Puzzle for June 7, 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 C to both sides of eq.5: C + D + C = A – C + E + C which becomes eq.5a) 2×C + D = A + E Subtract C from both sides of eq.2: F – C – C = A + C – C which becomes eq.2a) F – 2×C = A
Hint #2
In eq.5a, replace A with F – 2×C (from eq.2a): 2×C + D = F – 2×C + E Add 2×C to each side of the equation above: 2×C + D + 2×C = F – 2×C + E + 2×C which becomes eq.5b) 4×C + D = F + E
Hint #3
In eq.4, subtract F from both sides, and add A to both sides: D + F – A – F + A = A + B – F + A which becomes eq.4a) D = 2×A + B – F Substitute 2×A + B – F for D (from eq.4a) in eq.5b: 4×C + 2×A + B – F = F + E Subtract F from both sides of the equation above: 4×C + 2×A + B – F – F = F + E – F which becomes eq.5c) 4×C + 2×A + B – 2×F = E
Hint #4
Add F to both sides of eq.3: B – F + F = C + F + F which becomes eq.3a) B = C + 2×F Substitute C + 2×F for B (from eq.3a) in eq.5c: 4×C + 2×A + C + 2×F – 2×F = E which becomes eq.5d) 2×A + 5×C = E
Hint #5
Add C to both sides of eq.2: F – C + C = A + C + C which becomes eq.2b) F = A + 2×C Substitute (A + 2×C) for F (from eq.2b) in eq.3a: B = C + 2×(A + 2×C) which is the same as B = C + 2×A + 4×C which becomes eq.3b) B = 2×A + 5×C
Hint #6
Substitute 2×A + 5×C for E (from eq.5d) in eq.5: C + D = A – C + 2×A + 5×C which becomes C + D = 3×A + 4×C Subtract C from both sides of the above equation: C + D – C = 3×A + 4×C – C which becomes eq.5e) D = 3×A + 3×C
Hint #7
Substitute 2×A + 5×C for B (from eq.3b) and for E (from eq.5d), 3×A + 3×C for D (from eq.5e), and A + 2×C for F (from eq.2b) in eq.1: A + 2×A + 5×C + C + 3×A + 3×C + 2×A + 5×C + A + 2×C = 34 which simplifies to 9×A + 16×C = 34 Subtract 16×C from each side of the equation above: 9×A + 16×C – 16×C = 34 – 16×C which becomes 9×A = 34 – 16×C Divide both sides by 9: 9×A ÷ 9 = (34 – 16×C) ÷ 9 which means eq.1a) A = (34 – 16×C) ÷ 9
Solution
To make eq.1a true, check several possible values for C and A: If C = 0, then A = (34 – 16×0) ÷ 9 = (34 – 0) ÷ 9 = 34 ÷ 9 = 3.777777778 If C = 1, then A = (34 – 16×1) ÷ 9 = (34 – 16) ÷ 9 = 18 ÷ 9 = 2 If C = 2, then A = (34 – 16×2) ÷ 9 = (34 – 32) ÷ 9 = 2 ÷ 9 = 0.222222222 If C = 3, then A = (34 – 16×3) ÷ 9 = (34 – 48) ÷ 9 = –14 ÷ 9 = –1.555555556 If C > 3, then A < 0 Since A must be a non-negative integer, then A = 2 which makes C = 1 making B = 2×A + 5×C = 2×2 + 5×1 = 4 + 5 = 9 (from eq.3b) D = 3×A + 3×C = 3×2 + 3×1 = 6 + 3 = 9 (from eq.5e) E = 2×A + 5×C = 2×2 + 5×1 = 4 + 5 = 9 (from eq.5d) F = A + 2×C = 2 + 2×1 = 2 + 2 = 4 (from eq.2b) and ABCDEF = 291994