Puzzle for April 16, 2021  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 21 eq.2) F = D – E eq.3) C – A = A + B eq.4) D – B = A + C eq.5) C = average (B, D, E, F) eq.6) E ÷ D = A

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

Scratchpad

 

Help Area

Hint #1


Add E to both sides of eq.2: F + E = D – E + E which becomes F + E = D which may be written as eq.2a) E + F = D   Add B to both sides of eq.4: D – B + B = A + C + B which becomes D = A + C + B which may be written as eq.4a) D = A + B + C


  

Hint #2


In eq.1, replace A + B + C with D (from eq.4a), and E + F with D (from eq.2a): D + D + D = 21 which becomes 3×D = 21 Divide both sides of the above equation by 3: 3×D ÷ 3 = 21 ÷ 3 which becomes D = 7


  

Hint #3


In eq.6, replace D with 7: E ÷ 7 = A Multiply both sides of the equation above by 7: 7 × E ÷ 7 = 7 × A which makes E = 7×A


  

Hint #4


In eq.2, substitute 7 for D, and 7×A for E: eq.2a) F = 7 – 7×A


  

Hint #5


eq.5 may be written as: C = (B + D + E + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × C = 4 × (B + D + E + F) ÷ 4 which becomes eq.5a) 4×C = B + D + E + F


  

Hint #6


Add A to each side of eq.3: C – A + A = A + B + A which becomes eq.3a) C = 2×A + B   Substitute (2×A + B) for C (from eq.3a), and D for E + F (from eq.2a) in eq.5a: 4×(2×A + B) = B + D + D which becomes 8×A + 4×B = B + 2×D Subtract 8×A and B from each side of the equation above: 8×A + 4×B – 8×A – B = B + 2×D – 8×A – B which becomes eq.5b) 3×B = 2×D – 8×A


  

Hint #7


Substitute 7 for D in eq.5b: 3×B = 2×7 – 8×A which becomes 3×B = 14 – 8×A Divide both sides of the equation above by 3: B = (14 – 8×A) ÷ 3 which becomes eq.5c) B = 4⅔ – 2⅔×A


  

Hint #8


Substitute 4⅔ – 2⅔×A for B (from eq.5c) in eq.3a: C = 2×A + 4⅔ – 2⅔×A which makes eq.3b) C = 4⅔ – ⅔×A


  

Solution

Substitute 4⅔ – 2⅔×A for B (from eq.5c), 4⅔ – ⅔×A for C (from eq.3b), 7 for D, 7×A for E, and 7 – 7×A for F (from eq.2a) in eq.1: A + 4⅔ – 2⅔×A + 4⅔ – ⅔×A + 7 + 7×A + 7 – 7×A = 21 which simplifies to –2⅓×A + 23⅓ = 21 In the above equation, subtract 21 from each side, and add 2⅓×A to each side: –2⅓×A + 23⅓ – 21 + 2⅓×A = 21 – 21 + 2⅓×A which makes 2⅓ = 2⅓×A Divide both sides by 2⅓: 2⅓ ÷ 2⅓ = 2⅓×A ÷ 2⅓ which means 1 = A making B = 4⅔ – 2⅔×A = 4⅔ – 2⅔×1 = 4⅔ – 2⅔ = 2 (from eq.5c) C = 4⅔ – ⅔×A = 4⅔ – ⅔×1 = 4⅔ – ⅔ = 4 (from eq.3b) E = 7×A = 7×1 = 7 F = 7 – 7×A = 7 – 7×1 = 7 – 7 = 0 (from eq.2a) and ABCDEF = 124770