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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 _____

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7 ?

Let $x$ be the first digit.

Now, according to the question,

The third digit is equal to the first digit.

• third digit $=$ first digit $= x$
• third digit $= x$

The second digit is twice the first digit.

• second digit $= 2 \times$ first digit
• second digit  $= 2x$

The six digit is the sum of the first three digits.

• sixth digit $=$ first digit $+$ second digit $+$ third digit.
• sixth digit $= x + 2x + x$
• sixth digit $= 4x$

The fifth digit is the sum of first two digits.

• fifth digit $=$ first digit $+$ second digit
• fifth digit $= x + 2x$
• fifth digit $= 3x$

The fourth digit is the sum of the fifth and sixth digits.

• fourth digit $=$ fifth digit $+$ sixth digit
• fourth digit $= 3x + 4x$
• fourth digit $= 7x$

Therefore, the fourth digit will be a multiple of $7.$

The largest possible value of the fourth digit will be $7$ when the value of  $x$ will be $1$.

Because if we put $x$ equal to $2$ it will give the two-digit number equal to $\left(7\times 2 = 14\right)$ for the fourth digit, which is not possible.

We can also check for any value of $x.$

$\therefore$ The largest possible value of the fourth digit $= 7.$

Correct Answer $:7$

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