Puzzle for January 18, 2020 ( )
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 F and E to each side of eq.3: D – E + F + E = B – F + F + E which becomes eq.3a) D + F = B + E In eq.4, add D to both sides, and subtract E from both sides: E – F + D – E = C – D + D – E which becomes eq.4a) D – F = C – E
Hint #2
Add the left and right sides of eq.4a to the left and right sides of eq.3a, respectively: D + F + (D – F) = B + E + (C – E) which becomes eq.3b) 2×D = B + C In eq.2, replace B + C with 2×D (from eq.3b): 2×D = A + D Subtract D from each side of the equation above: 2×D – D = A + D – D which means D = A
Hint #3
Add C to both sides of eq.6: C + F + C = B – C + C eq.6a) 2×C + F = B In eq.3b, replace B with 2×C + F (from eq.6a): 2×D = 2×C + F + C which becomes eq.3c) 2×D = 3×C + F
Hint #4
In eq.5, substitute A + D for B + C (from eq.2): A + E + F = A + D + D which becomes A + E + F = A + 2×D Subtract A from each side of the equation above: A + E + F – A = A + 2×D – A which becomes eq.5a) E + F = 2×D
Hint #5
Substitute E + F for 2×D (from eq.5a) in eq.3c: E + F = 3×C + F Subtract F from each side of the above equation: E + F – F = 3×C + F – F which makes E = 3×C
Hint #6
Substitute 2×C + F for B (from eq.6a), and 3×C for E in eq.3a: D + F = 2×C + F + 3×C which becomes D + F = 5×C + F Subtract F from both sides of the above equation: D + F – F = 5×C + F – F which makes D = 5×C and also makes A = D = 5×C
Hint #7
Substitute 5×C for D, and 3×C for E in eq.4a: 5×C – F = C – 3×C which becomes 5×C – F = –2×C Add F and 2×C to both sides of the equation above: 5×C – F + F + 2×C = –2×C + F + 2×C which makes 7×C = F
Hint #8
Substitute 7×C for F in eq.6a: 2×C + 7×C = B which makes 9×C = B
Solution
Substitute 5×C for A and D, 9×C for B, 3×C for E, and 7×C for F in eq.1: 5×C + 9×C + C + 5×C + 3×C + 7×C = 30 which becomes 30×C = 30 Divide both sides of the equation above by 30: 30×C ÷ 30 = 30 ÷ 30 which means C = 1 making A = D = 5×C = 5 × 1 = 5 B = 9×C = 9 × 1 = 9 E = 3×C = 3 × 1 = 3 F = 7×C = 7 × 1 = 7 and ABCDEF = 591537