Puzzle for April 5, 2020 ( )
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 A and C to both sides of eq.4: C + F – A + A + C = A – C + D + A + C which becomes 2×C + F = 2×A + D In the above equation, replace F with C + D (from eq.2): 2×C + C + D = 2×A + D which becomes 3×C + D = 2×A + D Subtract D from both sides: 3×C + D – D = 2×A + D – D which makes eq.4a) 3×C = 2×A Divide both sides of eq.4a by 3: 3×C ÷ 3 = 2×A ÷ 3 which makes C = ⅔×A
Hint #2
In eq.3, replace F with C + D (from eq.2): B – C + E = C + C + D – B which becomes B – C + E = 2×C + D – B Add C and B to each side of the above equation: B – C + E + C + B = 2×C + D – B + C + B which becomes eq.3a) 2×B + E = 3×C + D
Hint #3
In eq.3a, substitute 2×A for 3×C (from eq.4a): 2×B + E = 2×A + D Subtract 2×B from each side of the equation above: 2×B + E – 2×B = 2×A + D – 2×B which becomes eq.3b) E = 2×A + D – 2×B
Hint #4
Substitute ⅔×A for C in eq.5: A – ⅔×A + E = B + ⅔×A which becomes ⅓×A + E = B + ⅔×A Subtract ⅓×A from both sides of the equation above: ⅓×A + E – ⅓×A = B + ⅔×A – ⅓×A which becomes eq.5a) E = B + ⅓×A
Hint #5
Add B and E to both sides of eq.6: E – B + B + E = B – E + F + B + E which becomes 2×E = 2×B + F Substitute C + D for F (from eq.2) in the equation above: eq.6a) 2×E = 2×B + C + D
Hint #6
Substitute ⅔×A for C in eq.6a: 2×E = 2×B + ⅔×A + D Divide both sides of the above equation by 2: 2×E ÷ 2 = (2×B + ⅔×A + D) ÷ 2 which becomes eq.6b) E = B + ⅓×A + ½×D
Hint #7
Substitute E for B + ⅓×A (from eq.5a) in eq.6b: E = E + ½×D Subtract E from each side of the above equation: E – E = E + ½×D – E which makes 0 = ½×D which means 0 = D
Hint #8
Substitute 0 for D in eq.2: F = C + 0 which makes F = C and also makes eq.2a) F = C = ⅔×A
Hint #9
Substitute 0 for D in eq.3b: E = 2×A + 0 – 2×B which becomes eq.3c) E = 2×A – 2×B
Hint #10
Substitute 2×A – 2×B for E (from eq.3c) in eq.5a: 2×A – 2×B = B + ⅓×A In the above equation, add 2×B to both sides, and subtract ⅓×A from both sides: 2×A – 2×B + 2×B – ⅓×A = B + ⅓×A + 2×B – ⅓×A which makes eq.5b) 1⅔×A = 3×B
Hint #11
Multiply both sides of eq.5a by 3: 3×E = 3×(B + ⅓×A) which becomes 3×E = 3×B + A Substitute 1⅔×A for 3×B (from eq.5b) in the above equation: 3×E = 1⅔×A + A which makes 3×E = 2⅔×A Divide both sides by 2⅔: 3×E ÷ 2⅔ = 2⅔×A ÷ 2⅔ which makes 1⅛×E = A
Hint #12
Substitute 1⅛×E for A in eq.2a: F = C = ⅔×(1⅛×E) which makes F = C = ¾×E
Hint #13
Substitute (1⅛×E) for A in eq.5a: E = B + ⅓×(1⅛×E) which becomes E = B + ⅜×E Subtract ⅜×E from both sides of the equation above: E – ⅜×E = B + ⅜×E – ⅜×E which makes ⅝×E = B
Solution
Substitute 1⅛×E for A, ⅝×E for B, ¾×E for C and F, and 0 for D in eq.1: 1⅛×E + ⅝×E + ¾×E + 0 + E + ¾×E = 34 which simplifies to 4¼×E = 34 Divide both sides of the above equation by 4¼: 4¼×E ÷ 4¼ = 34 ÷ 4¼ which means E = 8 making A = 1⅛×E = 1⅛ × 8 = 9 B = ⅝×E = ⅝ × 8 = 5 C = F = ¾×E = ¾ × 8 = 6 and ABCDEF = 956086