Puzzle for August 19, 2022  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 32 eq.2) B – D = D – E eq.3) E + C = B – C eq.4) A ÷ E = (D ÷ B) + (F ÷ D) eq.5) D ÷ E = average (C, F) eq.6)* BC + CD = DE

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  BC, CD, and DE are 2-digit numbers (not B×C, C×D, or D×E).

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Hint #1


Add D to both sides of eq.2: B – D + D = D – E + D which becomes eq.2a) B = 2×D – E   Add C to both sides of eq.3: E + C + C = B – C + C which becomes eq.3a) E + 2×C = B


  

Hint #2


In eq.2a, replace B with E + 2×C (from eq.3a): E + 2×C = 2×D – E Add E to both sides of the above equation: E + 2×C + E = 2×D – E + E which becomes 2×E + 2×C = 2×D Divide both sides by 2: (2×E + 2×C) ÷ 2 = 2×D ÷ 2 which becomes eq.2b) E + C = D


  

Hint #3


eq.6 may be written as: 10×B + C + 10×C + D = 10×D + E which becomes 10×B + 11×C + D = 10×D + E Subtract D from each side of the equation above: 10×B + 11×C + D – D = 10×D + E – D which becomes eq.6a) 10×B + 11×C = 9×D + E


  

Hint #4


In eq.6a, substitute (E + 2×C) for B (from eq.3a), and E + C for D (from eq.2b): 10×(E + 2×C) + 11×C = 9×(E + C) + E which becomes 10×E + 20×C + 11×C = 9×E + 9×C + E which becomes 10×E + 31×C = 10×E + 9×C Subtract 10×E and 9×C from both sides of the equation above: 10×E + 31×C – 10×E – 9×C = 10×E + 9×C – 10×E – 9×C which simplifies to 22×C = 0 which means C = 0


  

Hint #5


Substitute 0 for C in eq.2b: E + 0 = D which makes E = D


  

Hint #6


Substitute 0 for C in eq.3a: E + 2×0 = B which becomes E + 0 = B which makes E = B


  

Hint #7


Substitute E for B and D in eq.4: A ÷ E = (E ÷ E) + (F ÷ E) which becomes A ÷ E = 1 + (F ÷ E) Multiply both sides of the above equation by E: E × (A ÷ E) = E × (1 + (F ÷ E)) which becomes eq.4a) A = E + F


  

Hint #8


eq.5 may be written as: D ÷ E = (C + F) ÷ 2 Substitute E for D, and 0 for C in the above equation: E ÷ E = (0 + F) ÷ 2 which becomes 1 = F ÷ 2 Multiply both sides of the above equation by 2: 2 × 1 = 2 × (F ÷ 2) which makes 2 = F


  

Hint #9


Substitute 2 for F in eq.4a: eq.4b) A = E + 2


  

Solution

Substitute E + 2 for A (from eq.4b), E for B and D, 0 for C, and 2 for F in eq.1: E + 2 + E + 0 + E + E + 2 = 32 which becomes 4×E + 4 = 32 Subtract 4 from both sides of the equation above: 4×E + 4 – 4 = 32 – 4 which makes 4×E = 28 Divide both sides by 4: 4×E ÷ 4 = 28 ÷ 4 which means E = 7 making A = E + 2 = 7 + 2 = 9 (from eq.4b) B = D = E = 7 and ABCDEF = 970772