Puzzle for March 26, 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 re-written as: F = (B + C + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × F = 3 × (B + C + E) ÷ 3 which becomes eq.6a) 3×F = B + C + E
Hint #2
Add the left and right sides of eq.3 to the left and right sides of eq.2, respectively: B + D + A + C = A + F + D + E Subtract D and A from both sides of the above equation: B + D + A + C – D – A = A + F + D + E – D – A which becomes eq.2a) B + C = F + E
Hint #3
In eq.6a, replace B + C with F + E (from eq.2a): 3×F = F + E + E which becomes 3×F = F + 2×E Subtract F from each side of the equation above: 3×F – F = F + 2×E – F which makes 2×F = 2×E Divide both sides by 2: 2×F ÷ 2 = 2×E ÷ 2 which makes F = E
Hint #4
In eq.4, replace F with E: A + E = C + D – E Add E to both sides of the above equation: A + E + E = C + D – E + E which becomes eq.4a) A + 2×E = C + D
Hint #5
Subtract the left and right sides of eq.3 from the left and right sides of eq.4a, respectively: A + 2×E – (A + C) = C + D – (D + E) which becomes A + 2×E – A – C = C + D – D – E which becomes 2×E – C = C – E Add C and E to both sides of the equation above: 2×E – C + C + E = C – E + C + E which becomes 3×E = 2×C Divide both sides by 2: 3×E ÷ 2 = 2×C ÷ 2 which makes 1½×E = C
Hint #6
In eq.2a, substitute 1½×E for C, and E for F: B + 1½×E = E + E which becomes B + 1½×E = 2×E Subtract 1½×E from each side of the equation above: B + 1½×E – 1½×E = 2×E – 1½×E which makes B = ½×E
Hint #7
Substitute 1½×E for C in eq.3: A + 1½×E = D + E Subtract 1½×E from each side of the above equation: A + 1½×E – 1½×E = D + E – 1½×E which becomes eq.3a) A = D – ½×E
Hint #8
Substitute E for F, and 1½×E for C in eq.5: E + E – D = A – 1½×E + D which becomes 2×E – D = A – 1½×E + D Add D and 1½×E to both sides of the above equation: 2×E – D + D + 1½×E = A – 1½×E + D + D + 1½×E which becomes eq.5a) 3½×E = A + 2×D
Hint #9
Substitute D – ½×E for A (from eq.3a) in eq.5a: 3½×E = D – ½×E + 2×D which becomes 3½×E = 3×D – ½×E Add ½×E to both sides of the equation above: 3½×E + ½×E = 3×D – ½×E + ½×E which makes 4×E = 3×D Divide both sides by 3: 4×E ÷ 3 = 3×D ÷ 3 which makes 1⅓×E = D
Hint #10
Substitute 1⅓×E for D in eq.3a: A = 1⅓×E – ½×E which makes A = ⅚×E
Solution
Substitute ⅚×E for A, ½×E for B, 1½×E for C, 1⅓×E for D, and E for F in eq.1: ⅚×E + ½×E + 1½×E + 1⅓×E + E + E = 37 which simplifies to 6⅙×E = 37 Divide both sides of the above equation by 6⅙: 6⅙×E ÷ 6⅙ = 37 ÷ 6⅙ which means E = 6 making A = ⅚×E = ⅚ × 6 = 5 B = ½×E = ½ × 6 = 3 C = 1½×E = 1½ × 6 = 9 D = 1⅓×E = 1⅓ × 6 = 8 F = E = 6 and ABCDEF = 539866