Puzzle for January 19, 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.
* AB, BC, and DE are 2-digit numbers (not A×B, B×C, or D×E).
Scratchpad
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Hint #1
In eq.5, replace F with A + B (from eq.2): C - A = A + B - B which becomes C - A = A Add A to both sides of the above equation: C - A + A = A + A which makes C = 2×A
Hint #2
Substitute 2×A for C, and A + B for F (from eq.2) in eq.3: 2×A + D = B + A + B Subtract A from both sides of the above equation: 2×A + D - A = B + A + B - A which becomes eq.3a) A + D = 2×B
Hint #3
eq.4 may also be written as: E = B + F + D In the above equation, replace B + F with C + D (from eq.3): E = C + D + D which is equivalent to eq.4a) E = C + 2×D
Hint #4
In eq.4a, replace C with 2×A: E = 2×A + 2×D which may be written as E = 2×(A + D) In the above equation, substitute 2×B for A + D (from eq.3a): E = 2×(2×B) which means eq.4b) E = 4×B
Hint #5
eq.6 may be written as: A + 10×B + C = 10×D + E - (10×A + B) - F which is the same as A + 10×B + C = 10×D + E - 10×A - B - F Add (B - A) to both sides of the above equation: A + 10×B + C + (B - A) = 10×D + E - 10×A - B - F + (B - A) which becomes eq.6a) 11×B + C = 10×D + E - 11×A - F
Hint #6
Substitute 2×A for C, 4×B for E, and (A + B) for F (from eq.2) in eq.6a: 11×B + 2×A = 10×D + 4×B - 11×A - (A + B) which is the same as 11×B + 2×A = 10×D + 4×B - 11×A - A - B which becomes 11×B + 2×A = 10×D + 3×B - 12×A Add (12×A - 3×B) to both sides of the above equation: 11×B + 2×A + (12×A - 3×B) = 10×D + 3×B - 12×A + (12×A - 3×B) which becomes eq.6b) 8×B + 14×A = 10×D
Hint #7
eq.6b may be written as: 4×(2×B) + 14×A = 10×D Replace 2×B with A + D (from eq.3a): 4×(A + D) + 14×A = 10×D which is the same as 4×A + 4×D + 14×A = 10×D Subtract 4×D from both sides: 4×A + 4×D + 14×A - 4×D = 10×D - 4×D which simplifies to 18×A = 6×D Divide both sides by 6: 18×A ÷ 6 = 6×D ÷ 6 which makes 3×A = D
Hint #8
Substitute 3×A for D in eq.3a: A + 3×A = 2×B which becomes 4×A = 2×B Divide both sides of the above equation by 2: 4×A ÷ 2 = 2×B ÷ 2 which makes 2×A = B
Hint #9
In eq.4b, replace B with 2×A: E = 4×2×A which makes E = 8×A
Hint #10
In eq.2, replace B with 2×A: A + 2×A = F which makes 3×A = F
Solution
Substitute 2×A for B and C, 3×A for D and F, and 8×A for E in eq.1: A + 2×A + 2×A + 3×A + 8×A + 3×A = 19 which simplifies to 19×A = 19 Divide both sides of the above equation by 19: 19×A ÷ 19 = 19 ÷ 19 which means A = 1 making B = C = 2×A = 2 × 1 = 2 D = F = 3×A = 3 × 1 = 3 E = 8×A = 8 × 1 = 8 and ABCDEF = 122383