Puzzle for December 20, 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.
* EF and DE are 2-digit numbers (not E×F or D×E).
Scratchpad
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Hint #1
In eq.5, substitute (A – E) for F (from eq.3): A – D – (A – E) = D – C which is equivalent to A – D – A + E = D – C which becomes –D + E = D – C Add D and C to both sides of the above equation: –D + E + D + C = D – C + D + C which becomes eq.5a) E + C = 2×D
Hint #2
Add E to both sides of eq.4: C – D + E = F – E + E which becomes C – D + E = F which may be written as E + C – D = F In the equation above, replace E + C with 2×D (from eq.5a): 2×D – D = F which makes D = F
Hint #3
In eq.2, replace F with D: D – D = A – B which becomes 0 = A – B Add B to both sides of the equation above: 0 + B = A – B + B which makes B = A
Hint #4
Add D and E to both sides of eq.4: C – D + D + E = F – E + D + E which becomes C + E = F + D In the above equation, add A to the left side, and add B to the right side (since A = B): C + E + A = F + D + B which may be written as eq.4a) A + C + E = B + D + F
Hint #5
eq.1 may be written as: A + C + E + B + D + F = 34 In the equation above, substitute A + C + E for B + D + F (from eq.4a): A + C + E + A + C + E = 34 which is equivalent to 2×(A + C + E) = 34 Divide both sides by 2: 2×(A + C + E) ÷ 2 = 34 ÷ 2 which becomes eq.1a) A + C + E = 17
Hint #6
Substitute 17 for A + C + E (from eq.1a), and D for F in eq.4a: 17 = B + D + D which becomes 17 = B + 2×D Subtract 2×D from each side of the above equation: 17 – 2×D = B + 2×D – 2×D which makes 17 – 2×D = B and also makes eq.4b) A = B = 17 – 2×D
Hint #7
Substitute D for F, and 17 – 2×D for A (from eq.4b) in eq.3: D = 17 – 2×D – E In the above equation, add E to both sides, and subtract D from both sides: D + E – D = 17 – 2×D – E + E – D which becomes eq.3a) E = 17 – 3×D
Hint #8
Substitute 17 – 3×D for E (from eq.3a) in eq.5a: 17 – 3×D + C = 2×D In the above equation, subtract 17 from both sides, and add 3×D to both sides: 17 – 3×D + C – 17 + 3×D = 2×D – 17 + 3×D which becomes eq.5b) C = 5×D – 17
Hint #9
eq.6 may be written as: B = 10×E + F – (10×D + E) which is equivalent to B = 10×E + F – 10×D – E which makes eq.6a) B = 9×E + F – 10×D
Hint #10
Substitute 17 – 2×D for B (from eq.4b), (17 – 3×D) for E (from eq.3a), and D for F in eq.6a: 17 – 2×D = 9×(17 – 3×D) + D – 10×D which becomes 17 – 2×D = 153 – 27×D – 9×D which becomes 17 – 2×D = 153 – 36×D In the above equation, add 36×D to both sides, and subtract 17 from both sides: 17 – 2×D + 36×D – 17 = 153 – 36×D + 36×D – 17 which becomes 34×D = 136 Divide both sides by 34: 34×D ÷ 34 = 136 ÷ 34 which makes D = 4 and also makes F = D = 4
Hint #11
Substitute 4 for D in eq.4b: A = B = 17 – 2×4 which becomes A = B = 17 – 8 which makes A = B = 9
Hint #12
Substitute 4 for D in eq.5b: C = 5×4 – 17 which becomes C = 20 – 17 which makes C = 3
Solution
Substitute 4 for D in eq.3a: E = 17 – 3×4 which becomes E = 17 – 12 which makes E = 5 and ABCDEF = 993454