Puzzle for October 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.
* BC, DE, and EF are 2-digit numbers (not B×C, D×E, or E×F).
Scratchpad
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Hint #1
eq.5 may be written as: C + 10×E + F = A + D + B In the above equation, replace A + D with B + E + F (from eq.3): C + 10×E + F = B + E + F + B which becomes C + 10×E + F = 2×B + E + F Subtract 10×E and F from each side of the above equation: C + 10×E + F – 10×E – F = 2×B + E + F – 10×E – F which becomes eq.5a) C = 2×B – 9×E
Hint #2
eq.4 may be written as: 10×B + C – (10×D + E) = B + D + E + F which is equivalent to 10×B + C – 10×D – E = B + D + E + F In the above equation, subtract B from both sides, and add 10×D and E to both sides: 10×B + C – 10×D – E – B + 10×D + E = B + D + E + F – B + 10×D + E which becomes eq.4a) 9×B + C = 11×D + 2×E + F
Hint #3
In eq.4a, replace C with 2×B – 9×E (from eq.5a): 9×B + 2×B – 9×E = 11×D + 2×E + F which becomes 11×B – 9×E = 11×D + 2×E + F Add 9×E to each side of the equation above: 11×B – 9×E + 9×E = 11×D + 2×E + F + 9×E which becomes 11×B = 11×D + 11×E + F which may be written as eq.4b) 11×B = 11×(D + E) + F
Hint #4
In eq.4b, substitute B + C for D + E (from eq.2): 11×B = 11×(B + C) + F which becomes 11×B = 11×B + 11×C + F Subtract 11×B from each side of the equation above: 11×B – 11×B = 11×B + 11×C + F – 11×B which becomes 0 = 11×C + F Since C and F must be non-negative, the above equation makes: C = 0 and F = 0
Hint #5
Substitute 0 for C in eq.5a: 0 = 2×B – 9×E Add 9×E to both sides of the equation above: 0 + 9×E = 2×B – 9×E + 9×E which makes 9×E = 2×B Divide both sides by 2: 9×E ÷ 2 = 2×B ÷ 2 which makes 4½×E = B
Hint #6
Substitute 4½×E for B, and 0 for C in eq.2: D + E = 4½×E + 0 Subtract E from each side of the above equation: D + E – E = 4½×E + 0 – E which makes D = 3½×E
Hint #7
Substitute 3½×E for D, 4½×E for B, and 0 for F in eq.3: A + 3½×E = 4½×E + E + 0 which becomes A + 3½×E = 5½×E Subtract 3½×E from each side of the equation above: A + 3½×E – 3½×E = 5½×E – 3½×E which makes A = 2×E
Solution
Substitute 2×E for A, 4½×E for B, 0 for C and F, and 3½×E for D in eq.1: 2×E + 4½×E + 0 + 3½×E + E + 0 = 22 which simplifies to 11×E = 22 Divide both sides of the equation above by 11: 11×E ÷ 11 = 22 ÷ 11 which means E = 2 making A = 2×E = 2 × 2 = 4 B = 4½×E = 4½ × 2 = 9 D = 3½×E = 3½ × 2 = 7 and ABCDEF = 490720