Puzzle for July 17, 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.5, replace D + E with A + F (from eq.4): A + F = A + B – D Subtract A from both sides of the above equation: A + F – A = A + B – D – A which becomes eq.5a) F = B – D
Hint #2
In eq.2, replace F with B – D (from eq.5a): C = D + B – D which makes C = B
Hint #3
Subtract the left and right sides of eq.3 from the left and right sides of eq.4, respectively: A + F – (B + F) = D + E – (A + D) which is equivalent to A + F – B – F = D + E – A – D which becomes A – B = E – A Add A to both sides of the equation above: A – B + A = E – A + A which becomes eq.4a) 2×A – B = E
Hint #4
In eq.5, substitute 2×A – B for E (from eq.4a): D + 2×A – B = A + B – D Subtract A from both sides of the equation above: D + 2×A – B – A = A + B – D – A which becomes D + A – B = B – D Add B to both sides, and subtract D from both sides: D + A – B + B – D = B – D + B – D which becomes A = 2×B – 2×D which may be written as eq.5b) A = 2×(B – D)
Hint #5
In eq.5b, replace (B – D) with F (from eq.5a): eq.5c) A = 2×F
Hint #6
eq.6 may be written as: E = (A + B + C + D + F) ÷ 5 Multiply both sides of the equation above by 5: E × 5 = ((A + B + C + D + F) ÷ 5) × 5 which becomes eq.6a) 5×E = A + B + C + D + F
Hint #7
eq.1 may be written as: A + B + C + D + F + E = 42 Substitute 5×E for A + B + C + D + F (from eq.6a) in the above equation: 5×E + E = 42 which becomes 6×E = 42 Divide both sides by 6: 6×E ÷ 6 = 42 ÷ 6 which makes E = 7
Hint #8
Substitute 7 for E in eq.4a: 2×A – B = 7 In the above equation, add B to both sides, and subtract 7 from both sides: 2×A – B + B – 7 = 7 + B – 7 which becomes eq.4b) 2×A – 7 = B
Hint #9
Add D to both sides of eq.5a: F + D = B – D + D which becomes F + D = B which may be written as eq.5d) D + F = B Substitute 7 for E, B for C, and B for D + F (from eq.5d) in eq.6a: 5×7 = A + B + B + B which becomes eq.6b) 35 = A + 3×B
Hint #10
Substitute (2×A – 7) for B (from eq.4b) in eq.6b: 35 = A + 3×(2×A – 7) which becomes 35 = A + 6×A – 21 which becomes 35 = 7×A – 21 Add 21 to both sides of the above equation: 35 + 21 = 7×A – 21 + 21 which beomes 56 = 7×A Divide both sides by 7: 56 ÷ 7 = 7×A ÷ 7 which makes 8 = A making B = 2×8 – 7 = 16 – 7 = 9 (from eq.4b) C = B = 9
Hint #11
Substitute 8 for A in eq.5c: 8 = 2×F Divide both sides of the equation above by 2: 8 ÷ 2 = 2×F ÷ 2 which makes 4 = F
Solution
Substitute 4 for F, and 9 for B in eq.5d: D + 4 = 9 Subtract 4 from each side of the equation above: D + 4 – 4 = 9 – 4 which makes D = 5 and makes ABCDEF = 899574