Puzzle for May 3, 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 positive integer.
* F! is F-factorial. CD and DE are 2-digit numbers (not C×D or D×E).
Scratchpad
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Hint #1
In eq.4, replace A + E with B + C + D (from eq.1): B + C + D = C + F Subtract C from both sides of the above equation: B + C + D – C = C + F – C which becomes eq.4a) B + D = F
Hint #2
In eq.4, replace C with A + B (from eq.2), and replace F with B + D (from eq.4a): A + E = A + B + B + D which becomes A + E = A + 2×B + D Subtract A from both sides of the equation above: A + E – A = A + 2×B + D – A which becomes eq.4b) E = 2×B + D
Hint #3
In eq.3, substitute A + B for C (from eq.2), 2×B + D for E (from eq.4b), and B + D for F (from eq.4a): B + A + B + 2×B + D = A – B + D + B + D which simplifies to A + 4×B + D = A + 2×D Subtract A and D from each side of the above equation: A + 4×B + D – A – D = A + 2×D – A – D which makes 4×B = D
Hint #4
Substitute 4×B for D in eq.4a: B + 4×B = F which makes 5×B = F
Hint #5
Substitute 4×B for D in eq.4b: E = 2×B + 4×B which makes E = 6×B
Hint #6
eq.5 may be written as: F! = 10×C + D + 10×D + E which becomes F! = 10×C + 11×D + E Substitute 4×B for D, and 6×B for E in the equation above: F! = 10×C + 11×(4×B) + 6×B which becomes F! = 10×C + 44×B + 6×B which becomes eq.5a) F! = 10×C + 50×B
Hint #7
Substitute 5×B for F in eq.5a: (5×B)! = 10×C + 50×B Subtract 50×B from both sides of the equation above: (5×B)! – 50×B = 10×C + 50×B – 50×B which becomes (5×B)! – 50×B = 10×C Divide both sides by 10: ((5×B)! – 50×B) ÷ 10 = 10×C ÷ 10 which becomes eq.5b) ((5×B)! – 50×B) ÷ 10 = C
Hint #8
To make eq.5b true, check several possible values for B and C: If B = 1, then C = ((5×1)! – 50×1) ÷ 10 = (5! – 50) ÷ 10 = (120 – 50) ÷ 10 = 70 ÷ 10 = 7 If B = 2, then C = ((5×2)! – 50×2) ÷ 10 = (10! – 100) ÷ 10 = (3,628,800 – 100) ÷ 10 = 3,628,700 ÷ 10 = 362,870 If B > 2, then C > 362,870 Since B and C must be one-digit positive integers, then C ≤ 10 which means B < 2 and therefore makes B = 1 and C = 7 making D = 4×B = 4 × 1 = 4 E = 6×B = 6 × 1 = 6 F = 5×B = 5 × 1 = 5
Solution
Substitute 7 for C, and 1 for B in eq.2: 7 = A + 1 Subtract 1 from each side of the equation above: 7 – 1 = A + 1 – 1 which makes 6 = A and makes ABCDEF = 617465