Puzzle for April 3, 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
In eq.6, replace A + F with C + E (from eq.3): C + E = B + E – A In the above equation, subtract E from both sides, and add A to both sides: C + E – E + A = B + E – A – E + A which becomes eq.6a) C + A = B
Hint #2
In eq.4, replace B with C + A (from eq.6a): C + A + D = A + E Subtract A from each side of the above equation: C + A + D – A = A + E – A which becomes eq.4a) C + D = E
Hint #3
In eq.2, substitute C + D for E (from eq.4a): C + D = B + C Subtract C from each side of the equation above: C + D – C = B + C – C which makes B = D
Hint #4
Add A to both sides of eq.6: A + F + A = B + E – A + A which becomes eq.6b) 2×A + F = B + E Add F to both sides of eq.5: F – A + F = D – E – F + F which becomes eq.5a) 2×F – A = D – E
Hint #5
Add the left and right sides of eq.5a to the left and right sides of eq.6b, respectively: 2×A + F + 2×F – A = B + E + D – E which becomes eq.6c) A + 3×F = B + D
Hint #6
Substitute A + E for B + D (from eq.4) in eq.6c: A + 3×F = A + E Subtract A from both sides of the equation above: A + 3×F – A = A + E – A which makes 3×F = E
Hint #7
Substitute B for D, and 3×F for E in eq.5a: 2×F – A = B – 3×F In the above equation, add A and 3×F to both sides, and subtract B from both sides: 2×F – A + A + 3×F – B = B – 3×F + A + 3×F – B which simplifies to eq.5b) 5×F – B = A
Hint #8
Substitute 5×F – B for A (from eq.5b), and B for D in eq.6c: 5×F – B + 3×F = B + B which becomes 8×F – B = 2×B Add B to both sides of the equation above: 8×F – B + B = 2×B + B which makes 8×F = 3×B Divide both sides by 3: 8×F ÷ 3 = 3×B ÷ 3 which makes 2⅔×F = B and also makes D = B = 2⅔×F
Hint #9
Substitute 2⅔×F for B in eq.5b: 5×F – 2⅔×F = A which makes 2⅓×F = A
Hint #10
Substitute 3×F for E, and 2⅔×F for B in eq.2: 3×F = 2⅔×F + C Subtract 2⅔×F from each side of the above equation: 3×F – 2⅔×F = 2⅔×F + C – 2⅔×F which makes ⅓×F = C
Solution
Substitute 2⅓×F for A, 2⅔×F for B and D, ⅓×F for C, and 3×F for E in eq.1: 2⅓×F + 2⅔×F + ⅓×F + 2⅔×F + 3×F + F = 36 which simplifies to 12×F = 36 Divide both sides of the above equation by 12: 12×F ÷ 12 = 36 ÷ 12 which means F = 3 making A = 2⅓×F = 2⅓ × 3 = 7 B = D = 2⅔×F = 2⅔ × 3 = 8 C = ⅓×F = ⅓ × 3 = 1 E = 3×F = 3 × 3 = 9 and ABCDEF = 781893