simplifying this, we have:
(b + c + 31)
For the number to be divisible by 9, the sum of the digits should also be divisible by 9. So we have:
b + c + 31 ≡ 0 (mod 9)
Subtracting 31 from both sides, we get:
b + c ≡ -31 ≡ 4 (mod 9)
Now, let's look at the tens and units digit, which is formed by 89. For the number to be divisible by 11, the difference between the sum of the digits in the even positions and the sum of the digits in the odd positions should be divisible by 11.
The even-positioned digits are 8 and 9, while the odd-positioned digits are 1 and b. So we have:
(8 + 9) - (1 + b) ≡ 0 (mod 11)
Simplifying this, we get:
16 - b ≡ 0 (mod 11)
Now, we have a system of two congruences:
b + c ≡ 4 (mod 9)
16 - b ≡ 0 (mod 11)
To solve this system, we can find the values of b and c by trial and error.
We try different values for b and see if we can find a corresponding value for c that satisfies both congruences. Starting with b = 5, we have:
5 + c ≡ 4 (mod 9)
16 - 5 ≡ 0 (mod 11)
By trial and error, we find that when b = 5, c = 8 satisfies both congruences:
5 + 8 ≡ 4 (mod 9)
16 - 5 ≡ 0 (mod 11)
Therefore, a = 7, b = 5, and c = 8.
Finally, we calculate a+b+c:
7 + 5 + 8 = 20
So, a+b+c is equal to 20.