Puzzle for July 2, 2023 ( )
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.
Scratchpad
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Hint #1
Subtract B from each side of eq.6: A + D + E - B = B + C + F - B which becomes eq.6a) A + D + E - B = C + F Add C to both sides of eq.4: A - C + C = E + F - B + C which becomes A = E + F - B + C which may be written as eq.4a) A = E + C + F - B
Hint #2
In eq.4a, replace C + F with A + D + E - B (from eq.6a): A = E + A + D + E - B - B which becomes A = 2×E + A + D - 2×B In the above equation, subtract A from both sides, and add 2×B to both sides: A - A + 2×B = 2×E + A + D - 2×B - A + 2×B which becomes eq.4b) 2×B = 2×E + D
Hint #3
Add F and B to both sides of eq.5: B + E - F + F + B = A - B + F + F + B which becomes eq.5a) 2×B + E = A + 2×F In eq.5a, replace 2×B with 2×E + D (from eq.4b): 2×E + D + E = A + 2×F which becomes eq.5b) 3×E + D = A + 2×F
Hint #4
Subtract the left and right sides of eq.5b from the left and right sides of eq.2, respectively: D + F - (3×E + D) = A + E - (A + 2×F) which becomes D + F - 3×E - D = A + E - A - 2×F which becomes F - 3×E = E - 2×F Add 3×E and 2×F to both sides of the equation above: F - 3×E + 3×E + 2×F = E - 2×F + 3×E + 2×F which makes 3×F = 4×E Divide both sides by 4: 3×F ÷ 4 = 4×E ÷ 4 which makes ¾×F = E
Hint #5
In eq.2, substitute ¾×F for E: D + F = A + ¾×F Subtract ¾×F from each side of the equation above: D + F - ¾×F = A + ¾×F - ¾×F which becomes eq.2a) D + ¼×F = A
Hint #6
Substitute ¾×F for E, and D + ¼×F for A (from eq.2a) in eq.5a: 2×B + ¾×F = D + ¼×F + 2×F which becomes 2×B + ¾×F = D + 2¼×F Subtract ¾×F from each side of the above equation: 2×B + ¾×F - ¾×F = D + 2¼×F - ¾×F which becomes 2×B = D + 1½×F Divide both sides by 2: 2×B ÷ 2 = (D + 1½×F) ÷ 2 which becomes eq.5c) B = ½×D + ¾×F
Hint #7
Substitute ¾×F for E, and ½×D + ¾×F for B (from eq.5c) in eq.3: D + ¾×F = ½×D + ¾×F + C Subtract ¾×F and ½×D from each side of the equation above: D + ¾×F - ¾×F - ½×D = ½×D + ¾×F + C - ¾×F - ½×D which simplifies to eq.3a) ½×D = C
Hint #8
Substitute D + ¼×F for A (from eq.2a), ½×D for C (from eq.3a), ¾×F for E, and (½×D + ¾×F) for B (from eq.5c) in eq.4: D + ¼×F - ½×D = ¾×F + F - (½×D + ¾×F) which becomes ½×D + ¼×F = 1¾×F - ½×D - ¾×F which becomes ½×D + ¼×F = F - ½×D In the above equation, subtract ¼×F from both sides, and add ½×D to both sides: ½×D + ¼×F - ¼×F + ½×D = F - ½×D - ¼×F + ½×D which simplifies to D = ¾×F
Hint #9
Substitute (¾×F) for D in eq.3a: ½×(¾×F) = C which makes ⅜×F = C
Hint #10
Substitute (¾×F) for D in eq.5c: B = ½×(¾×F) + ¾×F which becomes B = ⅜×F + ¾×F which makes B = 1⅛×F
Hint #11
Substitute ¾×F for D in eq.2a: ¾×F + ¼×F = A which makes F = A
Solution
Substitute F for A, 1⅛×F for B, ⅜×F for C, and ¾×F for D and E in eq.1: F + 1⅛×F + ⅜×F + ¾×F + ¾×F + F = 40 which simplifies to 5×F = 40 Divide both sides of the above equation by 5: 5×F ÷ 5 = 40 ÷ 5 which means F = 8 making A = F = 8 B = 1⅛×F = 1⅛ × 8 = 9 C = ⅜×F = ⅜ × 8 = 3 D = E = ¾×F = ¾ × 8 = 6 and ABCDEF = 893668