Puzzle for November 29, 2019 ( )
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 D and E to both sides of eq.4: F – E + D + E = A + B – D + D + E which becomes F + D = A + B + E which may be written as D + F = A + B + E In eq.3, replace D + F with A + B + E: C + E = A + B + E Subtract E from each side of the equation above: C + E – E = A + B + E – E which becomes eq.4a) C = A + B
Hint #2
In eq.2, replace A + B with C (from eq.4a): E = C + C which makes eq.2a) E = 2×C
Hint #3
In eq.3, substitute 2×C for E: C + 2×C = D + F which becomes eq.3a) 3×C = D + F
Hint #4
eq.1 may be written as: A + B + C + D + F + E = 21 Substitute C for A + B (from eq.4a), 3×C for D + F (from eq.3a), and 2×C for E in the above equation: C + C + 3×C + 2×C = 21 which becomes 7×C = 21 Divide both sides by 7: 7×C ÷ 7 = 21 ÷ 7 which makes C = 3 making E = 2×C = 2×3 = 6
Hint #5
Add A and F to both sides of eq.5: D – F + A + F = C – A + A + F which becomes eq.5a) D + A = C + F eq.1 may be written as: D + A + B + E + C + F = 21 Substitute C + F for D + A (from eq.5a) and for B + E (from eq.6) in the equation above: C + F + C + F + C + F = 21 which may be written as eq.1a) 3×(C + F) = 21
Hint #6
Divide both sides of eq.1a by 3: 3×(C + F) ÷ 3 = 21 ÷ 3 which makes C + F = 7 Substitute 3 for C in the equation above: 3 + F = 7 Subtract 3 from each side: 3 + F – 3 = 7 – 3 which means F = 4
Hint #7
Substitute 3 for C, and 4 for F in eq.3a: 3×3 = D + 4 which becomes 9 = D + 4 Subtract 4 from both sides of the equation above: 9 – 4 = D + 4 – 4 which makes 5 = D
Hint #8
Substitute 5 for D, 4 for F, and 3 for C in eq.5: 5 – 4 = 3 – A which makes 1 = 3 – A In the above equation, add A to each side, and subtract 1 from each side: 1 + A – 1 = 3 – A + A – 1 which makes A = 2
Solution
Substitute 3 for C, and 2 for A in eq.4a: 3 = 2 + B Subtract 2 from both sides of the equation above: 3 – 2 = 2 + B – 2 which makes 1 = B and ABCDEF = 213564