Puzzle for April 16, 2020 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit non-negative integer.
* AB is a 2-digit number (not A×B).
Scratchpad
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Hint #1
eq.5 may be written as: 10×A + B – C = D + E + F In eq.1, replace D + E + F with 10×A + B – C: A + B + C + 10×A + B – C = 11 which becomes 11×A + 2×B = 11 Subtract 11×A from both sides of the equation above: 11×A + 2×B – 11×A = 11 – 11×A which becomes 2×B = 11 – 11×A Divide both sides by 2: 2×B ÷ 2 = (11 – 11×A) ÷ 2 which makes eq.1a) B = 5½ – 5½×A
Hint #2
To make eq.1a true, check several possible values for A and B: If A = 0, then B = 5½ – 5½×0 = 5½ – 0 = 5½ If A = 1, then B = 5½ – 5½×1 = 5½ – 5½ = 0 If A = 2, then B = 5½ – 5½×2 = 5½ – 11 = –5½ If A > 2, then B < –5½ Since B must be a non-negative integer, then B = 0 which means A = 1
Hint #3
In eq.2, replace A with 1, and replace B with 0: C – D = 1 + 0 Add D to each side of the above equation: C – D + D = 1 + 0 + D which makes eq.2a) C = 1 + D
Hint #4
In eq.3, substitute 0 for B, and 1 + D for C (from eq.2a): 0 + 1 + D = D + E Subtract D from both sides of the above equation: 0 + 1 + D – D = D + E – D which means 1 = E
Hint #5
Substitute (1 + D) for C (from eq.2a), and 1 for A in eq.4: F – (1 + D) = 1 + (1 + D) + D which becomes F – 1 – D = 2 + 2×D Add both 1 and D to each side of the above equation: F – 1 – D + 1 + D = 2 + 2×D + 1 + D which means eq.4a) F = 3 + 3×D
Solution
Substitute 1 for A and E, 0 for B, 1 + D for C (from eq.2a), and 3 + 3×D for F (from eq.4a) in eq.1: 1 + 0 + 1 + D + D + 1 + 3 + 3×D = 11 which becomes 6 + 5×D = 11 Subtract 6 from both sides of the above equation: 6 + 5×D – 6 = 11 – 6 which makes 5×D = 5 Divide both sides by 5: 5×D ÷ 5 = 5 ÷ 5 which means D = 1 making C = 1 + D = 1 + 1 = 2 (from eq.2a) F = 3 + 3×D = 3 + 3×1 = 3 + 3 = 6 (from eq.4a) and ABCDEF = 102116