Puzzle for November 7, 2021 ( )
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.
* BC and DE are 2-digit numbers (not B×C or D×E).
Scratchpad
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Hint #1
Add C, E, and F to both sides of eq.1: E – F – C + C + E + F = C – E + F + C + E + F which becomes 2×E = 2×C + 2×F Divide both sides of the above equation by 2: 2×E ÷ 2 = (2×C + 2×F) ÷ 2 which becomes eq.1a) E = C + F
Hint #2
eq.3 may be written as: 10×B + C + D = 10×D + E + B – A In the above equation, subtract D and B from both sides, and add A to both sides: 10×B + C + D – D – B + A = 10×D + E + B – A – D – B + A which becomes 9×B + C + A = 9×D + E which may be written as eq.3a) A + 9×B + C = 9×D + E
Hint #3
eq.4 may be written as: (A – B + 10×B + C) ÷ D = A which becomes (A + 9×B + C) ÷ D = A Multiply both sides of the above equation by D: ((A + 9×B + C) ÷ D) × D = A × D which becomes eq.4a) A + 9×B + C = A × D
Hint #4
In eq.3a, replace A + 9×B + C with A × D (from eq.4a): A × D = 9×D + E Subtract 9×D from each side of the equation above: A × D – 9×D = 9×D + E – 9×D which becomes A × D – 9×D = E which may be written as eq.3b) (A – 9) × D = E
Hint #5
A, D, and E are one-digit non-negative integers. D must also be non-zero (from eq.4). To make eq.3b true, check several possible values for A and E: If A = 9, then (9 – 9) × D = E making 0 × D = E which makes E = 0 If A = 8, then (8 – 9) × D = E making (–1) × D = E which makes E < 0 If A < 8, then E < 0 Since E ≥ 0, the only one-digit non-negative integer values for A and E that make eq.3b true are: A = 9 and E = 0
Hint #6
In eq.1a, substitute 0 for E: 0 = C + F Since C and F must be one-digit non-negative integers, the only value for C and F that makes the above equation true is: C = 0 and F = 0
Hint #7
Substitute 9 for A, and 0 for C and E in eq.3a: 9 + 9×B + 0 = 9×D + 0 which becomes 9×B + 9 = 9×D Divide both sides of the equation above by 9: (9×B + 9) ÷ 9 = 9×D ÷ 9 which makes eq.3c) B + 1 = D
Hint #8
Substitute 9 for A, and 0 for C and E and F in eq.2: D – (B – D) + 0 = 9 + B + 0 – D + 0 which becomes D – B + D = 9 + B – D which becomes 2×D – B = 9 + B – D Add B and D to both sides of the above equation: 2×D – B + B + D = 9 + B – D + B + D which becomes eq.2a) 3×D = 9 + 2×B
Hint #9
Substitute (B + 1) for D (from eq.3c) in eq.2a: 3×(B + 1) = 9 + 2×B which becomes 3×B + 3 = 9 + 2×B Subtract 3 and 2×B from both sides of the above equation: 3×B + 3 – 3 – 2×B = 9 + 2×B – 3 – 2×B which makes B = 6
Solution
Substitute 6 for B in eq.3c: 6 + 1 = D which makes 7 = D and makes ABCDEF = 960700