Puzzle for June 20, 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 non-negative integer.
* BC, CD, and EF are 2-digit numbers (not B×C, C×D, or E×F).
Scratchpad
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Hint #1
In eq.5, replace D with A + F (from eq.3): C – E + F = A + A + F which becomes C – E + F = 2×A + F Subtract F from each side of the equation above: C – E + F – F = 2×A + F – F eq.5a) C – E = 2×A
Hint #2
Subtract E from both sides of eq.2: E + F – E = A + B + C – E which becomes F = A + B + C – E In the equation above, replace C – E with 2×A (from eq.5a): F = A + B + 2×A which becomes eq.2a) F = 3×A + B
Hint #3
In eq.3, substitute 3×A + B for F (from eq.2a): D = A + 3×A + B which becomes eq.3a) D = 4×A + B
Hint #4
In eq.4, add C to both sides, and subtract A from both sides: B – C + C – A = A + C – A which becomes eq.4a) B – A = C In eq.5a, substitute B – A for C (from eq.4a): B – A – E = 2×A In the equation above, add E to both sides, and subtract 2×A from both sides: B – A – E + E – 2×A = 2×A + E – 2×A which becomes eq.5b) B – 3×A = E
Hint #5
eq.6 may be written as: A + 10×B + C = 10×C + D + 10×E + F Subtract C from each side of the above equation: A + 10×B + C – C = 10×C + D + 10×E + F – C which becomes eq.6a) A + 10×B = 9×C + D + 10×E + F
Hint #6
Substitute (B – A) for C (from eq.4a), 4×A + B for D (from eq.3a), (B – 3×A) for E (from eq.5b), and 3×A + B for F (from eq.2a) in eq.6a: A + 10×B = 9×(B – A) + 4×A + B + 10×(B – 3×A) + 3×A + B which is the same as A + 10×B = 9×B – 9×A + 4×A + B + 10×B – 30×A + 3×A + B which becomes A + 10×B = 21×B – 32×A In the above equation, add 32×A to each side, and subtract 10×B from each side: A + 10×B + 32×A – 10×B = 21×B – 32×A + 32×A – 10×B which simplifies to 33×A = 11×B Divide both sides by 11: 33×A ÷ 11 = 11×B ÷ 11 which makes 3×A = B
Hint #7
Substitute 3×A for B in eq.4a: 3×A – A = C which makes 2×A = C
Hint #8
Substitute 3×A for B in eq.3a: D = 4×A + 3×A which makes D = 7×A
Hint #9
Substitute 3×A for B in eq.5b: 3×A – 3×A = E which makes 0 = E
Hint #10
Substitute 3×A for B in eq.2a: F = 3×A + 3×A which makes F = 6×A
Solution
Substitute 3×A for B, 2×A for C, 7×A for D, 0 for E, and 6×A for F in eq.1: A + 3×A + 2×A + 7×A + 0 + 6×A = 19 which simplifies to 19×A = 19 Divide both sides of the above equation by 19: 19×A ÷ 19 = 19 ÷ 19 which means A = 1 making B = 3×A = 3 × 1 = 3 C = 2×A = 2 × 1 = 2 D = 7×A = 7 × 1 = 7 F = 6×A = 6 × 1 = 6 and ABCDEF = 132706