Puzzle for January 9, 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.
* BC, CD, DE, EF are 2-digit number (not B×C, C×D, D×E, or E×F).
Scratchpad
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Hint #1
In eq.3, add B to both sides, and subtract A from both sides: A + C + F – B + B - A = B + B - A which becomes eq.3a) C + F = 2×B – A eq.6 may be written as: A = (C + D + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × A = 3 × (C + D + F) ÷ 3 which becomes 3×A = C + D + F which may be written as eq.6a) 3×A = C + F + D
Hint #2
In eq.6a, replace C + F with 2×B – A (from eq.3a): 3×A = 2×B – A + D In the above equation, add A to both sides, and subtract 2×B from both sides: 3×A + A – 2×B = 2×B – A + D + A – 2×B which becomes eq.6b) 4×A – 2×B = D
Hint #3
In eq.2, substitute 4×A – 2×B for D (from eq.6b): A + E = B + 4×A – 2×B – E which becomes A + E = 4×A – B – E In the above equation, subtract A from both sides, and add E to both sides: A + E – A + E = 4×A – B – E – A + E which becomes 2×E = 3×A – B Divide both sides by 2: 2×E ÷ 2 = (3×A – B) ÷ 2 which becomes eq.2a) E = 1½×A – ½×B
Hint #4
eq.5 may be written as: A + B + C + E + F = 10×B + C – (10×D + E) which becomes A + B + C + E + F = 10×B + C – 10×D – E Subtract B, C, and E from both sides of the above equation: A + B + C + E + F – B – C – E = 10×B + C – 10×D – E – B – C – E which becomes eq.5a) A + F = 9×B – 10×D – 2×E
Hint #5
Substitute (4×A – 2×B) for D (from eq.6b), and (1½×A – ½×B) for E (from eq.2a) in eq.5a: A + F = 9×B – 10×(4×A – 2×B) – 2×(1½×A – ½×B) which becomes A + F = 9×B – 40×A + 20×B – 3×A + B which becomes A + F = 30×B – 43×A Subtract A from both sides of the equation above: A + F – A = 30×B – 43×A – A which becomes eq.5b) F = 30×B – 44×A
Hint #6
Substitute 30×B – 44×A for F (from eq.5b) in eq.3a: C + 30×B – 44×A = 2×B – A In the above equation, subtract 30×B from each side, and add 44×A to each side: C + 30×B – 44×A – 30×B + 44×A = 2×B – A – 30×B + 44×A which becomes eq.3b) C = 43×A – 28×B
Hint #7
eq.4 may be written as: A + B = 10×C + D – (10×E + F) which becomes A + B = 10×C + D – 10×E – F Substitute (43×A – 28×B) for C (from eq.3b), 4×A – 2×B for D (from eq.6b), (1½×A – ½×B) for E (from eq.2a), and (30×B – 44×A) for F (from eq.5b) in the above equation: A + B = 10×(43×A – 28×B) + 4×A – 2×B – 10×(1½×A – ½×B) – (30×B – 44×A) which becomes A + B = 430×A – 280×B + 4×A – 2×B – 15×A + 5×B – 30×B + 44×A which simplifies to eq.4a) A + B = 463×A – 307×B
Hint #8
In eq.4a, subtract A from both sides, and add 307×B to both sides: A + B – A + 307×B = 463×A – 307×B – A + 307×B which simplifies to 308×B = 462×A Divide both sides of the above equation by 308: 308×B ÷ 308 = 462×A ÷ 308 which makes B = 1½×A
Hint #9
Substitute (1½×A) for B in eq.3b: C = 43×A – 28×(1½×A) which becomes C = 43×A – 42×A which makes C = A
Hint #10
Substitute (1½×A) for B in eq.6b: 4×A – 2×(1½×A) = D which becomes 4×A – 3×A = D which makes A = D
Hint #11
Substitute (1½×A) for B in eq.2a: E = 1½×A – ½×(1½×A) which becomes E = 1½×A – ¾×A which makes E = ¾×A
Hint #12
Substitute (1½×A) for B in eq.5b: F = 30×(1½×A) – 44×A which becomes F = 45×A – 44×A which makes F = A
Solution
Substitute 1½×A for B, A for C and D and F, and ¾×A for E in eq.1: A + 1½×A + A + A + ¾×A + A = 25 which simplifies to 6¼×A = 25 Divide both sides of the above equation by 6¼: 6¼×A ÷ 6¼ = 25 ÷ 6¼ which means A = 4 making B = 1½×A = 1½ × 4 = 6 C = D = F = A = 4 E = ¾×A = ¾ × 4 = 3 and ABCDEF = 464434