Puzzle for February 3, 2020  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 31 eq.2) A = D + E eq.3) C + E = F eq.4) A + D = E + F eq.5) F – B = B + D eq.6) D – E + F = B + E

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

Scratchpad

 

Help Area

Hint #1


In eq.4, replace A with D + E (from eq.2): D + E + D = E + F which becomes 2×D + E = E + F Subtract E from both sides of the above equation: 2×D + E – E = E + F – E which makes eq.4a) 2×D = F


  

Hint #2


In eq.5, replace F with 2×D: 2×D – B = B + D In the equation above, add B to both sides, and subtract D from both sides: 2×D – B + B – D = B + D + B – D which makes D = 2×B


  

Hint #3


In eq.4a, substitute (2×B) for D: 2×(2×B) = F which makes 4×B = F


  

Hint #4


Substitute 2×B for D, and 4×B for F in eq.6: 2×B – E + 4×B = B + E which becomes 6×B – E = B + E In the equation above, add E to both sides, and subtract B from both sides: 6×B – E + E – B = B + E + E – B which makes 5×B = 2×E Divide both sides by 2: 5×B ÷ 2 = 2×E ÷ 2 which means 2½×B = E


  

Hint #5


Substitute 2×B for D, and 2½×B for E in eq.2: A = 2×B + 2½×B which makes A = 4½×B


  

Hint #6


Substitute 2½×B for E, and 4×B for F in eq.3: C + 2½×B = 4×B Subtract 2½×B from each side of the equation above: C + 2½×B – 2½×B = 4×B – 2½×B which makes C = 1½×B


  

Solution

Substitute 4½×B for A, 1½×B for C, 2×B for D, 2½×B for E, and 4×B for F in eq.1: 4½×B + B + 1½×B + 2×B + 2½×B + 4×B = 31 which simplifies to 15½×B = 31 Divide both sides of the above equation by 15½: 15½×B ÷ 15½ = 31 ÷ 15½ which means B = 2 making A = 4½×B = 4½ × 2 = 9 C = 1½×B = 1½ × 2 = 3 D = 2×B = 2 × 2 = 4 E = 2½×B = 2½ × 2 = 5 F = 4×B = 4 × 2 = 8 and ABCDEF = 923458