Puzzle for June 12, 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 C and B to both sides of eq.2: B – C + D + C + B = A – B + E + C + B which becomes eq.2a) 2×B + D = A + E + C Add B and E to both sides of eq.6: B – E + F + B + E = A – B + C + E + B + E which becomes eq.6a) 2×B + F = A + C + 2×E
Hint #2
Subtract the left and right sides of eq.2a from the left and right sides of eq.6a, respectively: 2×B + F – (2×B + D) = A + C + 2×E – (A + E + C) which becomes 2×B + F – 2×B – D = A + C + 2×E – A – E – C which simplifies to eq.6b) F – D = E
Hint #3
In eq.5, substitute (F – D) for E (from eq.6b): F – B – C = B + C – D – (F – D) which becomes F – B – C = B + C – D – F + D which becomes F – B – C = B + C – F Add B, C, and F to both sides of the above equation: F – B – C + B + C + F = B + C – F + B + C + F which becomes 2×F = 2×B + 2×C Divide both sides by 2: 2×F ÷ 2 = (2×B + 2×C) ÷ 2 which becomes eq.5a) F = B + C
Hint #4
In eq.6a, replace F with B + C (from eq.5a): 2×B + B + C = A + C + 2×E which becomes 3×B + C = A + C + 2×E Subtract C from each side of the above equation: 3×B + C – C = A + C + 2×E – C which becomes eq.6c) 3×B = A + 2×E
Hint #5
Add B, C, and F to both sides of eq.3: C + E – B + B + C + F = A + B – C – F + B + C + F which becomes eq.3a) 2×C + E + F = A + 2×B Add 2×E to both sides of eq.3a: 2×C + E + F + 2×E = A + 2×B + 2×E which becomes 2×C + 3×E + F = A + 2×B + 2×E which may be written as eq.3b) 2×C + 3×E + F = A + 2×E + 2×B
Hint #6
In eq.3b, replace F with B + C (from eq.5a), and A + 2×E with 3×B (from eq.6c): 2×C + 3×E + B + C = 3×B + 2×B which becomes 3×C + 3×E + B = 5×B Subtract 3×C and B from each side of the equation above: 3×C + 3×E + B – 3×C – B = 5×B – 3×C – B which becomes eq.3c) 3×E = 4×B – 3×C
Hint #7
Multiply both sides of eq.3a by 3: 3×(2×C + E + F) = 3×(A + 2×B) which becomes 6×C + 3×E + 3×F = 3×A + 6×B Substitute 4×B – 3×C for 3×E (from eq.3c), and (B + C) for F (from eq.5a) into the equation above: 6×C + 4×B – 3×C + 3×(B + C) = 3×A + 6×B which becomes 6×C + 4×B – 3×C + 3×B + 3×C = 3×A + 6×B which becomes 6×C + 7×B = 3×A + 6×B Subtract 6×B from each side: 6×C + 7×B – 6×B = 3×A + 6×B – 6×B which becomes eq.3d) 6×C + B = 3×A
Hint #8
Add B, C, and D to both sides of eq.5: F – B – C + B + C + D = B + C – D – E + B + C + D which becomes F + D = 2×B + 2×C – E which is the same as eq.5b) D + F = 2×B + 2×C – E Add F, A, and E to both sides of eq.4: A + E – F + F + A + E = C + D – A – E + F + A + E which becomes eq.4a) 2×A + 2×E = C + D + F
Hint #9
In eq.4a, substitute 2×B + 2×C – E for D + F (from eq.5b): 2×A + 2×E = C + 2×B + 2×C – E which becomes 2×A + 2×E = 3×C + 2×B – E Add E to both sides of the equation above: 2×A + 2×E + E = 3×C + 2×B – E + E which becomes eq.4b) 2×A + 3×E = 3×C + 2×B
Hint #10
Substitute 4×B – 3×C for 3×E (from eq.3c) into eq.4b: 2×A + 4×B – 3×C = 3×C + 2×B In the equation above, add 3×C to both sides, and subtract 4×B from both sides: 2×A + 4×B – 3×C + 3×C – 4×B = 3×C + 2×B + 3×C – 4×B which becomes 2×A = 6×C – 2×B Divide both sides by 2: 2×A ÷ 2 = (6×C – 2×B) ÷ 2 which becomes eq.4c) A = 3×C – B
Hint #11
Substitute (3×C – B) for A (from eq.4c) in eq.3d: 6×C + B = 3×(3×C – B) which becomes 6×C + B = 9×C – 3×B In the above equation, subtract 6×C from both sides, and add 3×B to both sides: 6×C + B – 6×C + 3×B = 9×C – 3×B – 6×C + 3×B which becomes 4×B = 3×C Divide both sides by 4: 4×B ÷ 4 = 3×C ÷ 4 which makes B = ¾×C
Hint #12
Substitute ¾×C for B in eq.4c: A = 3×C – ¾×C which makes A = 2¼×C
Hint #13
Substitute ¾×C for B in eq.5a: F = ¾×C + C which makes F = 1¾×C
Hint #14
Substitute (¾×C) for B in eq.3c: 3×E = 4×(¾×C) – 3×C which becomes 3×E = 3×C – 3×C which becomes 3×E = 0 which means E = 0
Hint #15
Substitute 1¾×C for F, and 0 for E in eq.6b: 1¾×C – D = 0 Add D to both sides of the equation above: 1¾×C – D + D = 0 + D which makes 1¾×C = D
Solution
Substitute 2¼×C for A, ¾×C for B, 1¾×C for D and F, and 0 for E in eq.1: 2¼×C + ¾×C + C + 1¾×C + 0 + 1¾×C = 30 which simplifies to 7½×C = 30 Divide both sides of the above equation by 7½: 7½×C ÷ 7½ = 30 ÷ 7½ which means C = 4 making A = 2¼×C = 2¼ × 4 = 9 B = ¾×C = ¾ × 4 = 3 D = F = 1¾×C = 1¾ × 4 = 7 and ABCDEF = 934707