Puzzle for August 23, 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.
* ABC, BCD, CDE, and DEF are 3-digit numbers (not A×B×C, B×C×D, C×D×E, or D×E×F).
Scratchpad
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Hint #1
In eq.5, replace A with B + C + D (from eq.3): B + C – E = B + C + D – D + E which becomes B + C – E = B + C + E In the above equation, add E to both sides: B + C – E + E = B + C + E + E which becomes B + C = B + C + 2×E Subtract B and C from both sides: B + C – B – C = B + C + 2×E – B – C which simplifies to 0 = 2×E which means 0 = E
Hint #2
In eq,2, replace E with 0: C + D + 0 = B which becomes eq.2a) C + D = B Add D to both sides of eq.5: B + C – E + D = A – D + E + D which becomes B + C + D – E = A + E In the equation above, replace C + D with B (from eq.2a), and replace E with 0: B + B – 0 = A + 0 which makes 2×B = A
Hint #3
In eq.4, substitute 0 for E, and add F to both sides: D – 0 + F + F = B – F + F which becomes eq.4a) D + 2×F = B In eq.2, substitute 0 for E, and D + 2×F for B (from eq.4a): C + D + 0 = D + 2×F Subtract D from each side: C + D + 0 – D = D + 2×F – D which makes C = 2×F
Hint #4
eq.6 may be written as: 100×A + 10×B + C – (100×C + 10×D + E) + F = 100×B + 10×C + D + 100×D + 10×E + F which is equivalent to 100×A + 10×B + C – 100×C – 10×D – E + F = 100×B + 10×C + 101×D + 10×E + F Subtract 10×B and F from both sides of the above equation: 100×A + 10×B + C – 100×C – 10×D – E + F – 10×B – F = 100×B + 10×C + 101×D + 10×E + F – 10×B – F which becomes 100×A – 99×C – 10×D – E = 90×B + 10×C + 101×D + 10×E Add 99×C, 10×D, and E to both sides: 100×A – 99×C – 10×D – E + 99×C + 10×D + E = 90×B + 10×C + 101×D + 10×E + 99×C + 10×D + E which becomes eq.6a) 100×A = 90×B + 109×C + 111×D + 11×E
Hint #5
Substitute (2×B) for A, (2×F) for C, and 0 for E in eq.6a: 100×(2×B) = 90×B + 109×(2×F) + 111×D + 11×0 which becomes 200×B = 90×B + 218×F + 111×D Subtract 90×B from each side: 200×B – 90×B = 90×B + 218×F + 111×D – 90×B which becomes eq.6b) 110×B = 218×F + 111×D
Hint #6
Substitute (D + 2×F) for B (from eq.4a) in eq.6b: 110×(D + 2×F) = 218×F + 111×D which is equivalent to 110×D + 220×F = 111×D + 218×F Subtract 110×D and 218×F from each side: 110×D + 220×F – 110×D – 218×F = 111×D + 218×F – 110×D – 218×F which makes 2×F = D
Hint #7
In eq.4a, replace D with 2×F: 2×F + 2×F = B which means 4×F = B
Hint #8
Substitute 4×F for B, and 2×F for both C and D in eq.3: A = 4×F + 2×F + 2×F which makes A = 8×F
Solution
Substitute 8×F for A, 4×F for B, 2×F for C and D, and 0 for E in eq.1: 8×F + 4×F + 2×F + 2×F + 0 + F = 17 which becomes 17×F = 17 Divide each side by 17: 17×F ÷ 17 = 17 ÷ 17 which means F = 1 making A = 8×F = 8 × 1 = 8 B = 4×F = 4 × 1 = 4 C = D = 2×F = 2 × 1 = 2 and ABCDEF = 842201