Puzzle for June 18, 2023 ( )
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 A, D, and F to both sides of eq.3: D - A + A + D + F = E - D - F + A + D + F which becomes 2×D + F = E + A which is equivalent to eq.3a) 2×D + F = A + E Add C, A, and D to both sides of eq.5: D - C + C + A + D = B - A - D + C + A + D which becomes eq.5a) 2×D + A = B + C
Hint #2
eq.6 may be written as: eq.6a) F - (A + E) = A - (B + C) - F
Hint #3
In eq.6a, replace A + E with 2×D + F (from eq.3a), and B + C with 2×D + A (from eq.5a): F - (2×D + F) = A - (2×D + A) - F which becomes F - 2×D - F = A - 2×D - A - F which becomes -2×D = -2×D - F Add 2×D to both sides of the above equation: -2×D + 2×D = -2×D - F + 2×D which becomes 0 = -F which means 0 = F
Hint #4
In eq.3a, substitute 0 for F: 2×D + 0 = A + E which becomes eq.3b) 2×D = A + E
Hint #5
In eq.4, substitute 0 for F: E - D = B - E - 0 which becomes E - D = B - E Add D and E to both sides of the above equation: E - D + D + E = B - E + D + E which becomes eq.4a) 2×E = B + D
Hint #6
Add the left and right sides of eq.4a to the left and right sides of eq.3b, respectively: 2×D + 2×E = A + E + B + D Subtract D and E from each side of the equation above: 2×D + 2×E - D - E = A + E + B + D - D - E which becomes eq.3c) D + E = A + B
Hint #7
Add A and D to both sides of eq.2: E - A + A + D = C - D + A + D which becomes E + D = C + A which may be written as eq.2a) D + E = C + A
Hint #8
Substitute A + B for D + E (from eq.3c) in eq.2a: A + B = C + A Subtract A from each side of the equation above: A + B - A = C + A - A which makes B = C
Hint #9
Add E to both sides of eq.4a: 2×E + E = B + D + E which becomes 3×E = B + D + E Substitute C + A for D + E (from eq.2a) in the above equation: eq.4b) 3×E = B + C + A
Hint #10
In eq.6, substitute 0 for F: 0 - A - E = A - B - C - 0 which becomes -A - E = A - B - C Add A, E, B, and C to both sides of the equation above: -A - E + A + E + B + C = A - B - C + A + E + B + C which becomes eq.6b) B + C = 2×A + E
Hint #11
Substitute 2×A + E for B + C (from eq.6b) into eq.4b: 3×E = 2×A + E + A which becomes 3×E = 3×A + E Subtract E from each side of the above equation: 3×E - E = 3×A + E - E which makes 2×E = 3×A Divide both sides by 2: 2×E ÷ 2 = 3×A ÷ 2 which makes E = 1½×A
Hint #12
Substitute 1½×A for E in eq.3b: 2×D = A + 1½×A which makes 2×D = 2½×A Divide both sides of the above equation by 2: 2×D ÷ 2 = 2½×A ÷ 2 which makes D = 1¼×A
Hint #13
Substitute B for C, and 1½×A for E in eq.6b: B + B = 2×A + 1½×A which makes 2×B = 3½×A Divide both sides of the above equation by 2: 2×B ÷ 2 = 3½×A ÷ 2 which makes B = 1¾×A and also makes B = C = 1¾×A
Solution
Substitute 1¾×A for B and C, 1¼×A for D, 1½×A for E, and 0 for F in eq.1: A + 1¾×A + 1¾×A + 1¼×A + 1½×A + 0 = 29 which simplifies to 7¼×A = 29 Divide both sides of the above equation by 7¼: 7¼×A ÷ 7¼ = 29 ÷ 7¼ which means A = 4 making B = C = 1¾×A = 1¾ × 4 = 7 D = 1¼×A = 1¼ × 4 = 5 E = 1½×A = 1½ × 4 = 6 and ABCDEF = 477560