Puzzle for February 12, 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.
* CD and AB are 2-digit numbers (not C×D or A×B).
Scratchpad
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Hint #1
eq.6 may be written as: A = (B + C + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × A = 3 × (B + C + F) ÷ 3 which becomes eq.6a) 3×A = B + C + F
Hint #2
eq.5 may be written as: 10×C + D - F = 10×A + B + E which may be written as eq.5a) 10×C + D - F = 9×A + A + B + E
Hint #3
In eq.5a, replace A + B + E with D + F (from eq.3): 10×C + D - F = 9×A + D + F Subtract D and F from both sides of the above equation: 10×C + D - F - D - F = 9×A + D + F - D - F which becomes eq.5b) 10×C - 2×F = 9×A which may be written as eq.5c) 10×C - 2×F = 3×(3×A)
Hint #4
In eq.5c, replace 3×A with B + C + F (from eq.6a): 10×C - 2×F = 3×(B + C + F) which becomes 10×C - 2×F = 3×B + 3×C + 3×F Subtract 3×C and 3×F from each side of the equation above: 10×C - 2×F - 3×C - 3×F = 3×B + 3×C + 3×F - 3×C - 3×F which becomes eq.5d) 7×C - 5×F = 3×B
Hint #5
Add B and A to both sides of eq.2: E - B + B + A = B - A + B + A which becomes eq.2a) E + A = 2×B
Hint #6
eq.3 may be written as: D + F = E + A + B In the equation above, substitute 2×B for E + A (from eq.2a): D + F = 2×B + B which becomes eq.3a) D + F = 3×B
Hint #7
Substitute D + F for 3×B (from eq.3a) into eq.5d: 7×C - 5×F = D + F Subtract F from each side of the above equation: 7×C - 5×F - F = D + F - F which becomes eq.5e) 7×C - 6×F = D
Hint #8
Multiply both sides of eq.2a by 9: 9×(E + A) = 9×(2×B) which becomes 9×E + 9×A = 18×B which is the same as eq.2b) 9×E + 9×A = 6×(3×B)
Hint #9
Substitute 10×C - 2×F for 9×A (from eq.5b), and 7×C - 5×F for 3×B (from eq.5d) in eq.2b: 9×E + 10×C - 2×F = 6×(7×C - 5×F) which becomes 9×E + 10×C - 2×F = 42×C - 30×F In the above equation, subtract 10×C from both sides, and add 2×F to both sides: 10×C - 2×F + 9×E - 10×C + 2×F = 42×C - 30×F - 10×C + 2×F which becomes eq.2c) 9×E = 32×C - 28×F
Hint #10
Add A to both sides of eq.4: A + B + A = C + D + E - A + A which becomes 2×A + B = C + D + E Multiply both sides of the above equation by 9: 9 × (2×A + B) = 9 × (C + D + E) which becomes 18×A + 9×B = 9×C + 9×D + 9×E which may be written as eq.4a) 2×(9×A) + 3×(3×B) = 9×C + 9×D + 9×E
Hint #11
Substitute 10×C - 2×F for 9×A (from eq.5b), 7×C - 5×F for 3×B (from eq.5d), (7×C - 6×F) for D (from eq.5e), and 32×C - 28×F for 9×E (from eq.2c) in eq.4a: 2×(10×C - 2×F) + 3×(7×C - 5×F) = 9×C + 9×(7×C - 6×F) + 32×C - 28×F which becomes 20×C - 4×F + 21×C - 15×F = 9×C + 63×C - 54×F + 32×C - 28×F which becomes 41×C - 19×F = 104×C - 82×F In the above equation, subtract 41×C from both sides, and add 82×F to both sides: 41×C - 19×F - 41×C + 82×F = 104×C - 82×F - 41×C + 82×F which makes 63×C = 63×F Divide both sides by 63: 63×C ÷ 63 = 63×F ÷ 63 which makes C = F
Hint #12
Substitute C for F in eq.5e: 7×C - 6×C = D which makes C = D
Hint #13
Substitute C for F in eq.2c: 9×E = 32×C - 28×C which becomes 9×E = 4×C Divide both sides of the above equation by 4: 9×E ÷ 4 = 4×C ÷ 4 which makes 2¼×E = C and also makes 2¼×E = C = D = F
Hint #14
Substitute (2¼×E) for C and F in eq.5b: 10×(2¼×E) - 2×(2¼×E) = 9×A which becomes 22½×E - 4½×E = 9×A which becomes 18×E = 9×A Divide both sides of the above equation by 9: 18×E ÷ 9 = 9×A ÷ 9 which makes 2×E = A
Hint #15
Substitute 2×E for A in eq.2a: E + 2×E = 2×B which becomes 3×E = 2×B Divide both sides of the above equation by 2: 3×E ÷ 2 = 2×B ÷ 2 which makes 1½×E = B
Solution
Substitute 2×E for A, 1½×E for B, and 2¼×E for C and D and F in eq.1: 2×E + 1½×E + 2¼×E + 2¼×E + E + 2¼×E = 45 which simplifies to 11¼×E = 45 Divide both sides of the above equation by 11¼: 11¼×E ÷ 11¼ = 45 ÷ 11¼ which means E = 4 making A = 2×E = 2 × 4 = 8 B = 1½×E = 1½ × 4 = 6 C = D = F = 2¼×E = 2¼ × 4 = 9 and ABCDEF = 869949