Puzzle for March 31, 2021  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 23 eq.2) B + D = C + F eq.3) D – E + B = C – B + E eq.4) F – B – E = B + E eq.5) A = (F ÷ E) – B – C eq.6) E ÷ C = F ÷ (A + B + C)

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

Scratchpad

 

Help Area

Hint #1


Add B and C to both sides of eq.5: A + B + C = (F ÷ E) – B – C + B + C which becomes A + B + C = F ÷ E In eq.6, replace A + B + C with F ÷ E: E ÷ C = F ÷ (F ÷ E) which is equivalent to E ÷ C = F × (E ÷ F) which becomes E ÷ C = E Multiply both sides of the above equation by C, and, since E ≠ 0 (from eq.5), divide both sides by E: (E ÷ C) × C ÷ E = E × C ÷ E which simplifies to 1 = C


  

Hint #2


In eq.3, replace C with 1: D – E + B = 1 – B + E In the equation above, add E to both sides, and subtract B from both sides: D – E + B + E – B = 1 – B + E + E – B which becomes eq.3a) D = 1 – 2×B + 2×E


  

Hint #3


Add B and E to both sides of eq.4: F – B – E + B + E = B + E + B + E which becomes eq.4a) F = 2×B + 2×E   In eq.2, substitute 1 – 2×B + 2×E for D (from eq.3a), 1 for C, and 2×B + 2×E for F (from eq.4a): B + 1 – 2×B + 2×E = 1 + 2×B + 2×E which becomes 1 – B + 2×E = 1 + 2×B + 2×E In the above equation, subtract 1 and 2×E from each side, and add B to each side: 1 – B + 2×E – 1 – 2×E + B = 1 + 2×B + 2×E – 1 – 2×E + B which simplifies to 0 = 3×B which means 0 = B


  

Hint #4


Substitute 0 for B in eq.4a: F = 2×0 + 2×E which makes F = 2×E


  

Hint #5


Substitute 0 for B in eq.3a: D = 1 – 2×0 + 2×E which makes eq.3b) D = 1 + 2×E


  

Hint #6


Substitute 2×E for F, 0 for B, and 1 for C in eq.5: A = (2×E ÷ E) – 0 – 1 which becomes A = 2 – 1 which makes A = 1


  

Solution

Substitute 1 for A and C, 0 for B, 1 + 2×E for D (from eq.3b), and 2×E for F in eq.1: 1 + 0 + 1 + 1 + 2×E + E + 2×E = 23 which simplifies to 3 + 5×E = 23 Subtract 3 from each side of the above equation: 3 + 5×E – 3 = 23 – 3 which makes 5×E = 20 Divide both sides by 5: 5×E ÷ 5 = 20 ÷ 5 which means E = 4 making D = 1 + 2×E = 1 + 2×4 = 1 + 8 = 9 (from eq.3b) F = 2×E = 2×4 = 8 and ABCDEF = 101948