Puzzle for June 7, 2019 ( )
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 EF are 2-digit numbers (not C×D or E×F).
Scratchpad
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Hint #1
Add both D and F to each side of eq.3: D - F + D + F = A - D + D + F which becomes eq.3a) 2×D = A + F Add D to each side of eq.4: B + D + D = A + C - D + D which becomes eq.4a) B + 2×D = A + C
Hint #2
In eq.4a, replace 2×D with A + F (from eq.3a): B + A + F = A + C Subtract A from each side of the above equation: B + A + F - A = A + C - A which becomes B + F = C In eq.2, replace C with B + F: F = B + B + F Subtract F from each side: F - F = B + B + F - F which becomes 0 = 2×B which means 0 = B
Hint #3
In eq.2, replace B with 0: F = 0 + C which makes F = C
Hint #4
Add E and F to both sides of eq.5: E - F + E + F = A + D - E + E + F which becomes 2×E = A + D + F which may be written as 2×E = A + F + D In the equation above, substitute 2×D for A + F (from eq.3a): 2×E = 2×D + D which means 2×E = 3×D Divide both sides by 2: 2×E ÷ 2 = 3×D ÷ 2 which makes eq.5a) E = 1½×D
Hint #5
eq.6 may be re-written as: A + C = 10×E + F - (10×C + D) which is the same as A + C = 10×E + F - 10×C - D Add 10×C + D to both sides of the equation above: A + C + 10×C + D = 10×E + F - 10×C - D + 10×C + D which becomes A + 11×C + D = 10×E + F Substitute 1½×D for E, and C for F: A + 11×C + D = 10×1½×D + C which becomes A + 11×C + D = 15×D + C Subtract C and D from both sides: A + 11×C + D - C - D = 15×D + C - C - D which becomes A + 10×C = 14×D which is equivalent to eq.6a) A + 10×C = 7×(2×D)
Hint #6
Substitute A + F for 2×D in eq.6a (from eq.3a): A + 10×C = 7×(A + F) which is the same as A + 10×C = 7×A + 7×F In the above equation, substitute C for F, and subtract A from both sides: A + 10×C - A = 7×A + 7×C - A which becomes 10×C = 6×A + 7×C Subtract 7×C from each side: 10×C - 7×C = 6×A + 7×C - 7×C which becomes 3×C = 6×A Divide both sides by 3: 3×C ÷ 3 = 6×A ÷ 3 which means C = 2×A which also means F = C = 2×A
Hint #7
Substitute 0 for B, and 2×A for C in eq.4a: 0 + 2×D = A + 2×A which means 2×D = 3×A Divide both sides by 2: 2×D ÷ 2 = 3×A ÷ 2 which makes D = 1½×A
Hint #8
Substitute (1½×A) for D in eq.5a: E = 1½×(1½×A) which makes E = 2¼×A
Solution
Substitute 0 for B, 2×A for C and F, 1½×A for D, and 2¼×A for E in eq.1: A + 0 + 2×A + 1½×A + 2¼×A + 2×A = 13 which simplifies to 8¾×A = 35 Divide both sides by 8¾: 8¾×A ÷ 8¾ = 35 ÷ 8¾ which means A = 4 making C = F = 2×A = 2 × 4 = 8 D = 1½×A = 1½ × 4 = 6 E = 2¼×A = 2¼ × 4 = 9 and ABCDEF = 408698