Puzzle for January 23, 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.
* DE and EF are 2-digit numbers (not D×E or E×F).
Scratchpad
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Hint #1
In eq.4, replace E – B with D + F (from eq.5): A – F = D + F Add F to both sides of the above equation: A – F + F = D + F + F which becomes eq.4a) A = D + 2×F
Hint #2
In eq.2, replace A with D + 2×F (from eq.4a): D + E = D + 2×F + C + F which becomes D + E = D + C + 3×F Subtract D from each side of the above equation: D + E – D = C + D + 3×F – D which becomes eq.2a) E = C + 3×F
Hint #3
In eq.3, substitute C + 3×F for E (from eq.2a): B + C + 3×F = C + D Subtract both C and 3×F from each side of the equation above: B + C + 3×F – C – 3×F = C + D – C – 3×F which simplifies to eq.3a) B = D – 3×F
Hint #4
Substitute (D – 3×F) for B (from eq.3a) in eq.5: E – (D – 3×F) = D + F which is the same as E – D + 3×F = D + F In the equation above, add D to both sides, and subtract 3×F from both sides: E – D + 3×F + D – 3×F = D + F + D – 3×F which becomes eq.5a) E = 2×D – 2×F
Hint #5
Substitute 2×D – 2×F for E (from eq.5a) in eq.2a: 2×D – 2×F = C + 3×F Subtract 3×F from each side of the above equation: 2×D – 2×F – 3×F = C + 3×F – 3×F which becomes eq.2b) 2×D – 5×F = C
Hint #6
eq.6 may be written as: A + C + 10×D + E + F = 10×E + F Subtract both E and F from each side of the equation above: A + C + 10×D + E + F – E – F = 10×E + F – E – F which becomes eq.6a) A + C + 10×D = 9×E
Hint #7
Substitute D + 2×F for A (from eq.4a), 2×D – 5×F for C (from eq.2b), and (2×D – 2×F) for E (from eq.5a) in eq.6a: D + 2×F + 2×D – 5×F + 10×D = 9×(2×D – 2×F) which is the same as 13×D – 3×F = 18×D – 18×F In the equation above, add 18×F to both sides, and subtract 13×D from both sides: 13×D – 3×F + 18×F – 13×D = 18×D – 18×F + 18×F – 13×D which simplifies to 15×F = 5×D Divide both sides by 5: 15×F ÷ 5 = 5×D ÷ 5 which makes 3×F = D
Hint #8
Substitute (3×F) for D in eq.2b: 2×(3×F) – 5×F = C which becomes 6×F – 5×F = C which makes F = C
Hint #9
Substitute (3×F) for D in eq.5a: E = 2×(3×F) – 2×F which becomes E = 6×F – 2×F which makes E = 4×F
Hint #10
Substitute 3×F for D in eq.3a: B = 3×F – 3×F which means B = 0
Hint #11
Substitute 3×F for D in eq.4a: A = 3×F + 2×F which makes A = 5×F
Solution
Substitute 5×F for A, 0 for B, F for C, 3×F for D, and 4×F for E in eq.1: 5×F + 0 + F + 3×F + 4×F + F = 14 which simplifies to 14×F = 14 Divide both sides of the above equation by 14: 14×F ÷ 14 = 14 ÷ 14 which means F = 1 making A = 5×F = 5 × 1 = 5 C = F = 1 D = 3×F = 3 × 1 = 3 E = 4×F = 4 × 1 = 4 and ABCDEF = 501341