Puzzle for April 20, 2019 ( )
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.
* CD and EF are 2-digit numbers (not C×D or E×F).
Scratchpad
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Hint #1
In eq.1, replace B + C with A + F (from eq.2), and replace D + E with A + F (from eq.5): A + A + F + A + F + F = 36 which becomes 3×A + 3×F = 36 Divide both sides of the above equation by 3: (3×A + 3×F) ÷ 3 = 36 ÷ 3 which makes eq.1a) A + F = 12
Hint #2
In eq.2, substitute 12 for A + F (from eq.1a): eq.2a) B + C = 12 In eq.5, substitute 12 for A + F (from eq.1a): eq.5a) D + E = 12
Hint #3
Subtract F from both sides of eq.1a: A + F - F = 12 - F which makes eq.1b) A = 12 - F Subtract E from both sides of eq.5a: D + E - E = 12 - E which makes eq.5b) D = 12 - E
Hint #4
Substitute 12 - F for A (from eq.1b), and 12 - E for D (from eq.5b) in eq.3: E + F = 12 - F + B + 12 - E Add (E + F - B) to both sides of the above equation: E + F + (E + F - B) = 12 - F + B + 12 - E + (E + F - B) which becomes eq.3a) 2×E + 2×F - B = 24
Hint #5
Subtract B from both sides of eq.2a: B + C - B = 12 - B which makes eq.2b) C = 12 - B
Hint #6
Substitute 12 - B for C (from eq.2b), 12 - E for D (from eq.5b), and 12 - F for A (from eq.1b) in eq.4: 12 - B + 12 - E - B = 12 - F + B which becomes 24 - 2×B - E = 12 - F + B Add (2×B + E - 12) to each side of the above equation: 24 - 2×B - E + (2×B + E - 12) = 12 - F + B + (2×B + E - 12) which becomes eq.4a) 12 = -F + 3×B + E
Hint #7
Multiply both sides of eq.4a by 2: 2 × 12 = 2 × (-F + 3×B + E) which becomes 24 = -2×F + 6×B + 2×E In eq.3a, substitute -2×F + 6×B + 2×E for 24: 2×E + 2×F - B = -2×F + 6×B + 2×E Add (B + 2×F - 2×E) to both sides of the equation above: 2×E + 2×F - B + (B + 2×F - 2×E) = -2×F + 6×B + 2×E + (B + 2×F - 2×E) which simplifies to 4×F = 7×B Divide each side by 4: 4×F ÷ 4 = 7×B ÷ 4 which makes F = 1¾×B
Hint #8
Substitute 1¾×B for F in eq.1b: eq.1c) A = 12 - 1¾×B
Hint #9
Substitute 1¾×B for F in eq.4a: 12 = -1¾×B + 3×B + E which becomes 12 = 1¼×B + E Subtract 1¼×B from both sides: 12 - 1¼×B = 1¼×B + E - 1¼×B which makes eq.4b) 12 - 1¼×B = E
Hint #10
Substitute (12 - 1¼×B) for E (from eq.4b) in eq.5b: D = 12 - (12 - 1¼×B) which becomes D = 12 - 12 + 1¼×B which means D = 1¼×B
Hint #11
eq.6 may be written as: 10×C + D = C + 10×E + F Subtract C from each side of the above equation: 10×C + D - C = C + 10×E + F - C which becomes eq.6a) 9×C + D = 10×E + F
Solution
Substitute (12 - B) for C (from eq.2b), 1¼×B for D, (12 - 1¼×B) for E (from eq.4b), and 1¾×B for F in eq.6a: 9×(12 - B) + 1¼×B = 10×(12 - 1¼×B) + 1¾×B which becomes 108 - 9×B + 1¼×B = 120 - 12½×B + 1¾×B which becomes 108 - 7¾×B = 120 - 10¾×B Add (10¾×B - 108) to both sides: 108 - 7¾×B + (10¾×B - 108) = 120 - 10¾×B + (10¾×B - 108) which simplifies to 3×B = 12 Divide both sides by 3: 3×B ÷ 3 = 12 ÷ 3 which means B = 4 making A = 12 - 1¾×B = 12 - 1¾×4 = 12 - 7 = 5 (from eq.1c) C = 12 - B = 12 - 4 = 8 (from eq.2b) D = 1¼×B = 1¼×4 = 5 E = 12 - 1¼×B = 12 - 1¼×4 = 12 - 5 = 7 (from eq.4b) F = 1¾×B = 1¾×4 = 7 and ABCDEF = 548577