Question

In a six-digit number, the sixth, that is, the rightmost, digit is the sum of the first three digits, the fifth digit is the sum of first two digits, the third digit is equal to the first digit, the second digit is twice the first digit and the fourth digit is the sum of fifth and sixth digits. Then, the largest possible value of the fourth digit is

• A. 8
• B. 7
• C. 9
• D. 4

Answer 1

The Correct Option is B

To solve the problem, we need to translate the conditions into equations based on the digits of the number, denoted as \( a, b, c, d, e, f \):

1. The sixth digit \( f \) is the sum of the first three digits: \( f = a + b + c \). 
2. The fifth digit \( e \) is the sum of the first two digits: \( e = a + b \).
3. The third digit \( c \) is equal to the first digit: \( c = a \).
4. The second digit \( b \) is twice the first digit: \( b = 2a \).
5. The fourth digit \( d \) is the sum of the fifth and sixth digits: \( d = e + f \).

Now let's express \( d \) in terms of \( a \):

• Since \( c = a \), we have \( f = a + b + c = a + 2a + a = 4a \).
• Since \( e = a + b \), we have \( e = a + 2a = 3a \).
• Thus, substituting for \( e \) and \( f \), we get \( d = e + f = 3a + 4a = 7a \).

To find the largest possible value of \( d \), remember \( d \) must be a valid single digit (0-9):

• For the largest \( d = 7a \leq 9 \), hence the largest integer value \( a \) can take is 1, which leads to:
• \( d = 7 \times 1 = 7 \)

Therefore, the largest possible value of the fourth digit \( d \) is 7.

Largest possible value of \( d \)

7