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

[TITA]

Accepted Answer

7. Let the 6-digit number be $d_1 d_2 d_3 d_4 d_5 d_6$. - $d_2 = 2 d_1$ - $d_3 = d_1$ - $d_5 = d_1 + d_2 = d_1 + 2d_1 = 3 d_1$ - $d_6 = d_1 + d_2 + d_3 = d_1 + 2d_1 + d_1 = 4 d_1$ - $d_4 = d_5 + d_6 = 3 d_1 + 4 d_1 = 7 d_1$ Since $d_4$ is a single digit ($0 \le d_4 \le 9$) and $d_1 \ge 1$: If $d_1 = 1 \implies d_4 = 7$. If $d_1 \ge 2 \implies d_4 \ge 14$ (not a single digit). Thus, the largest possible value of $d_4$ is 7.

Six-digit number condition

Explanation

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