Puzzle for October 31, 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 BC are 2-digit numbers (not A×B or B×C).
Scratchpad
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Hint #1
Add E to both sides of eq.3: A + E + E = C + D – E + E which becomes eq.3a) A + 2×E = C + D Add E, B, and F to both sides of eq.4: D – E + E + B + F = A – B + E – F + E + B + F which becomes eq.4a) D + B + F = A + 2×E
Hint #2
In eq.4a, replace A + 2×E with C + D (from eq.3a): D + B + F = C + D Subtract D from each side of the equation above: D + B + F – D = C + D – D which becomes eq.4b) B + F = C
Hint #3
eq.5 may be written as: E = (B + C + F) ÷ 3 Multiply both sides of the above equation by 3: E × 3 = ((B + C + F) ÷ 3) × 3 which becomes E × 3 = B + C + F which may be written as eq.5a) 3×E = B + F + C
Hint #4
In eq.5a, replace B + F with C (from eq.4b): 3×E = C + C which makes 3×E = 2×C Divide both sides of the above equation by 2: 3×E ÷ 2 = 2×C ÷ 2 which makes eq.5b) 1½×E = C
Hint #5
In eq.5a, replace C with B + F (from eq.4b): 3×E = B + F + B + F which becomes eq.5c) 3×E = 2×B + 2×F
Hint #6
Add B and F to both sides of eq.2: F – B + B + F = A – F + B + F which becomes 2×F = A + B In eq.5c, substitute A + B for 2×F: 3×E = 2×B + A + B which becomes 3×E = 3×B + A Subtract 3×B from each side of the equation above: 3×E – 3×B = 3×B + A – 3×B which becomes eq.5d) 3×E – 3×B = A
Hint #7
In eq.3a, substitute 1½×E for C (from eq.5b): A + 2×E = 1½×E + D Subtract 1½×E from each side of the equation above: A + 2×E – 1½×E = 1½×E + D – 1½×E which becomes eq.3b) A + ½×E = D
Hint #8
eq.6 may be written as: 10×A + B = A + 10×B + C + D Subtract A and B from both sides of the above equation: 10×A + B – A – B = A + 10×B + C + D – A – B which becomes eq.6a) 9×A = 9×B + C + D
Hint #9
Substitute 1½×E for C (from eq.5b), and A + ½×E for D (from eq.3b) in eq.6a: 9×A = 9×B + 1½×E + A + ½×E which becomes 9×A = 9×B + 2×E + A Subtract A from both sides of the equation above: 9×A – A = 9×B + 2×E + A – A which becomes eq.6b) 8×A = 9×B + 2×E
Hint #10
Substitute (3×E – 3×B) for A (from eq.5d) in eq.6b: 8×(3×E – 3×B) = 9×B + 2×E which becomes 24×E – 24×B = 9×B + 2×E In the above equation, add 24×B to both sides, and subtract 2×E from both sides: 24×E – 24×B + 24×B – 2×E = 9×B + 2×E + 24×B – 2×E which simplifies to 22×E = 33×B Divide both sides by 22: 22×E ÷ 22 = 33×B ÷ 22 which makes E = 1½×B
Hint #11
Substitute (1½×B) for E in eq.5b: 1½×(1½×B) = C which becomes 2¼×B = C
Hint #12
Substitute 2¼×B for C in eq.4b: B + F = 2¼×B Subtract B from both sides of the equation above: B + F – B = 2¼×B – B which makes F = 1¼×B
Hint #13
Substitute (1½×B) for E in eq.5d: 3×(1½×B) – 3×B = A which becomes 4½×B – 3×B = A which makes 1½×B = A
Hint #14
Substitute 1½×B for A and E in eq.3b: 1½×B + ½×(1½×B) = D which becomes 1½×B + ¾×B = D which makes 2¼×B = D
Solution
Substitute 1½×B for A and E, 2¼×B for C and D, and 1¼×B for F in eq.1: 1½×B + B + 2¼×B + 2¼×B + 1½×B + 1¼×B = 39 which simplifies to 9¾×B = 39 Divide both sides of the above equation by 9¾: 9¾×B ÷ 9¾ = 39 ÷ 9¾ which means B = 4 making A = E = 1½×B = 1½ × 4 = 6 C = D = 2¼×B = 2¼ × 4 = 9 F = 1¼×B = 1¼ × 4 = 5 and ABCDEF = 649965