Question

If a, b, and c are digits for which the following holds, then a +b+c= 9 a 1-7 2 b= c 8 2a+b+c=

Answer 1

The sum of the digits
`\( a \)`,
`\( b \)`, and
`\( c \)` is
`14`.

The image appears to show a column addition problem with missing digits represented by a, b, and c:

9 a 1

+ 7 2 b

------------

c 8 2

To solve this, we follow the standard procedure for addition, starting from the rightmost column:

1. In the rightmost column,
`\( b + 1 \)` must equal 2, or result in 2 when 10 is subtracted (if there's a carry-over from the middle column). Since there's no carry-over mentioned, and b is a single digit,
`\( b = 1 \).`

2. In the middle column,
`\( a + 2 \)` must give us the units digit of 8. Since
`\( b = 1 \)`, and there's no carry-over from the rightmost column,
`\( a \)` must be 6 to make
`\( 6 + 2 = 8 \).`

3. In the leftmost column, we add 9 and 7 to get 16. The digit 'c' in the sum must be 1 more than the 10's place of this sum due to the carry-over from the middle column (if any). Since 9 + 7 = 16, there is a carry-over, and so
`\( c = 6 + 1 = 7 \).`

Now we can sum a, b, and c:

`\( a + b + c = 6 + 1 + 7 \)`

Calculating this, we get:

The sum of the digits
`\( a \)`,
`\( b \)`, and
`\( c \)` is
`14`.

Answer 2

11

Step-by-step explanation:

Given the difference:

`\begin{gathered} \; \; 9\text{ a 1} \\ -7\text{ 2 b} \\ ---- \\ \; \; c\text{ 8 2} \end{gathered}`
`\begin{gathered} 11-9=2 \\ \text{Therefore, b=9} \end{gathered}`
`\begin{gathered} 10-2=8 \\ 10+1=11 \\ \text{Therefore, a=1} \end{gathered}`

Lastly:

`\begin{gathered} (9-1)-7=1 \\ \text{Therefore: c=1} \end{gathered}`

We then have:

`\begin{gathered} \; \; 9\text{ 1 1} \\ -7\text{ 2 }9 \\ ---- \\ \; \; 1\text{ 8 2} \end{gathered}`

Therefore:

a+b+c=9+1+1=11