Suppose that a, b, and c are digits for which the number
708,16b,8c9 is a multiple of 99. Find a+b+c.
There is no way to determine a because you have provided no mathematical statement in which it appears. Perhaps you copied the question incorrectly. This is the second time you have posted this, and my response is the same.
my teacher gave it to me an i strictly copied an pasted. i think that i have to solve the equation using algebra.by finding the value for A,plug in in an so forth..is that right???
I think your teacher made an error in the assignment. It doesnt make sense. You can't ask for an "a" value if there is no place to put it in the number.
To determine whether a number is a multiple of 99, we use its divisibility rule.
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 difference between the sum of its odd-numbered digits and the sum of its even-numbered digits is divisible by 11.
Let's apply this rule to the given number, 708,16b,8c9.
First, we find the sum of the digits:
7 + 0 + 8 + 1 + 6 + b + 8 + c + 9 = 39 + b + c.
To be divisible by 9, the sum of the digits must also be divisible by 9. Therefore, 39 + b + c must be divisible by 9.
Next, we consider the difference between the sum of the odd-numbered digits (7 + 8 + 1 + 8 + 9 = 33) and the sum of the even-numbered digits (0 + 6 + b + c = 6 + b + c).
To be divisible by 11, the difference must also be divisible by 11. Therefore, 33 - (6 + b + c) = 27 - (b + c) must be divisible by 11.
To find the values of b and c that satisfy both conditions, we can test different values until we find the pair that satisfies both conditions. We start with b = 1 and c = 1.
Substituting b = 1 and c = 1 into the two conditions:
39 + b + c = 39 + 1 + 1 = 41, which is not divisible by 9.
27 - (b + c) = 27 - (1 + 1) = 25, which is not divisible by 11.
Let's try another pair: b = 9 and c = 1.
Substituting b = 9 and c = 1 into the two conditions:
39 + b + c = 39 + 9 + 1 = 49, which is not divisible by 9.
27 - (b + c) = 27 - (9 + 1) = 17, which is not divisible by 11.
Now, let's try b = 6 and c = 3.
Substituting b = 6 and c = 3 into the two conditions:
39 + b + c = 39 + 6 + 3 = 48, which is divisible by 9.
27 - (b + c) = 27 - (6 + 3) = 18, which is divisible by 11.
Therefore, the values b = 6 and c = 3 satisfy both conditions.
Now, we need to find the value of a. Since the number 708,16b,8c9 is divisible by 99, we can conclude that 39 + b + c is divisible by 9.
Substituting b = 6 and c = 3:
39 + 6 + 3 = 48, which is divisible by 9.
This means that a can be any digit that makes 48 + a divisible by 9.
To find such a digit, we can calculate the remainders when (48 + a) is divided by 9.
For example, when a = 1, (48 + 1) divided by 9 yields a remainder of 4. When a = 2, (48 + 2) divided by 9 gives a remainder of 5. We continue this process until we find a remainder of 0, meaning 48 + a is divisible by 9.
Testing different values of a, we find that a = 3 satisfies the condition:
48 + 3 = 51, which is divisible by 9.
Therefore, a = 3, b = 6, and c = 3.
Finally, we find the sum of a, b, and c:
a + b + c = 3 + 6 + 3 = 12.
Therefore, the sum of a, b, and c is 12.