Puzzle for October 24, 2021 ( )
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 and C to both sides of eq.5: B – C + F + F + C = A + D – F + F + C which becomes B + 2×F = A + D + C which is the same as eq.5a) B + 2×F = A + C + D In eq.3, add A to both sides, and subtract C from both sides: E + F – A + A – C = A + C + A – C which becomes eq.3a) E + F – C = 2×A
Hint #2
In eq.5a, replace C + D with A + B + E (from eq.4): B + 2×F = A + A + B + E which becomes B + 2×F = 2×A + B + E Subtract B from each side of the above equation: B + 2×F – B = 2×A + B + E – B which becomes eq.5b) 2×F = 2×A + E
Hint #3
In eq.5b, replace 2×A with E + F – C (from eq.3a): 2×F = E + F – C + E which becomes 2×F = 2×E + F – C In the equation above, subtract F from both sides, and add C to both sides: 2×F – F + C = 2×E + F – C – F + C which makes eq.5c) F + C = 2×E
Hint #4
eq.6 may be written as: E = (B + C + D + F) ÷ 4 Multiply both sides of the above equation by 4: E × 4 = ((B + C + D + F) ÷ 4) × 4 which becomes eq.6a) 4×E = B + C + D + F
Hint #5
In eq.6a, substitute B + C for D + F (from eq.2): 4×E = B + C + B + C which becomes 4×E = 2×B + 2×C Divide both sides of the above equation by 2: 4×E ÷ 2 = (2×B + 2×C) ÷ 2 which becomes eq.6b) 2×E = B + C
Hint #6
In eq.5c, substitute B + C for 2×E (from eq.6b): F + C = B + C Subtract C from both sides of the above equation: F + C – C = B + C – C which makes F = B
Hint #7
Substitute B for F in eq.2: D + B = B + C Subtract B from each side of the above equation: D + B – B = B + C – B which makes D = C
Hint #8
Substitute F for B, and C for D in eq.5a: F + 2×F = A + C + C which becomes 3×F = A + 2×C Subtract 2×C from each side of the equation above: 3×F – 2×C = A + 2×C – 2×C which becomes eq.5d) 3×F – 2×C = A
Hint #9
Multiply both sides of eq.3a by 2: (E + F – C) × 2 = 2×A × 2 which becomes 2×E + 2×F – 2×C = 4×A Substitute F + C for 2×E (from eq.5c), and (3×F – 2×C) for A (from eq.5d) in the above equation: F + C + 2×F – 2×C = 4×(3×F – 2×C) which becomes 3×F – C = 12×F – 8×C Subtract 3×F from both sides, and add 8×C to both sides: 3×F – C – 3×F + 8×C = 12×F – 8×C – 3×F + 8×C which makes eq.3b) 7×C = 9×F
Hint #10
Multiply both sides of eq.5d by 3: (3×F – 2×C) × 3 = A × 3 which becomes 9×F – 6×C = 3×A Substitute 7×C for 9×F (from eq.3b) into the above equation: 7×C – 6×C = 3×A which makes C = 3×A and also makes D = C = 3×A
Hint #11
Substitute (3×A) for C in eq.5d: 3×F – 2×(3×A) = A which becomes 3×F – 6×A = A Add 6×A to both sides of the above equation: 3×F – 6×A + 6×A = A + 6×A which makes 3×F = 7×A Divide both sides by 3: 3×F ÷ 3 = 7×A ÷ 3 which makes F = 2⅓×A and also makes B = F = 2⅓×A
Hint #12
Substitute 2⅓×A for F, and 3×A for C in eq.5c: 2⅓×A + 3×A = 2×E which becomes 5⅓×A = 2×E Divide both sides of the above equation by 2: 5⅓×A ÷ 2 = 2×E ÷ 2 which makes 2⅔×A = E
Solution
Substitute 2⅓×A for B and F, 3×A for C and D, and 2⅔×A for E in eq.1: A + 2⅓×A + 3×A + 3×A + 2⅔×A + 2⅓×A = 43 which simplifies to 14⅓×A = 43 Divide both sides of the above equation by 14⅓: 14⅓×A ÷ 14⅓ = 43 ÷ 14⅓ which means A = 3 making B = F = 2⅓×A = 2⅓ × 3 = 7 C = D = 3×A = 3 × 3 = 9 E = 2⅔×A = 2⅔ × 3 = 8 and ABCDEF = 379987