Puzzle for January 16, 2021  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 27 eq.2) C – D = B – A eq.3) B + E = (A ÷ B) + D + F eq.4) F = (C ÷ D) + E eq.5)* A + C + E + F = (C ^ D) – C – A

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  "C ^ D" means "C raised to the power of D".

Scratchpad

 

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Hint #1


Add A and D to both sides of eq.2: C – D + A + D = B – A + A + D which becomes eq.2a) C + A = B + D   Add (C + A) to both sides of eq.5: A + C + E + F + (C + A) = (C ^ D) – C – A + (C + A) which becomes eq.5a) A + C + E + F + (C + A) = C ^ D


  

Hint #2


In eq.5a, replace (C + A) with B + D (from eq.2a): A + C + E + F + B + D = C ^ D which may be written as A + B + C + D + E + F = C ^ D In the above equation, replace A + B + C + D + E + F with 27 (from eq.1): eq.5b) 27 = C ^ D


  

Hint #3


To make eq.5b true, check several possible values for C and D (check only odd numbers for C, since an even number raised to the power of any positive integer will be an even number):   If C = 1, then 27 = 1 ^ D which means D = undefined If C = 3, then 27 = 3 ^ D which makes D = 3 If C = 5, then 27 = 5 ^ D which makes D = 2.04781858346 If C = 7, then 27 = 7 ^ D which makes D = 1.69372510216 If C = 9, then 27 = 9 ^ D which makes D = 1.5   Since C and D must be one-digit integers, the above equations make C = 3 and D = 3


  

Hint #4


In eq.2a, replace both C and D with 3: 3 + A = B + 3 Subtract 3 from both sides of the above equation: 3 + A – 3 = B + 3 – 3 which makes A = B


  

Hint #5


In eq.4, substitute 3 for C and D: F = (3 ÷ 3) + E which makes eq.4a) F = 1 + E


  

Hint #6


Substitute B for A, 3 for D, and 1 + E for F (from eq.4a) in eq.3: B + E = (B ÷ B) + 3 + 1 + E which becomes B + E = 1 + 4 + E which becomes B + E = 5 + E Subtract E from each side of the equation above: B + E – E = 5 + E – E which makes B = 5 and also makes A = B = 5


  

Solution

Substitute 5 for A and B, 3 for C and D, and 1 + E for F (from eq.4a) in eq.1: 5 + 5 + 3 + 3 + E + 1 + E = 27 which becomes 17 + 2×E = 27 Subtract 17 from both sides of the equation above: 17 + 2×E – 17 = 27 – 17 which makes 2×E = 10 Divide both sides by 2: 2×E ÷ 2 = 10 ÷ 2 which means E = 5 making F = 1 + 5 = 6 (from eq.4a) and ABCDEF = 553356