Puzzle for November 28, 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.
* AB and DE are 2-digit numbers (not A×B or D×E).
Scratchpad
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Hint #1
Subtract E and C from both sides of eq.2: B + E – E – C = C + D – E – C which becomes eq.2a) B – C = D – E Subtract E and A from both sides of eq.3: C + E – E – A = A + D – E – A which becomes eq.3a) C – A = D – E
Hint #2
In eq.2a, replace D – E with C – A (from eq.3a): B – C = C – A Add C and A to both sides of the equation above: B – C + C + A = C – A + C + A which becomes B + A = 2×C which may be written as eq.3b) A + B = 2×C
Hint #3
In eq.5, replace A + B with 2×C (from eq.3b): D + F = 2×C + C – D – E which becomes D + F = 3×C – D – E Add D and E to both sides of the above equation: D + F + D + E = 3×C – D – E + D + E which becomes eq.5a) 2×D + F + E = 3×C
Hint #4
eq.6 may be written as: 10×A + B – C = 10×D + E + F Add C to both sides of the above equation: 10×A + B – C + C = 10×D + E + F + C which becomes 10×A + B = 10×D + E + F + C which may be written as eq.6a) 9×A + A + B = 8×D + 2×D + F + E + C
Hint #5
In eq.6a, substitute 2×C for A + B (from eq.3b), and 3×C for 2×D + F + E (from eq.5a): 9×A + 2×C = 8×D + 3×C + C which becomes 9×A + 2×C = 8×D + 4×C Subtract 2×C and 8×D from both sides of the equation above: 9×A + 2×C – 2×C – 8×D = 8×D + 4×C – 2×C – 8×D which becomes eq.6b) 9×A – 8×D = 2×C Divide both sides of eq.6b by 2: (9×A – 8×D) ÷ 2 = 2×C ÷ 2 which becomes eq.6c) 4½×A – 4×D = C
Hint #6
Substitute A + B for 2×C (from eq.3b) into eq.6b: 9×A – 8×D = A + B Subtract A from each side of the equation above: 9×A – 8×D – A = A + B – A which becomes eq.6d) 8×A – 8×D = B
Hint #7
Subtract C from both sides of eq.3: C + E – C = A + D – C which becomes E = A + D – C Substitute (4½×A – 4×D) for C (from eq.6c) in the equation above: E = A + D – (4½×A – 4×D) which becomes E = A + D – 4½×A + 4×D which becomes E = –3½×A + 5×D which may be written as eq.3c) E = 5×D – 3½×A
Hint #8
Substitute 5×D – 3½×A for E (from eq.3c), and 8×A – 8×D for B (from eq.6d) in eq.4: 8×A – 8×D + F = D + 5×D – 3½×A – F which becomes 8×A – 8×D + F = 6×D – 3½×A – F In the equation above, subtract 8×A from both sides, and add 8×D and F to both sides: 8×A – 8×D + F – 8×A + 8×D + F = 6×D – 3½×A – F – 8×A + 8×D + F which simplifies to 2×F = 14×D – 11½×A Divide both sides by 2: 2×F ÷ 2 = (14×D – 11½×A) ÷ 2 which becomes eq.4a) F = 7×D – 5¾×A
Hint #9
Substitute 7×D – 5¾×A for F (from eq.4a), 8×A – 8×D for B (from eq.6d), 4½×A – 4×D for C (from eq.6c), and (5×D – 3½×A) for E (from eq.3c) in eq.5: D + 7×D – 5¾×A = A + 8×A – 8×D + 4½×A – 4×D – D – (5×D – 3½×A) which becomes 8×D – 5¾×A = 13½×A – 13×D – 5×D + 3½×A which becomes 8×D – 5¾×A = 17×A – 18×D Add 5¾×A and 18×D to both sides of the equation above: 8×D – 5¾×A + 5¾×A + 18×D = 17×A – 18×D + 5¾×A + 18×D which becomess 26×D = 22¾×A Divide both sides by 26: 26×D ÷ 26 = 22¾×A ÷ 26 which makes D = ⅞×A
Hint #10
Substitute (⅞×A) for D in eq.6d: 8×A – 8×(⅞×A) = B which becomes 8×A – 7×A = B which makes A = B
Hint #11
Substitute (⅞×A) for D in eq.6c: 4½×A – 4×(⅞×A) = C which becomes 4½×A – 3½×A = C which makes A = C
Hint #12
Substitute (⅞×A) for D in eq.3c: E = 5×(⅞×A) – 3½×A which becomes E = 4⅜×D – 3½×A which makes E = ⅞×A
Hint #13
Substitute (⅞×A) for D in eq.4a: F = 7×(⅞×A) – 5¾×A which becomes F = 6⅛×A – 5¾×A which makes F = ⅜×A
Solution
Substitute A for B and C, ⅞×A for D and E, and ⅜×A for F in eq.1: A + A + A + ⅞×A + ⅞×A + ⅜×A = 41 which simplifies to 5⅛×A = 41 Divide both sides of the above equation by 5⅛: 5⅛×A ÷ 5⅛ = 41 ÷ 5⅛ which means A = 8 making B = C = A = 8 D = E = ⅞×A = ⅞ × 8 = 7 F = ⅜×A = ⅜ × 8 = 3 and ABCDEF = 888773