Puzzle for May 2, 2020  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 26 eq.2) D + E + F = A + B + C eq.3) B + F = A + C eq.4) A = C + D eq.5) C – D – E = E – F eq.6) E + F – A = A + B

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

Scratchpad

 

Help Area

Hint #1


In eq.2, replace A + B with E + F – A (from eq.6): D + E + F = E + F – A + C Subtract E and F from each side of the equation above: D + E + F – E – F = E + F – A + C – E – F which becomes D = –A + C which may be written as eq.2a) D = C – A


  

Hint #2


In eq.4, replace D with C – A (from eq.2a): A = C + C – A which becomes A = 2×C – A Add A to each side of the above equation: A + A = 2×C – A + A which makes 2×A = 2×C Divide both sides by 2: 2×A ÷ 2 = 2×C ÷ 2 which makes A = C


  

Hint #3


In eq.2a, substitute A for C: D = A – A which makes D = 0


  

Hint #4


Substitute A for C, and 0 for D in eq.5: A – 0 – E = E – F which becomes A – E = E – F In the above equation, add E and F to both sides, and subtract A from each side: A – E + E + F – A = E – F + E + F – A which simplifies to eq.5a) F = 2×E – A


  

Hint #5


Substitute A for C in eq.3: B + F = A + A which becomes B + F = 2×A Subtract F from both sides of the equation above: B + F – F = 2×A – F which becomes eq.3a) B = 2×A – F


  

Hint #6


Substitute 0 for D, 2×A – F for B (from eq.3a), and A for C in eq.2: 0 + E + F = A + 2×A – F + A which becomes E + F = 4×A – F Subtract F from each side of the equation above: E + F – F = 4×A – F – F which becomes eq.2b) E = 4×A – 2×F


  

Hint #7


Substitute (2×E – A) for F (from eq.5a) in eq.2b: E = 4×A – 2×(2×E – A) which is equivalent to E = 4×A – 4×E + 2×A which becomes E = 6×A – 4×E Add 4×E to both sides of the equation above: E + 4×E = 6×A – 4×E + 4×E which makes 5×E = 6×A Divide both sides by 5: 5×E ÷ 5 = 6×A ÷ 5 which makes E = 1⅕×A


  

Hint #8


Substitute (1⅕×A) for E in eq.5a: F = 2×(1⅕×A) – A which becomes F = 2⅖×A – A which makes F = 1⅖×A


  

Hint #9


Substitute 1⅖×A for F in eq.3a: B = 2×A – 1⅖×A which makes B = ⅗×A


  

Solution

Substitute ⅗×A for B, A for C, 0 for D, 1⅕×A for E, and 1⅖×A for F in eq.1: A + ⅗×A + A + 0 + 1⅕×A + 1⅖×A = 26 which simplifies to 5⅕×A = 26 Divide both sides of the equation above by 5⅕: 5⅕×A ÷ 5⅕ = 26 ÷ 5⅕ which means A = 5 making B = ⅗×A = ⅗ × 5 = 3 C = A = 5 E = 1⅕×A = 1⅕ × 5 = 6 F = 1⅖×A = 1⅖ × 5 = 7 and ABCDEF = 535067