Puzzle for January 22, 2022 ( )
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, BC, and CD are 2-digit numbers (not A×B, B×C, or C×D).
Scratchpad
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Hint #1
Subtract E from each side of eq.2: B + C – E = A + E + F – E which becomes eq.2a) B + C – E = A + F Add A to both sides of eq.3: B + F + A = C + E – A + A which becomes B + F + A = C + E which is the same as eq.3a) B + A + F = C + E
Hint #2
In eq.3a, replace A + F with B + C – E (from eq.2a): B + B + C – E = C + E which becomes 2×B + C – E = C + E In the above equation, subtract C from both sides, and add E to both sides: 2×B + C – E – C + E = C + E – C + E which makes 2×B = 2×E Divide both sides by 2: 2×B ÷ 2 = 2×E ÷ 2 which makes B = E
Hint #3
In eq.4, substitute B for E: D + B + F = A + B – F In the equation above, subtract B from both sides, and add F to both sides: D + B + F – B + F = A + B – F – B + F which becomes eq.4a) D + 2×F = A
Hint #4
In eq.3, substitute B for E, and (D + 2×F) for A (from eq.4a): B + F = C + B – (D + 2×F) which becomes B + F = C + B – D – 2×F In the above equation, subtract B from both sides, and add D and 2×F to both sides: B + F – B + D + 2×F = C + B – D – 2×F – B + D + 2×F which simplifies to eq.3b) D + 3×F = C
Hint #5
eq.5 may be written as: 10×A + B + C = 10×C + D + E In the equation above, substitute B for E: 10×A + B + C = 10×C + D + B Subtract B and C from each side: 10×A + B + C – B – C = 10×C + D + B – B – C which becomes eq.5a) 10×A = 9×C + D
Hint #6
Substitute (D + 2×F) for A (from eq.4a), and (D + 3×F) for C (from eq.3b) in eq.5a: 10×(D + 2×F) = 9×(D + 3×F) + D which becomes 10×D + 20×F = 9×D + 27×F + D which becomes 10×D + 20×F = 10×D + 27×F Subtract 10×D and 20×F from both sides of the above equation: 10×D + 20×F – 10×D – 20×F = 10×D + 27×F – 10×D – 20×F which simplifies to 0 = 7×F which means 0 = F
Hint #7
Substitute 0 for F in eq.3b: D + 3×0 = C which makes D = C
Hint #8
Substitute 0 for F in eq.4a: D + 2×0 = A which makes D = A and also makes D = A = C
Hint #9
eq.6 may be written as: 10×C + D – (10×B + C) – F = B + E + F which becomes 10×C + D – 10×B – C – F = B + E + F which becomes 9×C + D – 10×B – F = B + E + F Add 10×B and F to both sides of the above equation: 9×C + D – 10×B – F + 10×B + F = B + E + F + 10×B + F which becomes eq.6a) 9×C + D = 11×B + E + 2×F
Hint #10
Substitute C for D, B for E, and 0 for F in eq.6a: 9×C + C = 11×B + B + 0 which becomes 10×C = 12×B Divide both sides of the above equation by 10: 10×C ÷ 10 = 12×B ÷ 10 which makes C = 1⅕×B and also makes D = A = C = 1⅕×B
Solution
Substitute 1⅕×B for A and C and D, B for E, and 0 for F in eq.1: 1⅕×B + B + 1⅕×B + 1⅕×B + B + 0 = 28 which simplifies to 5⅗×B = 28 Divide both sides of the above equation by 5⅗: 5⅗×B ÷ 5⅗ = 28 ÷ 5⅗ which means B = 5 making A = C = D = 1⅕×B = 1⅕ × 5 = 6 E = B = 5 and ABCDEF = 656650