Puzzle for June 26, 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, BC, and EF are 2-digit numbers (not A×B, B×C, or E×F).
Scratchpad
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Hint #1
In eq.4, replace B with D + E (from eq.2): F – C = D + E – D – E which becomes F – C = 0 Add C to both sides of the above equation: F – C + C = 0 + C which makes F = C
Hint #2
In eq.3, replace F with C: C – A = A + E – C In the above equation, subtract A from both sides, and add C to both sides: C – A – A + C = A + E – C – A + C which becomes eq.3a) 2×C – 2×A = E
Hint #3
In eq.5, substitute 2×C – 2×A for E (from eq.3a): A + C = B + 2×C – 2×A In the above equation, subtract 2×C from both sides, and add 2×A to both sides: A + C – 2×C + 2×A = B + 2×C – 2×A – 2×C + 2×A which becomes eq.5a) 3×A – C = B
Hint #4
Substitute 2×C – 2×A for E (from eq.3a), and 3×A – C for B (from eq.5a) in eq.2: D + 2×C – 2×A = 3×A – C In the equation above, subtract 2×C from both sides, and add 2×A to both sides: D + 2×C – 2×A – 2×C + 2×A = 3×A – C – 2×C + 2×A which becomes eq.2a) D = 5×A – 3×C
Hint #5
eq.6 may be written as: 10×B + C = 10×A + B + D – E + 10×E + F which becomes 10×B + C = 10×A + B + D + 9×E + F Substitute C for F in the equation above: 10×B + C = 10×A + B + D + 9×E + C Subtract B and C from each side: 10×B + C – B – C = 10×A + B + D + 9×E + C – B – C which becomes eq.6a) 9×B = 10×A + D + 9×E
Hint #6
Substitute (3×A – C) for B (from eq.5a), 5×A – 3×C for D (from eq.2a), and (2×C – 2×A) for E (from eq.3a) in eq.6a: 9×(3×A – C) = 10×A + 5×A – 3×C + 9×(2×C – 2×A) which becomes 27×A – 9×C = 15×A – 3×C + 18×C – 18×A which becomes 27×A – 9×C = –3×A + 15×C Add 9×C and 3×A to both sides of the above equation: 27×A – 9×C + 9×C + 3×A = –3×A + 15×C + 9×C + 3×A which becomes 30×A = 24×C Divide both sides by 24: 30×A ÷ 24 = 24×C ÷ 24 which makes 1¼×A = C and also makes F = C = 1¼×A
Hint #7
Substitute 1¼×A for C in eq.5a: 3×A – 1¼×A = B which makes 1¾×A = B
Hint #8
Substitute 1¼×A for C in eq.2a: D = 5×A – 3×(1¼×A) which becomes D = 5×A – 3¾×A which makes D = 1¼×A
Hint #9
Substitute 1¼×A for C in eq.3a: 2×(1¼×A) – 2×A = E which becomes 2½×A – 2×A = E which makes ½×A = E
Solution
Substitute 1¾×A for B, 1¼×A for C and D and F, and ½×A for E in eq.1: A + 1¾×A + 1¼×A + 1¼×A + ½×A + 1¼×A = 28 which simplifies to 7×A = 28 Divide both sides of the above equation by 7: 7×A ÷ 7 = 28 ÷ 7 which means A = 4 making B = 1¾×A = 1¾ × 4 = 7 C = D = F = 1¼×A = 1¼ × 4 = 5 E = ½×A = ½ × 4 = 2 and ABCDEF = 475525