Puzzle for July 10, 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 C + E with B + F (from eq.3): B + F + F = B + D which becomes B + 2×F = B + D Subtract B from both sides of the above equation: B + 2×F – B = B + D – B which makes 2×F = D
Hint #2
In eq.2, replace D with 2×F: 2×F + F = A + C which becomes eq.2a) 3×F = A + C
Hint #3
In eq.6, replace A + C with 3×F (from eq.2a): B + E = 3×F + F which becomes eq.6a) B + E = 4×F
Hint #4
eq.1 may be re–written as: A + C + B + E + D + F = 30 In the equation above, substitute 3×F for A + C (from eq.2a), 4×F for B + E (from eq.6a), and 2×F for D: 3×F + 4×F + 2×F + F = 30 which becomes 10×F = 30 Divide both sides by 10: 10×F ÷ 10 = 30 ÷ 10 which makes F = 3 and also makes D = 2×F = 2×3 = 6
Hint #5
Subtract the left and right sides of eq.3 from the left and right sides of eq.2, respectively: D + F – (B + F) = A + C – (C + E) which becomes D + F – B – F = A + C – C – E which becomes D – B = A – E Add B and E to both sides of the above equation: D – B + B + E = A – E + B + E which becomes eq.2b) D + E = A + B
Hint #6
Substitute 3 for F, and 6 for D in eq.4: A + 3 = B + 6 Subtract 3 and B from each side of the equation above: A + 3 – 3 – B = B + 6 – 3 – B which becomes A – B = 3 which may be written as eq.4a) 3 = A – B
Hint #7
Add the left and right sides of eq.4a to the left and right sides of eq.2b, respectively: D + E + 3 = A + B + A – B which becomes D + E + 3 = 2×A Substitute 6 for D in the equation above: 6 + E + 3 = 2×A which becomes E + 9 = 2×A Subtract 9 from both sides: E + 9 – 9 = 2×A – 9 which becomes eq.2c) E = 2×A – 9
Hint #8
Substitute 2×A – 9 for E (from eq.2c), and 3 for F in eq.6a: B + 2×A – 9 = 4×3 which becomes B + 2×A – 9 = 12 In the above equation, subtract 2×A from both sides, and add 9 to both sides: B + 2×A – 9 – 2×A + 9 = 12 – 2×A + 9 which becomes eq.6b) B = 21 – 2×A
Hint #9
Substitute (21 – 2×A) for B (from eq.6b) in eq.4a: 3 = A – (21 – 2×A) which becomes 3 = A – 21 + 2×A which becomes 3 = 3×A – 21 Add 21 to both sides of the equation above: 3 + 21 = 3×A – 21 + 21 which makes 24 = 3×A Divide both sides by 3: 24 ÷ 3 = 3×A ÷ 3 which makes 8 = A making B = 21 – 2×A = 21 – 2×8 = 21 – 16 = 5 (from eq.6b) E = 2×A – 9 = 2×8 – 9 = 16 – 9 = 7 (from eq.2c)
Solution
Substitute 3 for F, and 8 for A in eq.2a: 3×3 = 8 + C which becomes 9 = 8 + C Subtract 8 from each side of the above equation: 9 – 8 = 8 + C – 8 which becomes 1 = C and ABCDEF = 851673