Puzzle for September 13, 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 F to both sides of eq.3: C – A + F = A – D – F + F which becomes eq.3a) C – A + F = A – D eq.4 may be re-written as: eq.4a) D – E = A – D – C
Hint #2
In eq.4a, replace A – D with C – A + F (from eq.3a): D – E = C – A + F – C which becomes D – E = –A + F Add E and A to each side of the equation above: D – E + E + A = –A + F + E + A which becomes D + A = F + E which may be written as eq.4b) A + D = E + F
Hint #3
Add A, D, and E to each side of eq.2: B – A – D + A + D + E = A + D – E + A + D + E which becomes B + E = 2×A + 2×D which may be written as B + E = 2×(A + D) In the above equation, replace A + D with E + F (from eq.4b): B + E = 2×(E + F) which may be written as B + E = 2×E + 2×F Subtract E from both sides: B + E – E = 2×E + 2×F – E which becomes eq.2a) B = E + 2×F
Hint #4
In eq.5, substitute E + 2×F for B (from eq.2a): E – F = E + 2×F In the above equation, subtract E from both sides, and add F to both sides: E – F – E + F = E + 2×F – E + F which becomes 0 = 3×F which means 0 = F
Hint #5
Substitute 0 for F in eq.2a: B = E + 2×0 which becomes B = E + 0 which makes B = E
Hint #6
Substitute 0 for F in eq.4b: A + D = E + 0 which means eq.4c) A + D = E
Hint #7
Substitute 0 for F in eq.3: C – A = A – D – 0 which becomes C – A = A – D Add A to both sides of the equation above: C – A + A = A – D + A which becomes eq.3a) C = 2×A – D
Hint #8
Substitute E for B, and (2×A – D) for C (from eq.3a) in eq.6: E – (2×A – D) – D + E = A + (E ÷ E) which becomes 2×E – 2×A + D – D = A + 1 which becomes eq.6a) 2×E – 2×A = A + 1
Hint #9
Substitute (A + D) for E (from eq.4c) in eq.6a: 2×(A + D) – 2×A = A + 1 which is the same as 2×A + 2×D – 2×A = A + 1 which becomes 2×D = A + 1 Subtract 1 from both sides of the equation above: 2×D – 1 = A + 1 – 1 which becomes eq.6b) 2×D – 1 = A
Hint #10
Substitute (2×D – 1) for A (from eq.6b) in eq.3a: C = 2×(2×D – 1) – D which becomes C = 4×D – 2×1 – D which becomes eq.3b) C = 3×D – 2
Hint #11
Substitute 2×D – 1 for A (from eq.6b) in eq.4c: 2×D – 1 + D = E which means 3×D – 1 = E and also means eq.4d) B = E = 3×D – 1
Solution
Substitute 2×D – 1 for A (from eq.6b), 3×D – 1 for B and E (from eq.4d), 3×D – 2 for C (from eq.3b), and 0 for F in eq.1: 2×D – 1 + 3×D – 1 + 3×D – 2 + D + 3×D – 1 + 0 = 31 which simplifies to 12×D – 5 = 31 Add 5 to both sides of the equation above: 12×D – 5 + 5 = 31 + 5 which becomes 12×D = 36 Divide both sides by 12: 12×D ÷ 12 = 36 ÷ 12 which means D = 3 making A = 2×D – 1 = 2×3 – 1 = 6 – 1 = 5 (from eq.6b) B = E = 3×D – 1 = 3×3 – 1 = 9 – 1 = 8 (from eq.4d) C = 3×D – 2 = 3×3 – 2 = 9 – 2 = 7 (from eq.3b) and ABCDEF = 587380