Puzzle for February 4, 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
Add F to both sides of eq.3: E + F + F = C + D - F + F which becomes eq.3a) E + 2×F = C + D Add B to both sides of eq.5: B + F + B = A + C + D - B + B which becomes eq.5a) 2×B + F = A + C + D
Hint #2
In eq.5a, replace C + D with E + 2×F (from eq.3a): 2×B + F = A + E + 2×F Subtract F from each side of the equation above: 2×B + F - F = A + E + 2×F - F which becomes eq.5b) 2×B = A + E + F
Hint #3
Add F and B to both sides of eq.4: B - F + F + B = D + E - B + F + B which becomes eq.4a) 2×B = D + E + F
Hint #4
In eq.4a, replace 2×B with A + E + F (from eq.5b): A + E + F = D + E + F Subtract E and F from each side of the above equation: A + E + F - E - F = D + E + F - E - F which simplifies to A = D
Hint #5
In eq.2, substitute D for A: F - E = D - F Add E and F to both sides of the equation above: F - E + E + F = D - F + E + F which becomes eq.2a) 2×F = D + E
Hint #6
Substitute 2×F for D + E (from eq.2a) into eq.4a: 2×B = 2×F + F which becomes 2×B = 3×F Divide both sides of the above equation by 2: 2×B ÷ 2 = 3×F ÷ 2 which makes eq.4b) B = 1½×F
Hint #7
Substitute D + E for 2×F (from eq.2a) in eq.3a: E + D + E = C + D which becomes 2×E + D = C + D Subtract D from each side of the equation above: 2×E + D - D = C + D - D which makes 2×E = C
Hint #8
eq.6 may be re-written as: (A + B + E) ÷ 3 = (B + C + D + E + F) ÷ 5 Multiply both sides of the above equation by 3 and 5: 3 × 5 × (A + B + E) ÷ 3 = 3 × 5 × (B + C + D + E + F) ÷ 5 which becomes 5 × (A + B + E) = 3 × (B + C + D + E + F) which becomes 5×A + 5×B + 5×E = 3×B + 3×C + 3×D + 3×E + 3×F Subtract 3×B and 3×E from each side of the above equation: 5×A + 5×B + 5×E - 3×B - 3×E = 3×B + 3×C + 3×D + 3×E + 3×F - 3×B - 3×E which becomes eq.6a) 5×A + 2×B + 2×E = 3×C + 3×D + 3×F
Hint #9
Substitute 1½×F for B, (2×E) for C, and A for D in eq.6a: 5×A + 2×(1½×F) + 2×E = 3×(2×E) + 3×A + 3×F which becomes 5×A + 3×F + 2×E = 6×E + 3×A + 3×F Subtract 2×E, 3×A, and 3×F from each side of the above equation: 5×A + 3×F + 2×E - 2×E - 3×A - 3×F = 6×E + 3×A + 3×F - 2×E - 3×A - 3×F which becomes 2×A = 4×E Divide both sides by 2: 2×A ÷ 2 = 4×E ÷ 2 which makes A = 2×E and also makes D = A = 2×E
Hint #10
Substitute 2×E for D in eq.2a: 2×F = 2×E + E which makes 2×F = 3×E Divide both sides of the above equation by 2: 2×F ÷ 2 = 3×E ÷ 2 which makes F = 1½×E
Hint #11
Substitute (1½×E) for F in eq.4b: B = 1½×(1½×E) which makes B = 2¼×E
Solution
Substitute 2×E for A and C and D, 2¼×E for B, and 1½×E for F in eq.1: 2×E + 2¼×E + 2×E + 2×E + E + 1½×E = 43 which simplifies to 10¾×E = 43 Divide both sides of the above equation by 10¾: 10¾×E ÷ 10¾ = 43 ÷ 10¾ which means E = 4 making A = C = D = 2×E = 2 × 4 = 8 B = 2¼×E = 2¼ × 4 = 9 F = 1½×E = 1½ × 4 = 6 and ABCDEF = 898846