Puzzle for December 24, 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.
* AB and DE are 2-digit numbers (not A×B or D×E).
Scratchpad
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Hint #1
Add the left and right sides of eq.4 to the left and right sides of eq.5, respectively: A + B + E – F + F – B – D = F – C + A – C which becomes A + E – D = F – 2×C + A Subtract A from each side of the equation above: A + E – D – A = F – 2×C + A – A which becomes eq.5a) E – D = F – 2×C
Hint #2
In eq.5a, replace F with D + E (from eq.2): E – D = D + E – 2×C Subtract D and E from both sides of the above equation: E – D – D – E = D + E – 2×C – D – E which becomes –2×D = –2×C Divide both sides by (–2): –2×D ÷ (–2) = –2×C ÷ (–2) which makes D = C
Hint #3
Add C and E to both sides of eq.3: B – C + C + E = A – E + C + E which becomes eq.3a) B + E = A + C
Hint #4
In eq.5, substitute A + C for B + E (from eq.3a): A + A + C – F = F – C which becomes 2×A + C – F = F – C Add F and C to both sides of the above equation: 2×A + C – F + F + C = F – C + F + C which becomes 2×A + 2×C = 2×F Divide both sides by 2: (2×A + 2×C) ÷ 2 = 2×F ÷ 2 which becomes eq.5b) A + C = F
Hint #5
Substitute C for D, and A + C for F (from eq.5b) in eq.2: C + E = A + C Subtract C from each side of the above equation: C + E – C = A + C – C which makes E = A
Hint #6
Substitute A for E in eq.3a: B + A = A + C Subtract A from both sides of the equation above: B + A – A = A + C – A which makes B = C and also makes B = C = D
Hint #7
eq.6 may be written as: 10×A + B = 10×D + E + F Substitute C for B and D, A for E, and A + C for F (from eq.5b) in the equation above: 10×A + C = 10×C + A + A + C which becomes 10×A + C = 11×C + 2×A Subtract C and 2×A from both sides: 10×A + C – C – 2×A = 11×C + 2×A – C – 2×A which becomes 8×A = 10×C Divide both sides by 8: 8×A ÷ 8 = 10×B ÷ 8 which makes A = 1¼×B and also makes E = A = 1¼×B
Hint #8
Substitute B for D, and 1¼×B for E in eq.2: B + 1¼×B = F which makes 2¼×B = F
Solution
Substitute 1¼×B for A and E, B for C and D, and 2¼×B for F in eq.1: 1¼×B + B + B + B + 1¼×B + 2¼×B = 31 which simplifies to 7¾×B = 31 Divide both sides of the above equation by 7¾: 7¾×B ÷ 7¾ = 31 ÷ 7¾ which means B = 4 making A = E = 1¼×B = 1¼ × 4 = 5 C = D = B = 4 F = 2¼×B = 2¼ × 4 = 9 and ABCDEF = 544459