Puzzle for May 22, 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.
* EF and CD are 2-digit numbers (not E×F or C×D).
Scratchpad
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Hint #1
Subtract the left and right sides of eq.2 from the left and right sides of eq.3, respectively: B – D – F – (B + D) = C + D + E – (C – D + E + F) which becomes B – D – F – B – D = C + D + E – C + D – E – F which becomes –2×D – F = 2×D – F Add 2×D and F to both sides of the above equation: –2×D – F + 2×D + F = 2×D – F + 2×D + F which makes 0 = 4×D which means 0 = D
Hint #2
Subtract the left and right sides of eq.6 from the left and right sides of eq.5, respectively: EF – (A + C) – (B + EF – A) = A + B – (A – B + CD) which becomes EF – A – C – B – EF + A = A + B – A + B – CD which becomes –C – B = 2×B – CD which may be written as eq.5a) –C – B = 2×B – (10×C + D)
Hint #3
In eq.5a, substitute 0 for D: –C – B = 2×B – (10×C + 0) which becomes –C – B = 2×B – 10×C Add B and 10×C to both sides of the above equation: –C – B + B + 10×C = 2×B – 10×C + B + 10×C which becomes 9×C = 3×B Divide both sides by 3: 9×C ÷ 3 = 3×B ÷ 3 which makes 3×C = B
Hint #4
In eq.4, replace D with 0, and B with 3×C: A + 0 – E = 3×C + C + E + F – A – 0 which becomes A – E = 4×C + E + F – A Add E and A to both sides of the above equation: A – E + E + A = 4×C + E + F – A + E + A which becomes eq.4a) 2×A = 4×C + 2×E + F
Hint #5
eq.6 may be written as: B + 10×E + F – A = A – B + 10×C + D Add A and B to both sides of the above equation: B + 10×E + F – A + A + B = A – B + 10×C + D + A + B which becomes eq.6a) 2×B + 10×E + F = 2×A + 10×C + D
Hint #6
In eq.6a, replace B with (3×C), D with 0, and 2×A with 4×C + 2×E + F (from eq.4a): 2×(3×C) + 10×E + F = 4×C + 2×E + F + 10×C + 0 which becomes 6×C + 10×E + F = 14×C + 2×E + F Subtract 6×C, 2×E, and F from both sides of the equation above: 6×C + 10×E + F – 6×C – 2×E – F = 14×C + 2×E + F – 6×C – 2×E – F which simplifies to 8×E = 8×C Divide both sides by 8: 8×E ÷ 8 = 8×C ÷ 8 which makes E = C
Hint #7
In eq.2, substitute 3×C for B, 0 for D, and C for E: 3×C + 0 = C – 0 + C + F which becomes 3×C = 2×C + F Subtract 2×C from each side of the above equation: 3×C – 2×C = 2×C + F – 2×C which makes C = F
Hint #8
Substitute C for E and F in eq.4a: 2×A = 4×C + 2×C + C which makes 2×A = 7×C Divide both sides of the above equation by 2: 2×A ÷ 2 = 7×C ÷ 7 which makes A = 3½×C
Solution
Substitute 3½×C for A, 3×C for B, 0 for D, and C for E and F in eq.1: 3½×C + 3×C + C + 0 + C + C = 19 which simplifies to 9½×C = 19 Divide both sides of the above equation by 9½: 9½×C ÷ 9½ = 19 ÷ 9½ which means C = 2 making A = 3½×C = 3½ × 2 = 7 B = 3×C = 3 × 2 = 6 E = F = C = 2 and ABCDEF = 762022