Puzzle for January 20, 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.
Scratchpad
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Hint #1
eq.6 may be written as: F = (A + C + D + E) ÷ 4 Multiply both sides of the above equation by 4: 4 × F = 4 × (A + C + D + E) ÷ 4 which becomes eq.6a) 4×F = A + C + D + E
Hint #2
Subtract A from each side of eq.4: E – A – A = A – C – D – A which becomes E – 2×A = –C – D Multiply both sides of the above equation by (–1): (–1) × (E – 2×A) = (–1) × (–C – D) which becomes –E + 2×A = C + D which may be written as eq.4a) 2×A – E = C + D
Hint #3
In eq.6a, replace C + D with 2×A – E (from eq.4a): 4×F = A + 2×A – E + E which becomes 4×F = 3×A Divide both sides by 4: 4×F ÷ 4 = 3×A ÷ 4 which makes F = ¾×A
Hint #4
In eq.3, replace F with ¾×A: D – ¾×A = A – D Add ¾×A and D to both sides of the equation above: D – ¾×A + ¾×A + D = A – D + ¾×A + D which makes 2×D = 1¾×A Divide both sides by 2: 2×D ÷ 2 = 1¾×A ÷ 2 which makes D = ⅞×A
Hint #5
Substitute C + F for B (from eq.2) into eq.5: C + F – C + E = A + D which becomes eq.5a) F + E = A + D
Hint #6
Substitute ¾×A for F, and ⅞×A for D in eq.5a: ¾×A + E = A + ⅞×A which becomes ¾×A + E = 1⅞×A Subtract ¾×A from both sides of the above equation: ¾×A + E – ¾×A = 1⅞×A – ¾×A which makes E = 1⅛×A
Hint #7
In eq.4a, substitute 1⅛×A for E, and ⅞×A for D: 2×A – 1⅛×A = C + ⅞×A which becomes ⅞×A = C + ⅞×A Subtract ⅞×A from each side of the equation above: ⅞×A – ⅞×A = C + ⅞×A – ⅞×A which makes 0 = C
Hint #8
Substitute 0 for C, and ¾×A for F in eq.2: 0 + ¾×A = B which makes ¾×A = B
Solution
Substitute ¾×A for B and F, 0 for C, ⅞×A for D, and 1⅛×A for E in eq.1: A + ¾×A + 0 + ⅞×A + 1⅛×A + ¾×A = 36 which simplifies to 4½×A = 36 Divide both sides of the above equation by 4½: 4½×A ÷ 4½ = 36 ÷ 4½ which means A = 8 making B = F = ¾×A = ¾ × 8 = 6 D = ⅞×A = ⅞ × 8 = 7 E = 1⅛×A = 1⅛ × 8 = 9 and ABCDEF = 860796