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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
Correct answer is '7'. Can you explain this answer?
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In a six-digit number, the sixth, that is, the rightmost, digit is th...
Let the six-digit number be ABCDEF
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F = A+B+C, E= A+B, C=A, B= 2A, D= E+F.
Therefore D = 2A+2B+C = 2A + 4A + A= 7A.
A cannot be 0 as the number is a 6-digit number.
A cannot be 2 as D would become 2 digit number.
Therefore A is 1 and D is 7.
Most Upvoted Answer
In a six-digit number, the sixth, that is, the rightmost, digit is th...
Understanding the Problem
To solve this problem, we need to establish the relationships between the digits of the six-digit number, represented as \( ABCDEF \).
Setting Up the Equations
1. From the problem statement:
- \( F = A + B + C \) (The sixth digit)
- \( E = A + B \) (The fifth digit)
- \( C = A \) (The third digit)
- \( B = 2A \) (The second digit)
- \( D = E + F \) (The fourth digit)
Substituting the Values
2. Substituting \( B \) and \( C \):
- Since \( C = A \) and \( B = 2A \), we can express \( E \) and \( F \):
- \( E = A + B = A + 2A = 3A \)
- \( F = A + B + C = A + 2A + A = 4A \)
3. Finding the Fourth Digit:
- Now substituting \( E \) and \( F \) into the equation for \( D \):
- \( D = E + F = 3A + 4A = 7A \)
Constraints on the Digits
4. Considering Digit Values:
- The digits \( A, B, C, D, E, F \) must be between 0 and 9.
- Since \( B = 2A \), \( 2A \) must be ≤ 9, giving \( A \) ≤ 4.
- Since \( D = 7A \), \( 7A \) must be ≤ 9, giving \( A \) ≤ 1.
Maximizing the Fourth Digit
5. Finding Maximum A:
- The maximum value for \( A \) that satisfies both conditions is \( A = 1 \).
- Therefore, substituting \( A = 1 \):
- \( B = 2 \cdot 1 = 2 \)
- \( C = 1 \)
- \( D = 7 \cdot 1 = 7 \)
- \( E = 3 \cdot 1 = 3 \)
- \( F = 4 \cdot 1 = 4 \)
Conclusion
The largest possible value of the fourth digit \( D \) is 7.
To solve this problem, we need to establish the relationships between the digits of the six-digit number, represented as \( ABCDEF \).
Setting Up the Equations
1. From the problem statement:
- \( F = A + B + C \) (The sixth digit)
- \( E = A + B \) (The fifth digit)
- \( C = A \) (The third digit)
- \( B = 2A \) (The second digit)
- \( D = E + F \) (The fourth digit)
Substituting the Values
2. Substituting \( B \) and \( C \):
- Since \( C = A \) and \( B = 2A \), we can express \( E \) and \( F \):
- \( E = A + B = A + 2A = 3A \)
- \( F = A + B + C = A + 2A + A = 4A \)
3. Finding the Fourth Digit:
- Now substituting \( E \) and \( F \) into the equation for \( D \):
- \( D = E + F = 3A + 4A = 7A \)
Constraints on the Digits
4. Considering Digit Values:
- The digits \( A, B, C, D, E, F \) must be between 0 and 9.
- Since \( B = 2A \), \( 2A \) must be ≤ 9, giving \( A \) ≤ 4.
- Since \( D = 7A \), \( 7A \) must be ≤ 9, giving \( A \) ≤ 1.
Maximizing the Fourth Digit
5. Finding Maximum A:
- The maximum value for \( A \) that satisfies both conditions is \( A = 1 \).
- Therefore, substituting \( A = 1 \):
- \( B = 2 \cdot 1 = 2 \)
- \( C = 1 \)
- \( D = 7 \cdot 1 = 7 \)
- \( E = 3 \cdot 1 = 3 \)
- \( F = 4 \cdot 1 = 4 \)
Conclusion
The largest possible value of the fourth digit \( D \) is 7.
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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 isCorrect answer is '7'. Can you explain this answer?
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