Puzzle for June 6, 2020  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 30 eq.2) B + C + D = A + E + F eq.3) C + D = E – F eq.4) A + C = E + F eq.5) A – C + E = B + C – E eq.6) 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


Subtract B, E, and F from both sides of eq.2: B + C + D – B – E – F = A + E + F – B – E – F which becomes eq.2a) C + D – E – F = A – B   In eq.5, subtract B and E from both sides, and add C to each side: A – C + E – B – E + C = B + C – E – B – E + C which becomes eq.5a) A – B = 2×C – 2×E


  

Hint #2


In eq.2a, replace A – B with 2×C – 2×E (from eq.5a): C + D – E – F = 2×C – 2×E In the equation above, subtract C and D from each side, and add 2×E to each side: C + D – E – F – C – D + 2×E = 2×C – 2×E – C – D + 2×E which simplifies to –F + E = C – D which may be written as eq.2b) E – F = C – D


  

Hint #3


In eq.3, replace E – F with C – D (from eq.2b): C + D = C – D In the equation above, subtract C from each side, and add D to each side: C + D – C + D = C – D – C + D which makes 2×D = 0 which means D = 0


  

Hint #4


Subtract the left and right sides of eq.3 from the left and right sides of eq.4, respectively: A + C – (C + D) = E + F – (E – F) which is equivalent to A + C – C – D = E + F – E + F which becomes A – D = 2×F In the above equation, substitute 0 for D: A – 0 = 2×F which makes A = 2×F


  

Hint #5


Substitute E – F for C + D (from eq.3) in eq.2: B + E – F = A + E + F In the equation above, subtract A and E from each side, and add F to each side: B + E – F – A – E + F = A + E + F – A – E + F which simplifies to eq.2c) B – A = 2×F


  

Hint #6


Substitute 2×F for A in eq.2c: B – 2×F = 2×F Add 2×F to both sides of the above equation: B – 2×F + 2×F = 2×F + 2×F which makes B = 4×F


  

Hint #7


Substitute 2×F for A, and 4×F for B in eq.6: C – F = 2×F + 4×F – C which becomes C – F = 6×F – C Add C and F to both sides of the equation above: C – F + C + F = 6×F – C + C + F which makes 2×C = 7×F Divide both sides by 2: 2×C ÷ 2 = 7×F ÷ 2 which makes C = 3½×F


  

Hint #8


Substitute 3½×F for C, and 0 for D in eq.3: 3½×F + 0 = E – F Add F to each side of the equation above: 3½×F + 0 + F = E – F + F which makes 4½×F = E


  

Solution

Substitute 2×F for A, 4×F for B, 3½×F for C, 0 for D, and 4½×F for E in eq.1: 2×F + 4×F + 3½×F + 0 + 4½×F + F = 30 which simplifies to 15×F = 30 Divide each side of the equation above by 15: 15×F ÷ 15 = 30 ÷ 15 which means F = 2 making A = 2×F = 2 × 2 = 4 B = 4×F = 4 × 2 = 8 C = 3½×F = 3½ × 2 = 7 E = 4½×F = 4½ × 2 = 9 and ABCDEF = 487092