Puzzle for January 18, 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 positive integer.
Scratchpad
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Hint #1
Add (B + D) to both sides of eq.4: A - B + (B + D) = C - D + (B + D) which becomes A + D = C + B In eq.2, replace A + D with C + B: C + E = C + B Subtract C from both sides of the equation above: C + E - C = C + B - C which makes E = B
Hint #2
Add B to both sides of eq.6: (E × F) - B + B = B × E + B which becomes E × F = B × (E + 1) In the equation above, replace B with E: E × F = E × (E + 1) Divide both sides of the above equation by E: E × F ÷ E = E × (E + 1) ÷ E which makes eq.6a) F = E + 1
Hint #3
Substitute E for B, and (E + 1) for F (from eq.6a) in eq.5: C - E = E + E - (E + 1) which is the same as C - E = E + E - E - 1 which becomes C - E = E - 1 Add E to each side of the above equation: C - E + E = E - 1 + E which makes eq.5a) C = 2×E - 1
Hint #4
Substitute 2×E - 1 for C (from eq.5a), E for B, and E + 1 for F (from eq.6a) in eq.3: 2×E - 1 + D - E = E + E + 1 which becomes E - 1 + D = 2×E + 1 In the above equation, add 1 to each side, and subtract E from each side: E - 1 + D + 1 - E = 2×E + 1 + 1 - E which makes eq.3a) D = E + 2
Hint #5
Substitute 2×E - 1 for C (from eq.5a), and E + 2 for D (from eq.3a) in eq.2: 2×E - 1 + E = A + E + 2 which becomes 3×E - 1 = A + E + 2 Subtract both E and 2 from each side of the above equation: 3×E - 1 - E - 2 = A + E + 2 - E - 2 which makes eq.2a) 2×E - 3 = A
Hint #6
Substitute E for B, and E + 2 for D (from eq.3a) in eq.1: E + E + 2 = A + E Subtract E from each side of the above equation: E + E + 2 - E = A + E - E which becomes eq.1a) E + 2 = A
Solution
Substitute E + 2 for A (from eq.1a) in eq.2a: 2×E - 3 = E + 2 In the above equation, subtract E from each side, and add 3 to each side: 2×E - 3 - E + 3 = E + 2 - E + 3 which makes E = 5 making A = E + 2 = 5 + 2 = 7 (from eq.1a) B = E = 5 C = 2×E - 1 = 2×5 - 1 = 10 - 1 = 9 (from eq.5a) D = E + 2 = 5 + 2 = 7 (from eq.3a) F = 5 + 1 = 6 (from eq.6a) and ABCDEF = 759756