Puzzle for August 25, 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.
Scratchpad
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Hint #1
Add D to both sides of eq.4: D – C + D = B – D + D which becomes eq.4a) 2×D – C = B In eq.3, replace B with 2×D – C (from eq.4a): 2×D – C – E = A – C Add C and E to both sides of the above equation: 2×D – C – E + C + E = A – C + C + E which becomes eq.3a) 2×D = A + E
Hint #2
Add F to both sides of eq.5: A + D + E – F + F = B + F + F which becomes A + D + E = B + 2×F which may be written as A + E + D = B + 2×F In the above equation, replace A + E with 2×D (from eq.3a): 2×D + D = B + 2×F which becomes eq.5a) 3×D = B + 2×F
Hint #3
Add C and F to both sides of eq.2: F – C + C + F = A – F + C + F which becomes 2×F = A + C In eq.5a, substitute A + C for 2×F, and 2×D – C for B (from eq.4a): 3×D = 2×D – C + A + C which becomes 3×D = 2×D + A Subtract 2×D from each side: 3×D – 2×D = 2×D + A – 2×D which makes D = A
Hint #4
Substitute D for A in eq.3a: 2×D = D + E Subtract D from both sides of the equation above: 2×D – D = D + E – D which makes D = E
Hint #5
Add C to both sides of eq.4a: 2×D – C + C = B + C which becomes eq.4b) 2×D = B + C Substitute D for A and E, and 2×D for B + C (from eq.4b) in eq.1: D + 2×D + D + D + F = 41 which becomes 5×D + F = 41 Subtract 5×D from both sides of the above equation: eq.1a) F = 41 – 5×D
Hint #6
Substitute (41 – 5×D) for F in eq.5a: 3×D = B + 2×(41 – 5×D) which becomes 3×D = B + 82 – 10×D In the equation above, add 10×D to both sides, and subtract 82 from each side: 3×D + 10×D – 82 = B + 82 – 10×D + 10×D – 82 which becomes eq.5b) 13×D – 82 = B
Hint #7
Substitute 13×D – 82 for B (from eq.5b) in eq.4b: 2×D = 13×D – 82 + C In the above equation, add 82 to both sides, and subtract 13×D from each side: 2×D + 82 – 13×D = 13×D – 82 + C + 82 – 13×D which becomes eq.4c) 82 – 11×D = C
Hint #8
Substitute (82 – 11×D) for C (from eq.4c), (41 – 5×D) for F (from eq.1a), D for A and E, and 13×D – 82 for B (from eq.5b) in eq.6: (82 – 11×D) × (41 – 5×D) = D + 13×D – 82 + D + D which becomes 3362 – 451×D – 410×D + 55×D² = 16×D – 82 which becomes 3362 – 861×D + 55×D² = 16×D – 82 In the above equation, add 82 to both sides, and subtract 16×D from each side: 3362 – 861×D + 55×D² + 82 – 16×D = 16×D – 82 + 82 – 16×D which becomes 3444 – 877×D + 55×D² = 0 which may be written eq.6a) 55×D² – 877×D + 3444 = 0
Solution
eq.6a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for D in eq.6a yields: D = { (–1)×(–877) ± sq.rt.[(–877)² – (4 × (55) × (3444))] } ÷ (2 × (55)) which becomes D = {877 ± sq.rt.(769129 – 757680)} ÷ 110 which becomes D = {877 ± sq.rt.(11449)} ÷ 110 which becomes D = {877 ± 107} ÷ 110 In the above equation, either D = {877 + 107} ÷ 110 = 984 ÷ 110 = 8.94545454545 or D = {877 – 107} ÷ 110 = 770 ÷ 110 = 7 Since D must be an integer, then D ≠ 8.94545454545 and therefore makes D = 7 making A = E = D = 7 B = 13×D – 82 = 13 × 7 – 82 = 91 – 82 = 9 (from eq.5b) C = 82 – 11×D = 82 – 11 × 7 = 82 – 77 = 5 (from eq.4c) F = 41 – 5×D = 41 – 5 × 7 = 41 – 35 = 6 (from eq.1a) and ABCDEF = 795776