Puzzle for November 23, 2019 ( )
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 and EF are 2-digit numbers (not B×C or E×F).
Scratchpad
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Hint #1
In eq.2, replace C + E with A – B – F (from eq.5): A + D = B + A – B – F + F which becomes A + D = A Subtract A from each side of the above equation: A + D – A = A – A which means D = 0
Hint #2
In eq.3, replace D with 0: B – C = 0 – E Add C and E to each side of the equation above: B – C + C + E = 0 – E + C + E which becomes eq.3a) B + E = C
Hint #3
In eq.4, substitute 0 for D: 0 + E + F = B + C – E Add E to both sides of the above equation: 0 + E + F + E = B + C – E + E which becomes eq.4a) 2×E + F = B + C
Hint #4
In eq.4a, substitute B + E for C (from eq.3a): 2×E + F = B + B + E Subtract 2×E from each side of the equation above: 2×E + F – 2×E = B + B + E – 2×E which becomes eq.4b) F = 2×B – E
Hint #5
In eq.5, add B to both sides, and subtract E from each side: C + E + B – E = A – B – F + B – E which becomes C + B = A – F – E which may be written as B + C = A – E – F Substitute A – E – F for B + C in eq.4a: 2×E + F = A – E – F Add E and F to both sides of the above equation: 2×E + F + E + F = A – E – F + E + F which becomes eq.5a) 3×E + 2×F = A
Hint #6
eq.6 may be written as: 10×B + C – E = A + 10×E + F Add E to each side of the above equation: 10×B + C – E + E = A + 10×E + F + E which becomes 10×B + C = A + 11×E + F Substitute B + E for C (from eq.3a), and 3×E + 2×F for A (from eq.5a) in the above equation: 10×B + B + E = 3×E + 2×F + 11×E + F which becomes 11×B + E = 14×E + 3×F Subtract E from each side: 11×B + E – E = 14×E + 3×F – E which becomes eq.6a) 11×B = 13×E + 3×F
Hint #7
Substitute (2×B – E) for F (from eq.4b) in eq.6a: 11×B = 13×E + 3×(2×B – E) which is equivalent to 11×B = 13×E + 6×B – 3×E which becomes 11×B = 10×E + 6×B Subtract 6×B from both sides of the above equation: 11×B – 6×B = 10×E + 6×B – 6×B which means 5×B = 10×E Divide both sides by 5: 5×B ÷ 5 = 10×E ÷ 5 which makes B = 2×E
Hint #8
Substitute (2×E) for B in eq.4b: F = 2×(2×E) – E which becomes F = 4×E – E which makes F = 3×E
Hint #9
Substitute (3×E) for F in eq.5a: 3×E + 2×(3×E) = A which becomes 3×E + 6×E = A which makes 9×E = A
Hint #10
Substitute 2×E for B in eq.3a: 2×E + E = C which makes 3×E = C
Solution
Substitute 9×E for A, 2×E for B, 3×E for C and F, and 0 for D in eq.1: 9×E + 2×E + 3×E + 0 + E + 3×E = 18 which simplifies to 18×E = 18 Divide both sides of the above equation by 18: 18×E ÷ 18 = 18 ÷ 18 which means E = 1 making A = 9×E = 9×1 = 9 B = 2×E = 2×1 = 2 C = F = 3×E = 3×1 = 3 and ABCDEF = 923013