Puzzle for April 28, 2021  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 16 eq.2) B + C + F = D – B – C eq.3) C + E + F = D eq.4) D – C = A + B + C eq.5) E = D – A eq.6) F = B + C

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

Scratchpad

 

Help Area

Hint #1


In eq.4, replace B + C with F (from eq.6): D – C = A + F Add C to both sides of the equation above: D – C + C = A + F + C which becomes eq.4a) D = A + F + C


  

Hint #2


In eq.3, replace D with A + F + C (from eq.4a): C + E + F = A + F + C Subtract C and F from each side of the equation above: C + E + F – C – F = A + F + C – C – F which simplifies to E = A


  

Hint #3


In eq.5, substitute A for E: A = D – A Add A to both sides of the equation above: A + A = D – A + A which makes 2×A = D


  

Hint #4


eq.2 may be written as: B + C + F = D – (B + C) Substitute F for B + C (from eq.6) in the above equation: F + F = D – (F) which makes 2×F = D – F Add F to both sides: 2×F + F = D – F + F which makes eq.2a) 3×F = D


  

Hint #5


Substitute 2×A for D in eq.2a: 3×F = 2×A Divide both sides of the equation above by 2: 3×F ÷ 2 = 2×A ÷ 2 which makes 1½×F = A and also makes E = A = 1½×F


  

Hint #6


Substitute 3×F for D, and 1½×F for A in eq.4a: 3×F = 1½×F + F + C which becomes 3×F = 2½×F + C Subtract 2½×F from each side of the equation above: 3×F – 2½×F = 2½×F + C – 2½×F which makes ½×F = C


  

Hint #7


Substitute ½×F for C in eq.6: F = B + ½×F Subtract ½×F from both sides of the above equation: F – ½×F = B + ½×F – ½×F which makes ½×F = B


  

Solution

Substitute 1½×F for A and E, ½×F for B and C, and 3×F for D in eq.1: 1½×F + ½×F + ½×F + 3×F + 1½×F + F = 16 which simplifies to 8×F = 16 Divide both sides of the above equation by 8: 8×F ÷ 8 = 16 ÷ 8 which means F = 2 making A = E = 1½×F = 1½ × 2 = 3 B = C = ½×F = ½ × 2 = 1 D = 3×F = 3 × 2 = 6 and ABCDEF = 311632