Puzzle for January 29, 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.
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Hint #1
eq.5 may be written as: B = (A + C) ÷ 2 Multiply both sides of the above equation by 2: 2 × B = 2 × (A + C) ÷ 2 which becomes eq.5a) 2×B = A + C
Hint #2
eq.2 may be written as: F = (A + B + C + D) ÷ 4 Multiply both sides of the above equation by 4: 4 × F = 4 × (A + B + C + D) ÷ 4 which becomes 4×F = A + B + C + D which is the same as eq.2a) 4×F = A + C + B + D
Hint #3
In eq.2a, replace A + C with 2×B (from eq.5a): 4×F = 2×B + B + D which becomes 4×F = 3×B + D Subtract 3×B from each side of the equation above: 4×F - 3×B = 3×B + D - 3×B which becomes eq.2b) 4×F - 3×B = D
Hint #4
eq.6 may be written as: A = ((B + D) ÷ 2) + E Subtract E from each side of the equation above: A - E = ((B + D) ÷ 2) + E - E which becomes A - E = (B + D) ÷ 2 Multiply both sides of the above equation by 2: 2 × (A - E) = 2 × (B + D) ÷ 2 which becomes eq.6a) 2×A - 2×E = B + D
Hint #5
In eq.6a, replace D with 4×F - 3×B (from eq.2b): 2×A - 2×E = B + 4×F - 3×B which becomes 2×A - 2×E = 4×F - 2×B Divide both sides of the above equation by 2: (2×A - 2×E) ÷ 2 = (4×F - 2×B) ÷ 2 which becomes eq.6b) A - E = 2×F - B
Hint #6
eq.4 may be written as: E = (A + D + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × E = 3 × (A + D + F) ÷ 3 which becomes 3×E = A + D + F Subtract E from both sides: 3×E - E = A + D + F - E which becomes eq.4a) 2×E = A - E + D + F
Hint #7
In eq.4a, substitute 2×F - B for A - E (from eq.6b), and 4×F - 3×B for D (from eq.2b): 2×E = 2×F - B + 4×F - 3×B + F which becomes 2×E = 7×F - 4×B Divide both sides of the above equation by 2: 2×E ÷ 2 = (7×F - 4×B) ÷ 2 which becomes eq.4b) E = 3½×F - 2×B
Hint #8
Substitute (3½×F - 2×B) for E (from eq.4b) in eq.6b: A - (3½×F - 2×B) = 2×F - B which becomes A - 3½×F + 2×B = 2×F - B In the above equation, add 3½×F to both sides, and subtract 2×B from both sides: A - 3½×F + 2×B + 3½×F - 2×B = 2×F - B + 3½×F - 2×B which becomes eq.6c) A = 5½×F - 3×B
Hint #9
Substitute 5½×F - 3×B for A (from eq.6c) in eq.5a: 2×B = 5½×F - 3×B + C In the above equation, subtract 5½×F from both sides, and add 3×B to both sides: 2×B - 5½×F + 3×B = 5½×F - 3×B + C - 5½×F + 3×B which becomes eq.5b) 5×B - 5½×F = C
Hint #10
eq.3 may be written as: C = (A + B + E + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × C = 4 × (A + B + E + F) ÷ 4 which becomes eq.3a) 4×C = A + B + E + F
Hint #11
Substitute 5×B - 5½×F for C (from eq.5b), 5½×F - 3×B for A (from eq.6c), and 3½×F - 2×B for E (from eq.4b) in eq.3a: 4×(5×B - 5½×F) = 5½×F - 3×B + B + 3½×F - 2×B + F which becomes 20×B - 22×F = 10×F - 4×B Add 22×F and 4×B to both sides of the above equation: 20×B - 22×F + 22×F + 4×B = 10×F - 4×B + 22×F + 4×B which simplifies to 24×B = 32×F Divide both sides by 32: 24×B ÷ 32 = 32×F ÷ 32 which makes ¾×B = F
Hint #12
Substitute (¾×B) for F in eq.5b: 5×B - 5½×(¾×B) = C which becomes 5×B - 4⅛×B = C which makes ⅞×B = C
Hint #13
Substitute (¾×B) for F in eq.6c: A = 5½×(¾×B) - 3×B which becomes A = 4⅛×B - 3×B which makes A = 1⅛×B
Hint #14
Substitute (¾×B) for F in eq.4b: E = 3½×(¾×B) - 2×B which becomes E = 2⅝×B - 2×B which makes E = ⅝×B
Hint #15
Substitute (¾×B) for F in eq.2b: 4×(¾×B) - 3×B = D which becomes 3×B - 3×B = D which makes 0 = D
Solution
Substitute 1⅛×B for A, ⅞×B for C, 0 for D, ⅝×B for E, and ¾×B for F in eq.1: 1⅛×B + B + ⅞×B + 0 + ⅝×B + ¾×B = 35 which simplifies to 4⅜×B = 35 Divide both sides of the above equation by 4⅜: 4⅜×B ÷ 4⅜ = 35 ÷ 4⅜ which means B = 8 making A = 1⅛×B = 1⅛ × 8 = 9 C = ⅞×B = ⅞ × 8 = 7 E = ⅝×B = ⅝ × 8 = 5 F = ¾×B = ¾ × 8 = 6 and ABCDEF = 987056