Puzzle for October 2, 2022 ( )
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
eq.6 may be written as: D = (B + C + E + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × D = 4 × (B + C + E + F) ÷ 4 which becomes 4×D = B + C + E + F which may be written as eq.6a) 4×D = B + C + F + E
Hint #2
eq.5 may be written as: E = (A + C + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × E = 3 × (A + C + F) ÷ 3 which becomes eq.5a) 3×E = A + C + F Subtract A from each side of eq.5a: 3×E - A = A + C + F - A which becomes eq.5b) 3×E - A = C + F
Hint #3
In eq.6a, replace C + F with 3×E - A (from eq.5b): 4×D = B + 3×E - A + E which becomes 4×D = B + 4×E - A Add A to both sides of the above equation: 4×D + A = B + 4×E - A + A which becomes eq.6b) 4×D + A = B + 4×E
Hint #4
Add D and F to both sides of eq.2: E - D + D + F = D - F + D + F which becomes eq.2a) E + F = 2×D
Hint #5
Add F to both sides of eq.4: B + F + F = A + E - F + F which becomes eq.4a) B + 2×F = A + E Add E to both sides of eq.6b: 4×D + A + E = B + 4×E + E which becomes eq.6c) 4×D + A + E = B + 5×E
Hint #6
In eq.6c, substitute B + 2×F for A + E (from eq.4a): 4×D + B + 2×F = B + 5×E Subtract B from each side of the equation above: 4×D + B + 2×F - B = B + 5×E - B which becomes 4×D + 2×F = 5×E which may be written as eq.6d) 2×(2×D) + 2×F = 5×E
Hint #7
Substitute E + F for 2×D (from eq.2a): 2×(E + F) + 2×F = 5×E which becomes 2×E + 2×F + 2×F = 5×E which becomes 2×E + 4×F = 5×E Subtract 2×E from each side of the above equation: 2×E + 4×F - 2×E = 5×E - 2×E which makes 4×F = 3×E Divide both sides by 4: 4×F ÷ 4 = 3×E ÷ 4 which makes F = ¾×E
Hint #8
Substitute ¾×E for F in eq.2a: E + ¾×E = 2×D which makes 1¾×E = 2×D Divide both sides of the above equation by 2: 1¾×E ÷ 2 = 2×D ÷ 2 which makes ⅞×E = D
Hint #9
Add B to both sides of eq.3: B + E + B = A - B + C + B which becomes eq.3a) 2×B + E = A + C
Hint #10
Substitute 2×B + E for A + C (from eq.3a), and ¾×E for F in eq.5a: 3×E = 2×B + E + ¾×E which becomes 3×E = 2×B + 1¾×E Subtract 1¾×E from each side of the above equation: 3×E - 1¾×E = 2×B + 1¾×E - 1¾×E which makes 1¼×E = 2×B Divide both sides of the above equation by 2: 1¼×E ÷ 2 = 2×B ÷ 2 which makes ⅝×E = B
Hint #11
Substitute ⅝×E for B, and (¾×E) for F in eq.4a: A + E = ⅝×E + 2×(¾×E) which becomes A + E = ⅝×E + 1½×E which becomes A + E = 2⅛×E Subtract E from each side of the equation above: A + E - E = 2⅛×E - E which makes A = 1⅛×E
Hint #12
Substitute 1⅛×E for A, and (⅝×E) for B in eq.3a: 1⅛×E + C = 2×(⅝×E) + E which becomes 1⅛×E + C = 1¼×E + E which becomes 1⅛×E + C = 2¼×E Subtract 1⅛×E from each side of the above equation: 1⅛×E + C - 1⅛×E = 2¼×E - 1⅛×E which makes C = 1⅛×E
Solution
Substitute 1⅛×E for A and C, ⅝×E for B, ⅞×E for D, and ¾×E for F in eq.1: 1⅛×E + ⅝×E + 1⅛×E + ⅞×E + E + ¾×E = 44 which simplifies to 5½×E = 44 Divide both sides of the above equation by 5½: 5½×E ÷ 5½ = 44 ÷ 5½ which means E = 8 making A = C = 1⅛×E = 1⅛ × 8 = 9 B = ⅝×E = ⅝ × 8 = 5 D = ⅞×E = ⅞ × 8 = 7 F = ¾×E = ¾ × 8 = 6 and ABCDEF = 959786