Puzzle for January 26, 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.
Scratchpad
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Hint #1
Subtract B, C, and E from both sides of eq.3: B + D – B – C – E = A + C + E – B – C – E which becomes eq.3a) D – C – E = A – B In eq.5, add C to both sides, and subtract B from both sides: A – C + C – B = B – D + C – B which becomes eq.5a) A – B = –D + C
Hint #2
In eq.3a, replace A – B with –D + C (from eq.5a): D – C – E = –D + C In the above equation, add D and E to each side, and subtract C from each side: D – C – E + D + E – C = –D + C + D + E – C which becomes eq.3b) 2×D – 2×C = E
Hint #3
Add A and E to both sides of eq.6: C – A – E + A + E = A + E + A + E which becomes eq.6a) C = 2×A + 2×E In eq.3b, substitute (2×A + 2×E) for C (from eq.6a): 2×D – 2×(2×A + 2×E) = E which becomes 2×D – 4×A – 4×E = E Add 4×A and 4×E to each side of the above equation: 2×D – 4×A – 4×E + 4×A + 4×E = E + 4×A + 4×E which becomes 2×D = 4×A + 5×E Divide both sides by 2: 2×D ÷ 2 = (4×A + 5×E) ÷ 2 which becomes eq.3c) D = 2×A + 2½×E
Hint #4
Add F to both sides of eq.4: C – F + F = E + F + F which becomes eq.4a) C = E + F + F Substitute A + B for E + F (from eq.2) in eq.4a: eq.4b) C = A + B + F
Hint #5
Substitute A + B + F for C (from eq.4b) in eq.3: B + D = A + A + B + F + E which becomes B + D = 2×A + B + F + E Subtract B from both sides of the above equation: B + D – B = 2×A + B + F + E – B which becomes eq.3d) D = 2×A + F + E
Hint #6
Substitute 2×A + 2½×E for D (from eq.3c) in eq.3d: 2×A + 2½×E = 2×A + F + E Subtract 2×A and E from both sides of the above equation: 2×A + 2½×E – 2×A – E = 2×A + F + E – 2×A – E which makes 1½×E = F
Hint #7
Substitute 1½×E for F in eq.4a: C = E + 1½×E + 1½×E which makes C = 4×E
Hint #8
Substitute 4×E for C in eq.6a: 4×E = 2×A + 2×E Subtract 2×E from both sides of the equation above: 4×E – 2×E = 2×A + 2×E – 2×E which makes 2×E = 2×A Divide both sides by 2: 2×E ÷ 2 = 2×A ÷ 2 which makes E = A
Hint #9
Substitute E for A in eq.3c: D = 2×E + 2½×E which makes D = 4½×E
Hint #10
Substitute E for A, 4×E for C, and 4½×E for D in eq.5: E – 4×E = B – 4½×E which becomes –3×E = B – 4½×E Add 4½×E to both sides of the equation above: –3×E + 4½×E = B – 4½×E + 4½×E which makes 1½×E = B
Solution
Substitute E for A, 1½×E for B and F, 4×E for C, and 4½×E for D in eq.1: E + 1½×E + 4×E + 4½×E + E + 1½×E = 27 which simplifies to 13½×E = 27 Divide both sides of the equation above by 13½: 13½×E ÷ 13½ = 27 ÷ 13½ which means E = 2 making A = E = 2 B = F = 1½×E = 1½ × 2 = 3 C = 4×E = 4 × 2 = 8 D = 4½×E = 4½ × 2 = 9 and ABCDEF = 238923