Puzzle for May 22, 2022 ( )
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, BC, DE, and EF are all 2-digit numbers (not A×B, B×C, D×E, E×F).
Scratchpad
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Hint #1
In eq.1, replace A + B + C with D + E + F (from eq.4): D + E + F + D + E + F = 24 which becomes 2×(D + E + F) = 24 Divide both sides of the above equation by 2: 2×(D + E + F) ÷ 2 = 24 ÷ 2 which becomes eq.1a) D + E + F = 12
Hint #2
In eq.4, replace D + E + F with 12 (from eq.1a): eq.4a) A + B + C = 12
Hint #3
Add D to both sides of eq.2: D + F + D = A + B – D + D which becomes eq.2a) 2×D + F = A + B Add A and F to both sides of eq.3: F – A + D + A + F = B + E – F + A + F which becomes 2×F + D = B + E + A which may be written as eq.3a) 2×F + D = A + B + E
Hint #4
In eq.3a, substitute 2×D + F for A + B (from eq.2a): 2×F + D = 2×D + F + E Subtract D and F from each side of the equation above: 2×F + D – D – F = 2×D + F + E – D – F which simplifies to eq.3b) F = D + E
Hint #5
Substitute F for D + E (from eq.3b) into eq.1a: F + F = 12 which makes 2×F = 12 Divide both sides of the above equation by 2: 2×F ÷ 2 = 12 ÷ 2 which makes F = 6
Hint #6
Substitute 6 for F in eq.3b: 6 = D + E Subtract E from each side of the above equation: 6 – E = D + E – E which becomes eq.3c) 6 – E = D
Hint #7
Substitute 2×D + F for A + B (from eq.2a) in eq.4a: eq.4b) 2×D + F + C = 12
Hint #8
Substitute (6 – E) for D (from eq.3c), and 6 for F in eq.4b: 2×(6 – E) + 6 + C = 12 which becomes 12 – 2×E + 6 + C = 12 which becomes 18 – 2×E + C = 12 In the above equation, subtract 18 from both sides, and add 2×E to both sides: 18 – 2×E + C – 18 + 2×E = 12 – 18 + 2×E which becomes C = –6 + 2×E which may be written as eq.4c) C = 2×E – 6
Hint #9
eq.6 may be written as: 10×B + C + 10×D + E = A – D + 10×E + F In the above equation, subtract E from both sides, and add D to both sides: 10×B + C + 10×D + E – E + D = A – D + 10×E + F – E + D which becomes eq.6a) 10×B + C + 11×D = A + 9×E + F
Hint #10
Substitute 2×E – 6 for C (from eq.4c), (6 – E) for D (from eq.3c), and 6 for F in eq.6a: 10×B + 2×E – 6 + 11×(6 – E) = A + 9×E + 6 which becomes 10×B – 9×E + 60 = A + 9×E + 6 In the above equation, add 9×E to both sides, and subtract 6 from both sides: 10×B – 9×E + 60 + 9×E – 6 = A + 9×E + 6 + 9×E – 6 which becomes eq.6b) 10×B + 54 – 18×E = A
Hint #11
Substitute 6 for F, and 6 – E for D (from eq.3c) in eq.3a: 2×6 + 6 – E = A + B + E which becomes 12 + 6 – E = A + B + E which becomes 18 – E = A + B + E Add E to both sides of the equation above: 18 – E + E = A + B + E + E which becomes eq.3d) 18 = A + B + 2×E
Hint #12
Substitute 10×B + 54 – 18×E for A (from eq.6b) into eq.3d: 18 = 10×B + 54 – 18×E + B + 2×E which becomes 18 = 11×B + 54 – 16×E In the above equation, add 16×E to both sides, and subtract 54 from both sides: 18 + 16×E – 54 = 11×B + 54 – 16×E + 16×E – 54 which becomes 16×E – 36 = 11×B Divide both sides by 11: (16×E – 36) ÷ 11 = 11×B ÷ 11 which becomes eq.3e) (16×E – 36) ÷ 11 = B
Hint #13
Substitute ((16×E – 36) ÷ 11) for B in eq.3d: 18 = A + ((16×E – 36) ÷ 11) + 2×E Multiply both sides of the above equation by 11: 11 × 18 = 11 × (A + ((16×E – 36) ÷ 11) + 2×E) which becomes 198 = 11×A + 16×E – 36 + 22×E which becomes 198 = 11×A + 38×E – 36 Subtract 38×E from both sides, and add 36 to both sides: 198 – 38×E + 36 = 11×A + 38×E – 38×E – 36 + 36 which becomes 234 – 38×E = 11×A Divide both sides by 11: (234 – 38×E) ÷ 11 = 11×A ÷ 11 which becomes eq.3f) (234 – 38×E) ÷ 11 = A
Hint #14
eq.5 may be written as: 10×A + B = 10×E + F – (A + B + C) Substitute ((234 – 38×E) ÷ 11) for A (from eq.3f), ((16×E – 36) ÷ 11) for B (from eq.3e), 6 for F, and 12 for A + B + C (from eq.4a): 10×((234 – 38×E) ÷ 11) + ((16×E – 36) ÷ 11) = 10×E + 6 – (12) which becomes eq.5a) ((2340 – 380×E) ÷ 11) + ((16×E – 36) ÷ 11) = 10×E – 6
Hint #15
Multiply both sides of eq.5a by 11: 11 × (((2340 – 380×E) ÷ 11) + ((16×E – 36) ÷ 11)) = 11 × (10×E – 6) which becomes 2340 – 380×E + 16×E – 36 = 110×E – 66 which becomes 2304 – 364×E = 110×E – 66 Add 364×E and 66 to both sides of the above equation: 2304 – 364×E + 364×E + 66 = 110×E – 66 + 364×E + 66 which simplifies to 2370 = 474×E Divide both sides of the above equation by 474: 2370 ÷ 474 = 474×E ÷ 474 which makes 5 = E
Solution
Since E = 5, then: A = (234 – 38×5) ÷ 11 = (234 – 190) ÷ 11 = 44 ÷ 11 = 4 (from eq.3f) B = (16×5 – 36) ÷ 11 = (80 – 36) ÷ 11 = 44 ÷ 11 = 4 (from eq.3e) C = 2×5 – 6 = 10 – 6 = 4 (from eq.4c) D = 6 – 5 = 1 (from eq.3c) and ABCDEF = 444156