Puzzle for April 7, 2022  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 41 eq.2) D = C × E eq.3) C × D = (D × E) ÷ C

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

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Hint #1


Multiply both sides of eq.3 by C: C × C × D = C × (D × E) ÷ C which becomes C² × D = D × E Divide both sides of the above equation by D: C² × D ÷ D = D × E ÷ D which makes eq.3a) C² = E


  

Hint #2


In eq.2, replace E with C²: D = C × C² which makes eq.2a) D = C³


  

Hint #3


Check several possible one-digit positive integers for C, D, and E in eq.2a and eq.3a:   If C = 1, then D = C³ = 1³ = 1, and E = C² = 1² = 1 If C = 2, then D = C³ = 2³ = 8, and E = C² = 2² = 4 If C = 3, then D = C³ = 3³ = 27, and E = C² = 3² = 9 If C > 3, then D > 27, and E > 9   The only values for C, D, and E that make eq.2a and eq.3a true are:   C = 1, and D = 1, and E = 1 or C = 2, and D = 8, and E = 4


  

Hint #4


Begin checking: C = 1, D = 1, and E = 1 ...   Substituting 1 for C, D, and E in eq.1 would yield: A + B + 1 + 1 + 1 + F = 41 which becomes A + B + 3 + F = 41 Subtract 3 from each side of the above equation: A + B + 3 + F – 3 = 41 – 3 which becomes eq.1a) A + B + F = 38


  

Hint #5


Continue checking: C = 1, D = 1, and E = 1 ...   Since A, B, and F are one-digit positive integers, then: ie.1a) A ≤ 9 ie.1b) B ≤ 9 ie.1c) F ≤ 9 Combining the above three inequalities yields: A + B + F ≤ 9 + 9 + 9 which becomes ie.1d) A + B + F ≤ 27


  

Hint #6


Finish checking: C = 1, D = 1, and E = 1 ...   Substituting 38 for A + B + F (from eq.1a) in ie.1d would make: 38 ≤ 27 Since the above inequality cannot be true, then: C ≠ 1, and D ≠ 1, and E ≠ 1 and therefore makes C = 2, and D = 8, and E = 4


  

Hint #7


Substitute 2 for C, 8 for D, and 4 for E in eq.1: A + B + 2 + 8 + 4 + F = 41 which becomes A + B + 14 + F = 41 Subtract 14 from both sides of the above equation: A + B + 14 + F – 14 = 41 – 14 which becomes eq.1b) A + B + F = 27


  

Solution

Combining inequalities ie.1a, ie.1b, and ie.1c with eq.1b makes: A = 9 B = 9 F = 9 and makes ABCDEF = 992849