Question

2nd digit + 1st digit = 12
3rd digit + 4th digit = 15
4th digit + 2nd digit = 13
1st digit + 4th digit = 11
3rd digit + 2nd digit = 16

Explanation:

Let the 1st digit be \( a \), 2nd digit be \( b \), 3rd digit be \( c \), and 4th digit be \( d \). We have the following equations:

Step 1: List the equations

1. \( a + b = 12 \)
2. \( c + d = 15 \)
3. \( d + b = 13 \)
4. \( a + d = 11 \)
5. \( c + b = 16 \)

Step 2: Solve for \( b \) from equation 1 and 3

From equation 1: \( a = 12 - b \)
Substitute \( a \) into equation 4: \( (12 - b) + d = 11 \) ⇒ \( d = b - 1 \)
Substitute \( d = b - 1 \) into equation 3: \( (b - 1) + b = 13 \) ⇒ \( 2b - 1 = 13 \) ⇒ \( 2b = 14 \) ⇒ \( b = 7 \)

Step 3: Solve for \( a \)

From equation 1: \( a + 7 = 12 \) ⇒ \( a = 5 \)

Step 4: Solve for \( d \)

From \( d = b - 1 \): \( d = 7 - 1 = 6 \)

Step 5: Solve for \( c \)

From equation 5: \( c + 7 = 16 \) ⇒ \( c = 9 \)

Answer:

1st digit \( a = 5 \), 2nd digit \( b = 7 \), 3rd digit \( c = 9 \), 4th digit \( d = 6 \)
The number is \( 5796 \) (if we consider the digits in order).