Question

1st digit + 2nd digit = 3
3rd digit + 2nd digit = 6
4th digit + 1st digit = 9
4th digit + 2nd digit = 10
1st digit + 3rd digit = 5

Explanation:

Define variables for the digits

Let the four digits be represented by variables:

• Let \(a\) be the 1st digit.
• Let \(b\) be the 2nd digit.
• Let \(c\) be the 3rd digit.
• Let \(d\) be the 4th digit.

Write the system of equations

Translate the given clues into algebraic equations:

1. \(a + b = 3\)
2. \(c + b = 6\)
3. \(d + a = 9\)
4. \(d + b = 10\)
5. \(a + c = 5\)

Solve for the first two variables

Using equations (3) and (4):
Subtract equation (3) from equation (4):

$$(d + b) - (d + a) = 10 - 9$$

$$b - a = 1$$

Now, use equation (1), \(a + b = 3\), which is equivalent to:

$$b + a = 3$$

Add these two equations together:

$$(b - a) + (b + a) = 1 + 3$$

$$2b = 4 \implies b = 2$$

Substitute \(b = 2\) back into \(b - a = 1\):

$$2 - a = 1 \implies a = 1$$

Solve for the remaining variables

Substitute the known values of \(a\) and \(b\) to find \(c\) and \(d\):
Using equation (2) with \(b = 2\):

$$c + 2 = 6 \implies c = 4$$

Using equation (3) with \(a = 1\):

$$d + 1 = 9 \implies d = 8$$

Verify all equations

Check the values \(a = 1\), \(b = 2\), \(c = 4\), and \(d = 8\) against all clues:

• 1st + 2nd: \(1 + 2 = 3\) (Correct)
• 3rd + 2nd: \(4 + 2 = 6\) (Correct)
• 4th + 1st: \(8 + 1 = 9\) (Correct)
• 4th + 2nd: \(8 + 2 = 10\) (Correct)
• 1st + 3rd: \(1 + 4 = 5\) (Correct)

The digits are \(1, 2, 4, 8\).

Answer:

The four digits are:

• 1st digit: 1
• 2nd digit: 2
• 3rd digit: 4
• 4th digit: 8