Puzzle for March 8, 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, CD, DE, and EF are 2-digit numbers (not B×C, C×D, D×E, or E×F).
Scratchpad
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Hint #1
Subtract both D and E from each side of eq.2: A + D - D - E = C + E - D - E which becomes eq.2a) A - E = C - D
Hint #2
Add F to both sides of eq.4: A + E + F + F = C + D - F + F which becomes eq.4a) A + E + 2×F = C + D
Hint #3
Add the left and right sides of eq.4a to the left and right sides of eq.2a, respectively: A - E + A + E + 2×F = C - D + C + D which becomes 2×A + 2×F = 2×C Divide both sides by 2: (2×A + 2×F) ÷ 2 = 2×C ÷ 2 which becomes eq.2b) A + F = C
Hint #4
In eq.3, replace C with A + F (from eq.2b): A + F - E = A - F Subtract A from each side of the equation above: A + F - E - A = A - F - A which becomes F - E = -F Add E + F to both sides: F - E + E + F = -F + E + F which means 2×F = E
Hint #5
eq.4 may be written as: A + F + E = C + D - F In the equation above, substitute C for A + F (from eq.2b): C + E = C + D - F Subtract C from both sides: C + E - C = C + D - F - C which becomes E = D - F Substitute 2×F for E, and add F to both sides: 2×F + F = D - F + F which makes 3×F = D
Hint #6
eq.6 may be written as: 10×D + E + F = 10×C + D Substitute 3×F for D, and 2×F for E in the equation above: 10×3×F + 2×F + F = 10×C + 3×F which becomes 33×F = 10×C + 3×F Subtract 3×F from each side: 33×F - 3×F = 10×C + 3×F - 3×F which becomes 30×F = 10×C Divide both sides by 3: 30×F ÷ 3 = 10×C ÷ 3 which makes 3×F = C
Hint #7
Substitute 3×F for C in eq.2b: A + F = 3×F Subtract F from each side: A + F - F = 3×F - F which makes A = 2×F
Hint #8
eq.5 may be written as: 10×B + C = B + C + 10×E + F Subtract both B and C from each side of the above equation: 10×B + C - B - C = B + C + 10×E + F - B - C which simplifies to 9×B = 10×E + F Substitute 2×F for E: 9×B = 10×2×F + F which is the same as 9×B = 20×F + F which becomes 9×B = 21×F Divide both sides by 9: 9×B ÷ 9 = 21×F ÷ 9 which makes B = 2⅓×F
Solution
Substitute 2×F for A and E, 2⅓×F for B, and 3×F for C and D in eq.1: 2×F + 2⅓×F + 3×F + 3×F + 2×F + F = 40 which simplifies to 13⅓×F = 40 Divide each side by 13⅓: 13⅓×F ÷ 13⅓ = 40 ÷ 13⅓ which makes F = 3 making A = E = 2×F = 2 × 3 = 6 B = 2⅓×F = 2⅓ × 3 = 7 C = D = 3×F = 3 × 3 = 9 and ABCDEF = 679963