Puzzle for October 15, 2022  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 35 eq.2) A = E + F eq.3) C + E = B + F eq.4) C + D = A + F eq.5) B - D = A - C + D - F eq.6) D = average (B, E, F)

A, B, C, D, E, and F each represent a one-digit non-negative integer.

Scratchpad

 

Help Area

Hint #1


eq.6 may be written as: D = (B + E + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × D = 3 × (B + E + F) ÷ 3 which becomes eq.6a) 3×D = B + E + F


  

Hint #2


Add D, C, and F to both sides of eq.5: B - D + D + C + F = A - C + D - F + D + C + F which becomes B + C + F = A + 2×D which my be written as eq.5a) A + 2×D = B + C + F


  

Hint #3


Subtract the left and right sides of eq.6a from the left and right sides of eq.5a, respectively: A + 2×D - 3×D = B + C + F - (B + E + F) which becomes A - D = B + C + F - B - E - F which becomes A - D = C - E Add D and E to both sides of the above equation above: A - D + D + E = C - E + D + E which becomes eq.5b) A + E = C + D


  

Hint #4


In eq.4, replace C + D with A + E (from eq.5b): A + E = A + F Subtract A from each side of the above equation: A + E - A = A + F - A which makes E = F


  

Hint #5


In eq.2, substitute E for F: A = E + E which makes A = 2×E


  

Hint #6


In eq.3, substitute E for F: C + E = B + E Subtract E from each side of the equation above: C + E - E = B + E - E which makes C = B


  

Hint #7


Substitute 2×E for A, B for C, and E for F in eq.5a: 2×E + 2×D = B + B + E which becomes 2×E + 2×D = 2×B + E Subtract 2×D and E from both sides of the above equation: 2×E + 2×D - 2×D - E = 2×B + E - 2×D - E which makes eq.5c) E = 2×B - 2×D and also makes eq.5d) F = E = 2×B - 2×D


  

Hint #8


Substitute 2×B - 2×D for E and F (from eq.5d) in eq.6a: 3×D = B + 2×B - 2×D + 2×B - 2×D which becomes 3×D = 5×B - 4×D Add 4×D to both sides of the above equation: 3×D + 4×D = 5×B - 4×D + 4×D which makes eq.6b) 7×D = 5×B


  

Hint #9


Multiply both sides of eq.5c by 5: 5 × E = 5 × (2×B - 2×D) which becomes 5×E = 10×B - 10×D which may be written as eq.5e) 5×E = 2×(5×B) - 10×D


  

Hint #10


Substitute 7×D for 5×B (from eq.6b) in eq.5e: 5×E = 2×(7×D) - 10×D which becomes 5×E = 14×D - 10×D which makes 5×E = 4×D Divide both sides of the above equation by 4: 5×E ÷ 4 = 4×D ÷ 4 which makes 1¼×E = D


  

Hint #11


Substitute 1¼×E for D in eq.6b: 7×(1¼×E) = 5×B which becomes 8¾×E = 5×B Divide both sides of the above equation by 5: 8¾×E ÷ 5 = 5×B ÷ 5 which makes 1¾×E = B and also makes 1¾×E = B = C


  

Solution

Substitute 2×E for A, 1¾×E for B and C, 1¼×E for D, and E for F in eq.1: 2×E + 1¾×E + 1¾×E + 1¼×E + E + E = 35 which simplifies to 8¾×E = 35 Divide both sides of the above equation by 8¾: 8¾×E ÷ 8¾ = 35 ÷ 8¾ which means E = 4 making A = 2×E = 2 × 4 = 8 B = C = 1¾×E = 1¾ × 4 = 7 D = 1¼×E = 1¼ × 4 = 5 F = E = 4 and ABCDEF = 877544