Puzzle for July 31, 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 to both sides of eq.2: E – A + A = A + F + A which becomes eq.2a) E = 2×A + F In eq.5, replace E with 2×A + F (from eq.2a): D + F = A + 2×A + F which becomes D + F = 3×A + F Subtract F from both sides of the above equation: D + F – F = 3×A + F – F which makes D = 3×A
Hint #2
Add B and E to both sides of eq.3: F – E + B + E = A – B + B + E which becomes eq.3a) F + B = A + E In eq.5, replace A + E with F + B (from eq.3a): D + F = F + B Subtract F from both sides of the equation above: D + F – F = F + B – F which makes D = B and also makes B = D = 3×A
Hint #3
Add A and F to both sides of eq.4: C – A + F + A + F = E – F + A + F which becomes C + 2×F = E + A which may be written as C + 2×F = A + E In eq.5, substitute 3×A for D, and C + 2×F for A + E: 3×A + F = C + 2×F Subtract F from both sides: 3×A + F – F = C + 2×F – F which becomes eq.5a) 3×A = C + F
Hint #4
Add C to both sides of eq.5: D + F + C = A + E + C which may be written as D + C + F = A + E + C Substitute 3×A for D, and 3×A for C + F (from eq.5a) in the equation above: 3×A + 3×A = A + E + C which becomes 6×A = A + E + C Subtract A and C from both sides: 6×A – A – C = A + E + C – A – C which becomes eq.5b) 5×A – C = E
Hint #5
Substitute 3×A for B, and (5×A – C) for E (from eq.5b) in eq.6: 3×A – C = C – (5×A – C) which is the same as 3×A – C = C – 5×A + C which becomes 3×A – C = 2×C – 5×A Add C and 5×A to both sides of the above equation: 3×A – C + C + 5×A = 2×C – 5×A + C + 5×A which makes 8×A = 3×C Divide both sides by 3: 8×A ÷ 3 = 3×C ÷ 3 which makes 2⅔×A = C
Hint #6
Substitute 2⅔×A for C in eq.5b: 5×A – 2⅔×A = E which makes 2⅓×A = E
Hint #7
Substitute 2⅓×A for E, and 3×A for B in eq.3: F – 2⅓×A = A – 3×A which becomes F – 2⅓×A = –2×A Add 2⅓×A to both sides of the equation above: F – 2⅓×A + 2⅓×A = –2×A + 2⅓×A which makes F = ⅓×A
Solution
Substitute 3×A for B and D, 2⅔×A for C, 2⅓×A for E, and ⅓×A for F in eq.1: A + 3×A + 2⅔×A + 3×A + 2⅓×A + ⅓×A = 37 which simplifies to 12⅓×A = 37 Divide both sides of the equation above by 12⅓: 12⅓×A ÷ 12⅓ = 37 ÷ 12⅓ which means A = 3 making B = D = 3×A = 3 × 3 = 9 C = 2⅔×A = 2⅔ × 3 = 8 E = 2⅓×A = 2⅓ × 3 = 7 F = ⅓×A = ⅓ × 3 = 1 and ABCDEF = 398971