Puzzle for November 27, 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
Add the left and right sides of eq.5 to the left and right sides of eq.3, respectively: A + F + B + E = C + E + C + D + F which becomes A + F + B + E = 2×C + E + D + F Subtract F and E from each side of the equation above: A + F + B + E - F - E = 2×C + E + D + F - F - E which makes eq.3a) A + B = 2×C + D
Hint #2
Add B and C to both sides of eq.4: C - B + B + C = A - C + D + B + C which becomes 2×C = A + D + B which may be written as eq.4a) 2×C = A + B + D
Hint #3
In eq.4a, replace A + B with 2×C + D (from eq.3a): 2×C = 2×C + D + D which becomes 2×C = 2×C + 2×D Subtract 2×C from each side of the above equation: 2×C - 2×C = 2×C + 2×D - 2×C which makes 0 = 2×D which means 0 = D
Hint #4
In eq.2, replace D with 0: B + 0 = A + E which becomes eq.2a) B = A + E
Hint #5
eq.6 may be written as: F = (A + D + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × F = 3 × (A + D + E) ÷ 3 which becomes 3×F = A + D + E which may be written as eq.6a) 3×F = A + E + D
Hint #6
In eq.6a, substitute B for A + E (from eq.2a), and 0 for D: 3×F = B + 0 which makes 3×F = B
Hint #7
Substitute 3×F for B, and 0 for D in eq.5: 3×F + E = C + 0 + F which becomes 3×F + E = C + F Subtract F from both sides of the equation above: 3×F + E - F = C + F - F which becomes eq.5a) 2×F + E = C
Hint #8
Substitute 2×F + E for C (from eq.5a) in eq.3: A + F = 2×F + E + E which becomes A + F = 2×F + 2×E Subtract F from both sides of the above equation: A + F - F = 2×F + 2×E - F which becomes eq.3b) A = F + 2×E
Hint #9
Substitute 3×F for B, and F + 2×E for A (from eq.3b) in eq.2: 3×F + D = F + 2×E + E which becomes 3×F + D = F + 3×E Subtract 3×F from each side of the above equation: 3×F + D - 3×F = F + 3×E - 3×F which becomes eq.2b) D = 3×E - 2×F
Hint #10
Substitute (2×F + E) for C (from eq.5a), F + 2×E for A (from eq.3b), 3×F for B, and 3×E - 2×F for D (from eq.2b) in eq.4a: 2×(2×F + E) = F + 2×E + 3×F + 3×E - 2×F which becomes 4×F + 2×E = 5×E + 2×F Subtract 2×E and 2×F from each side of the equation above: 4×F + 2×E - 2×E - 2×F = 5×E + 2×F - 2×E - 2×F which becomes 2×F = 3×E Divide both sides by 3: 2×F ÷ 3 = 3×E ÷ 3 which makes ⅔×F = E
Hint #11
Substitute (⅔×F) for E in eq.3b: A = F + 2×(⅔×F) which becomes A = F + 1⅓×F which makes A = 2⅓×F
Hint #12
Substitute ⅔×F for E in eq.5a: 2×F + ⅔×F = C which makes 2⅔×F = C
Hint #13
Substitute (⅔×F) for E in eq.2b: D = 3×(⅔×F) - 2×F which becomes D = 2×F - 2×F which makes D = 0
Solution
Substitute 2⅓×F for A, 3×F for B, 2⅔×F for C, 0 for D, and ⅔×F for E in eq.1: 2⅓×F + 3×F + 2⅔×F + 0 + ⅔×F + F = 29 which simplifies to 9⅔×F = 29 Divide both sides of the above equation by 9⅔: 9⅔×F ÷ 9⅔ = 29 ÷ 9⅔ which means F = 3 making A = 2⅓×F = 2⅓ × 3 = 7 B = 3×F = 3 × 3 = 9 C = 2⅔×F = 2⅔ × 3 = 8 E = ⅔×F = ⅔ × 3 = 2 and ABCDEF = 798023