Puzzle for April 17, 2021 ( )
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
Help Area
Hint #1
Add the left and right sides of eq.6 to the left and right sides of eq.4, respectively: A – B – C + D + B – A = B – D + C – D which becomes –C + D = B + C – 2×D Add 2×D to both sides of the equation above: –C + D + 2×D = B + C – 2×D + 2×D which becomes eq.4a) –C + 3×D = B + C
Hint #2
In eq.3, replace B + C with –C + 3×D (from eq.4a): D + F – C = –C + 3×D – D which becomes D + F – C = –C + 2×D In the above equation, add C to both sides, and subtract D from both sides: D + F – C + C – D = –C + 2×D + C – D which makes F = D
Hint #3
In eq.3, substitute D for F: D + D – C = B + C – D which becomes 2×D – C = B + C – D In the equation above, subtract C from both sides, and add D to both sides: 2×D – C – C + D = B + C – D – C + D which becomes eq.3a) 3×D – 2×C = B
Hint #4
Substitute 3×D – 2×C for B (from eq.3a) in eq.6: 3×D – 2×C – A = C – D In the equation above, add A and D to both sides, and subtract C from both sides: 3×D – 2×C – A + A + D – C = C – D + A + D – C which simplifies to eq.6a) 4×D – 3×C = A
Hint #5
Substitute 3×D – 2×C for B (from eq.3a), D for F, and 4×D – 3×C for A (from eq.6a) in eq.2: 3×D – 2×C – E + D = 4×D – 3×C + C + E which becomes 4×D – 2×C – E = 4×D – 2×C + E In the above equation, subtract 4×D from each side, and add 2×C to each side: 4×D – 2×C – E – 4×D + 2×C = 4×D – 2×C + E – 4×D + 2×C which makes –E = E Add E to both sides: –E + E = E + E which makes 0 = 2×E which means 0 = E
Hint #6
Substitute 0 for E, D for F, 4×D – 3×C for A (from eq.6a), and 3×D – 2×C for B (from eq.3a) in eq.5: C + 0 + D = 4×D – 3×C + 3×D – 2×C + D – 0 which becomes C + D = 8×D – 5×C In the above equation, subtract D from each side, and add 5×C to each side: C + D – D + 5×C = 8×D – 5×C – D + 5×C which makes eq.5a) 6×C = 7×D
Hint #7
Multiply both sides of eq.3a by 3: 3 × (3×D – 2×C) = 3 × B which becomes 9×D – 6×C = 3×B In the equation above, substitute 7×D for 6×C (from eq.5a): 9×D – 7×D = 3×B which makes 2×D = 3×B Divide both sides by 2: 2×D ÷ 2 = 3×B ÷ 2 which makes D = 1½×B and also makes F = D = 1½×B
Hint #8
Substitute (1½×B) for D in eq.5a: 6×C = 7×(1½×B) which makes 6×C = 10½×B Divide both sides of the above equation by 6: 6×C ÷ 6 = 10½×B ÷ 6 which makes C = 1¾×B
Hint #9
Substitute (1½×B) for D, and (1¾×B) for C in eq.6a: 4×(1½×B) – 3×(1¾×B) = A which becomes 6×B – 5¼×B = A which makes ¾×B = A
Solution
Substitute ¾×B for A, 1¾×B for C, 1½×B for D and F, and 0 for E in eq.1: ¾×B + B + 1¾×B + 1½×B + 0 + 1½×B = 26 which simplifies to 6½×B = 26 Divide both sides of the equation above by 6½: 6½×B ÷ 6½ = 26 ÷ 6½ which means B = 4 making A = ¾×B = ¾ × 4 = 3 C = 1¾×B = 1¾ × 4 = 7 D = F = 1½×B = 1½ × 4 = 6 and ABCDEF = 347606