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

Entered answer:

✅ Correct Answer: 7

Let us denote the six-digit number as: $d_1d_2d_3d_4d_5d_6$

From the problem, we have these conditions:

$d_6 = d_1 + d_2 + d_3$ (sixth digit = sum of first three digits)

$d_5 = d_1 + d_2$ (fifth digit = sum of first two digits)

$d_3 = d_1$ (third digit = first digit)

$d_2 = 2d_1$ (second digit = twice the first digit)

$d_4 = d_5 + d_6$ (fourth digit = sum of fifth and sixth digits)

Since $d_3 = d_1$ and $d_2 = 2d_1$ , we can substitute:

For $d_6$ : $d_6 = d_1 + d_2 + d_3 = d_1 + 2d_1 + d_1 = 4d_1$

For $d_5$ : $d_5 = d_1 + d_2 = d_1 + 2d_1 = 3d_1$

For $d_4$ : $d_4 = d_5 + d_6 = 3d_1 + 4d_1 = 7d_1$

So our six-digit number becomes: $d_1, 2d_1, d_1, 7d_1, 3d_1, 4d_1$

Since each position must be a single digit (0-9), we need:

$d_1 \leq 9$

$2d_1 \leq 9$

$7d_1 \leq 9$

$3d_1 \leq 9$

$4d_1 \leq 9$

The most restrictive constraint is $7d_1 \leq 9$ , which gives us $d_1 \leq \tfrac{9}{7} \approx 1.29$

Since $d_1$ must be a positive integer (as the first digit of a six-digit number), we have $d_1 = 1$ .

With $d_1 = 1$ , the maximum value of $d_4 = 7d_1 = 7 \times 1 = 7$ .

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