Puzzle for December 17, 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
eq.5 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 eq.5a) 3×E = A + D + F
Hint #2
In eq.2, add A to both sides, and subtract B from both sides: F – A + A – B = B + D + A – B which becomes F – B = D + A which may be written as eq.2a) F – B = A + D
Hint #3
In eq.5a, replace A + D with F – B (from eq.2a): 3×E = F – B + F which becomes 3×E = 2×F – B In the above equation, add B to both sides, and subtract 3×E from both sides: 3×E + B – 3×E = 2×F – B + B – 3×E which becomes eq.5b) B = 2×F – 3×E
Hint #4
In eq.3, substitute (2×F – 3×E) for B (from eq.5b): (2×F – 3×E) – F = E – (2×F – 3×E) which is equivalent to F – 3×E = E – 2×F + 3×E which becomes F – 3×E = 4×E – 2×F Add 3×E and 2×F to both sides of the equation above: F – 3×E + 3×E + 2×F = 4×E – 2×F + 3×E + 2×F which becomes 3×F = 7×E Divide both sides by 3: 3×F ÷ 3 = 7×E ÷ 3 which makes F = 2⅓×E
Hint #5
Substitute (2⅓×E) for F in eq.5b: B = 2×(2⅓×E) – 3×E which becomes B = 4⅔×E – 3×E which makes B = 1⅔×E
Hint #6
Substitute 2⅓×E for F, and 1⅔×E for B in eq.4: A + 2⅓×E = 1⅔×E + E which becomes A + 2⅓×E = 2⅔×E Subtract 2⅓×E from each side of the equation above: A + 2⅓×E – 2⅓×E = 2⅔×E – 2⅓×E which makes A = ⅓×E
Hint #7
Substitute 2⅓×E for F, ⅓×E for A, and 1⅔×E for B in eq.2: 2⅓×E – ⅓×E = 1⅔×E + D which becomes 2×E = 1⅔×E + D Subtract 1⅔×E from both sides of the above equation: 2×E – 1⅔×E = 1⅔×E + D – 1⅔×E which makes ⅓×E = D
Hint #8
eq.6 may be written as: D + E = (B + C + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × (D + E) = 3 × (B + C + F) ÷ 3 which becomes eq.6a) 3×D + 3×E = B + C + F
Hint #9
Substitute ⅓×E for D, 1⅔×E for B, and 2⅓×E for F in eq.6a: 3×(⅓×E) + 3×E = 1⅔×E + C + 2⅓×E which becomes E + 3×E = 4×E + C which becomes 4×E = 4×E + C Subtract 4×E from each side of the above equation: 4×E – 4×E = 4×E + C – 4×E which makes 0 = C
Solution
Substitute ⅓×E for A and D, 1⅔×E for B, 0 for C, and 2⅓×E for F in eq.1: ⅓×E + 1⅔×E + 0 + ⅓×E + E + 2⅓×E = 17 which simplifies to 5⅔×E = 17 Divide both sides of the above equation by 5⅔: 5⅔×E ÷ 5⅔ = 17 ÷ 5⅔ which means E = 3 making A = D = ⅓×E = ⅓ × 3 = 1 B = 1⅔×E = 1⅔ × 3 = 5 F = 2⅓×E = 2⅓ × 3 = 7 and ABCDEF = 150137