Solved on Dec 09, 2023

Find the digits A, B, and C that solve the addition problem.

STEP 1

Assumptions
1. The given problem is a simple addition problem with missing digits represented by A, B, and C.
2. The digits A, B, and C are each a single digit from 0 to 9.
3. The sum of the two numbers is 1C63, where C is one of the digits we need to find.
4. The problem requires carrying over digits if the sum of any column exceeds 9.

STEP 2

First, we will look at the units column. Since there is no digit below the 7 in the units column, we can infer that A + 7 must end with a 3.
$A + 7 \equiv 3 \mod{10}$

STEP 3

Solve for A by considering the possible values that satisfy the congruence relation from the previous step.
$A = 3 - 7 \mod{10}$ $A = -4 \mod{10}$ $A = 6$

STEP 4

Now that we have the value of A, we can place it in the addition problem and look at the tens column.
$\begin{array}{r} 54 6 \\ +3 B 7 \\ \hline C 63 \\ \end{array}$

STEP 5

In the tens column, we have 4 + B, and since there is no carry from the units column, this sum must end with a 6.
$4 + B \equiv 6 \mod{10}$

STEP 6

Solve for B by considering the possible values that satisfy the congruence relation from the previous step.
$B = 6 - 4 \mod{10}$ $B = 2 \mod{10}$ $B = 2$

STEP 7

Now that we have the value of B, we can place it in the addition problem and look at the hundreds column.
$\begin{array}{r} 546 \\ +327 \\ \hline C 63 \\ \end{array}$

STEP 8

In the hundreds column, we have 5 + 3, which equals 8. Since there is no carry from the tens column, C must be 8.
$C = 5 + 3$ $C = 8$

STEP 9

Now that we have the value of C, we can place it in the addition problem to verify our solution.
$\begin{array}{r} 546 \\ +327 \\ \hline 863 \\ \end{array}$

STEP 10

Verify that the sum of each column is correct and that it matches the given sum of 1C63.
$6 + 7 = 13$ (3 in the units place and carry over 1 to the tens place) $4 + 2 + 1 = 7$ (7 in the tens place) $5 + 3 = 8$ (8 in the hundreds place)

STEP 11

Since the sum of each column is correct and matches the given sum, we have found the correct digits for A, B, and C.
The digits that replace A, B, and C are 6, 2, and 8 respectively.

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Find the digits A, B, and C that solve the addition problem.

STEP 1

STEP 2

First, we will look at the units column. Since there is no digit below the 7 in the units column, we can infer that A + 7 must end with a 3.
$A + 7 \equiv 3 \mod{10}$

STEP 3

Solve for A by considering the possible values that satisfy the congruence relation from the previous step.
$A = 3 - 7 \mod{10}$ $A = -4 \mod{10}$ $A = 6$

STEP 4

Now that we have the value of A, we can place it in the addition problem and look at the tens column.
$\begin{array}{r} 54 6 \\ +3 B 7 \\ \hline C 63 \\ \end{array}$

STEP 5

In the tens column, we have 4 + B, and since there is no carry from the units column, this sum must end with a 6.
$4 + B \equiv 6 \mod{10}$

STEP 6

Solve for B by considering the possible values that satisfy the congruence relation from the previous step.
$B = 6 - 4 \mod{10}$ $B = 2 \mod{10}$ $B = 2$

STEP 7

Now that we have the value of B, we can place it in the addition problem and look at the hundreds column.
$\begin{array}{r} 546 \\ +327 \\ \hline C 63 \\ \end{array}$

STEP 8

In the hundreds column, we have 5 + 3, which equals 8. Since there is no carry from the tens column, C must be 8.
$C = 5 + 3$ $C = 8$

STEP 9

Now that we have the value of C, we can place it in the addition problem to verify our solution.
$\begin{array}{r} 546 \\ +327 \\ \hline 863 \\ \end{array}$

STEP 10

Verify that the sum of each column is correct and that it matches the given sum of 1C63.
$6 + 7 = 13$ (3 in the units place and carry over 1 to the tens place) $4 + 2 + 1 = 7$ (7 in the tens place) $5 + 3 = 8$ (8 in the hundreds place)

STEP 11

Since the sum of each column is correct and matches the given sum, we have found the correct digits for A, B, and C.
The digits that replace A, B, and C are 6, 2, and 8 respectively.

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Find the digits A, B, and C that solve the addition problem.

STEP 1

STEP 2

First, we will look at the units column. Since there is no digit below the 7 in the units column, we can infer that A + 7 must end with a 3.A+7≡3mod 10 A + 7 \equiv 3 \mod{10} A+7≡3mod10

STEP 3

Solve for A by considering the possible values that satisfy the congruence relation from the previous step.A=3−7mod 10 A = 3 - 7 \mod{10} A=3−7mod10 A=−4mod 10 A = -4 \mod{10} A=−4mod10 A=6 A = 6 A=6

STEP 4

Now that we have the value of A, we can place it in the addition problem and look at the tens column.546+3B7C63 \begin{array}{r} 54 6 \\ +3 B 7 \\ \hline C 63 \\ \end{array} 546+3B7C63​​

STEP 5

In the tens column, we have 4 + B, and since there is no carry from the units column, this sum must end with a 6.4+B≡6mod 10 4 + B \equiv 6 \mod{10} 4+B≡6mod10

STEP 6

Solve for B by considering the possible values that satisfy the congruence relation from the previous step.B=6−4mod 10 B = 6 - 4 \mod{10} B=6−4mod10 B=2mod 10 B = 2 \mod{10} B=2mod10 B=2 B = 2 B=2

STEP 7

Now that we have the value of B, we can place it in the addition problem and look at the hundreds column.546+327C63 \begin{array}{r} 546 \\ +327 \\ \hline C 63 \\ \end{array} 546+327C63​​

STEP 8

In the hundreds column, we have 5 + 3, which equals 8. Since there is no carry from the tens column, C must be 8.C=5+3 C = 5 + 3 C=5+3 C=8 C = 8 C=8

STEP 9

Now that we have the value of C, we can place it in the addition problem to verify our solution.546+327863 \begin{array}{r} 546 \\ +327 \\ \hline 863 \\ \end{array} 546+327863​​

STEP 10

Verify that the sum of each column is correct and that it matches the given sum of 1C63.6+7=13 6 + 7 = 13 6+7=13 (3 in the units place and carry over 1 to the tens place) 4+2+1=7 4 + 2 + 1 = 7 4+2+1=7 (7 in the tens place) 5+3=8 5 + 3 = 8 5+3=8 (8 in the hundreds place)

STEP 11

Since the sum of each column is correct and matches the given sum, we have found the correct digits for A, B, and C.The digits that replace A, B, and C are 6, 2, and 8 respectively.

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First, we will look at the units column. Since there is no digit below the 7 in the units column, we can infer that A + 7 must end with a 3.
$A + 7 \equiv 3 \mod{10}$

Solve for A by considering the possible values that satisfy the congruence relation from the previous step.
$A = 3 - 7 \mod{10}$ $A = -4 \mod{10}$ $A = 6$

Now that we have the value of A, we can place it in the addition problem and look at the tens column.
$\begin{array}{r} 54 6 \\ +3 B 7 \\ \hline C 63 \\ \end{array}$

In the tens column, we have 4 + B, and since there is no carry from the units column, this sum must end with a 6.
$4 + B \equiv 6 \mod{10}$

Solve for B by considering the possible values that satisfy the congruence relation from the previous step.
$B = 6 - 4 \mod{10}$ $B = 2 \mod{10}$ $B = 2$

Now that we have the value of B, we can place it in the addition problem and look at the hundreds column.
$\begin{array}{r} 546 \\ +327 \\ \hline C 63 \\ \end{array}$

In the hundreds column, we have 5 + 3, which equals 8. Since there is no carry from the tens column, C must be 8.
$C = 5 + 3$ $C = 8$

Now that we have the value of C, we can place it in the addition problem to verify our solution.
$\begin{array}{r} 546 \\ +327 \\ \hline 863 \\ \end{array}$

Verify that the sum of each column is correct and that it matches the given sum of 1C63.
$6 + 7 = 13$ (3 in the units place and carry over 1 to the tens place) $4 + 2 + 1 = 7$ (7 in the tens place) $5 + 3 = 8$ (8 in the hundreds place)

Since the sum of each column is correct and matches the given sum, we have found the correct digits for A, B, and C.
The digits that replace A, B, and C are 6, 2, and 8 respectively.