Puzzle for December 8, 2018  ( )

Scratchpad

Find the 6-digit number ABCDEF by solving the following equations:

eq.1) A + B + C + D + E + F = 23 eq.2) B + C = D + E eq.3) B - C = D eq.4) B + D - F = A - B + F eq.5)* DE - F = AB + F eq.6)** BCD + F = B + C + DEF

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  AB and DE are 2-digit numbers (not A×B or D×E).
**  BCD and DEF are 3-digit numbers (not B×C×D or D×E×F).

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Hint #1


From eq.3, substitute B - C for D in eq.2: B + C = B - C + E Add (C - B) to each side of the above equation: B + C + (C - B) = B - C + E + (C - B) which makes eq.2a) 2×C = E


  

Hint #2


eq.6 can be re-stated as: 100×B + 10×C + D + F = B + C + 100×D + 10×E + F Subtract B, C, D, and F from each side of the above equation: 100×B + 10×C + D + F - B - C - D - F = B + C + 100×D + 10×E + F - B - C - D - F which simplifies to 99×B + 9×C = 99×D + 10×E Substitute (2×C) for E: 99×B + 9×C = 99×D + 10×(2×C) which is the same as 99×B + 9×C = 99×D + 20×C Subtract 9×C from each side: 99×B + 9×C - 9×C = 99×D + 20×C - 9×C which becomes eq.6a) 99×B = 99×D + 11×C


  

Hint #3


From eq.3, substitute (B - C) for D in eq.6a: 99×B = 99×(B - C) + 11×C which is equivalent to 99×B = 99×B - 99×C + 11×C which becomes 99×B = 99×B - 88×C Subtract 99×B from both sides: 99×B - 99×B = 99×B - 88×C - 99×B which simplifies to 0 = -88×C which means 0 = C


  

Hint #4


In eq.2a, replace C with 0: 2×0 = E which means 0 = E


  

Hint #5


In eq.3, replace C with 0: B - 0 = D which means B = D


  

Hint #6


Substitute B for D in eq.4: B + B - F = A - B + F Add (F + B - A) to both sides: B + B - F + (F + B - A) = A - B + F + (F + B - A) which simplifies to eq.4a) 3×B - A = 2×F


  

Hint #7


eq.5 can be re-written as: 10×D + E - F = 10×A + B + F Substitute B for D, and 0 for E in the equation above: 10×B + 0 - F = 10×A + B + F Add (F - 10×A - B) to both sides: 10×B + 0 - F + (F - 10×A - B) = 10×A + B + F + (F - 10×A - B) which becomes eq.5a) 9×B - 10×A = 2×F


  

Hint #8


Substitute 9×B - 10×A for 2×F (from eq.5a) in eq.4a: 3×B - A = 9×B - 10×A Add (10×A - 3×B) to each side: 3×B - A + (10×A - 3×B) = 9×B - 10×A + (10×A - 3×B) which simplifies to 9×A = 6×B Divide both sides by 6: 9×A ÷ 6 = 6×B ÷ 6 which means 1½×A = B


  

Hint #9


Substitute (1½×A) for B in eq.4a: 3×(1½×A) - A = 2×F which is the same as 4½×A - A = 2×F which becomes 3½×A = 2×F Divide both sides by 2: 3½×A ÷ 2 = 2×F ÷ 2 which makes 1¾×A = F


  

Solution

Substitute 1½×A for B and D, 0 for C and E, and 1¾×A for F in eq.1: A + 1½×A + 0 + 1½×A + 0 + 1¾×A = 23 which becomes 5¾×A = 23 Divide both sides by 5¾: 5¾×A ÷ 5¾ = 23 ÷ 5¾ which makes A = 4 making B = D = 1½×A = 1½ × 4 = 6 F = 1¾×A = 1¾ × 6 = 7 and ABCDEF = 460607