Puzzle for July 11, 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
Subtract C and D from both sides of eq.2: D + E – C – D = A + B + C – C – D which becomes E – C = A + B – D Subtract the left and right sides of the equation above from the left and right sides of eq.5, respectively: E + F – (E – C) = A + B + D – (A + B – D) which is equivalent to E + F – E + C = A + B + D – A – B + D which becomes F + C = 2×D which may be written as eq.5a) C + F = 2×D
Hint #2
Add B and F to both sides of eq.6: C – B + B + F = B – F + B + F which becomes C + F = 2×B In eq.5a, replace C + F with 2×B: 2×B = 2×D Divide both sides of the above equation by 2: 2×B ÷ 2 = 2×D ÷ 2 which makes B = D
Hint #3
In eq.3, replace B with D: D + F = A + E Subtract E and F from both sides of the above equation: D + F – E – F = A + E – E – F which becomes eq.3a) D – E = A – F
Hint #4
In eq.5, replace B with D: E + F = A + D + D which becomes E + F = A + 2×D Subtract F and 2×D from each side of the equation above: E + F – F – 2×D = A + 2×D – F – 2×D which becomes eq.5b) E – 2×D = A – F
Hint #5
In eq.5b, substitute D – E for A – F (from eq.3a): E – 2×D = D – E Add E and 2×D to both sides of the above equation: E – 2×D + E + 2×D = D – E + E + 2×D which makes 2×E = 3×D Divide both sides by 2: 2×E ÷ 2 = 3×D ÷ 2 which makes eq.5c) E = 1½×D
Hint #6
Substitute D for B, and 1½×D for E in eq.3: D + F = A + 1½×D Subtract D from each side of the equation above: D + F – D = A + 1½×D – D which becomes eq.3b) F = A + ½×D
Hint #7
Substitute A + ½×D for F (from eq.3b) in eq.4: A + A + ½×D = D – A which becomes 2×A + ½×D = D – A In the above equation, subtract ½×D from each side, and add A to each side: 2×A + ½×D – ½×D + A = D – A – ½×D + A which makes 3×A = ½×D Multiply both sides by 2: 3×A × 2 = ½×D × 2 which makes 6×A = D and also makes B = D = 6×A
Hint #8
Substitute (6×A) for D in eq.3b: F = A + ½×(6×A) which becomes F = A + 3×A which makes F = 4×A
Hint #9
Substitute (6×A) for D in eq.5c: E = 1½×(6×A) which makes E = 9×A
Hint #10
Substitute 4×A for F, and (6×A) for D in eq.5a: C + 4×A = 2×(6×A) which becomes C + 4×A = 12×A Subtract 4×A from both sides of the above equation: C + 4×A – 4×A = 12×A – 4×A which makes C = 8×A
Solution
Substitute 6×A for B and D, 8×A for C, 9×A for E, and 4×A for F in eq.1: A + 6×A + 8×A + 6×A + 9×A + 4×A = 34 which simplifies to 34×A = 34 Divide both sides of the above equation by 34: 34×A ÷ 34 = 34 ÷ 34 which means A = 1 making B = D = 6×A = 6 × 1 = 6 C = 8×A = 8 × 1 = 8 E = 9×A = 9 × 1 = 9 F = 4×A = 4 × 1 = 4 and ABCDEF = 168694