Puzzle for July 1, 2019  ( )

Scratchpad

Find the 5-digit number ABCDE by solving the following equations:

eq.1) A + B + C + D + E = 21 eq.2) B + E = C eq.3) E - A = B - E eq.4) D + E = C - B eq.5)* C ^ D = E - A

A, B, C, D, and E 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 B to both sides of eq.4: D + E + B = C - B + B which becomes D + E + B = C which may also be written as D + B + E = C In the above equation, replace B + E with C (from eq.2): D + C = C Subtract C from both sides: D + C - C = C - C which means D = 0


  

Hint #2


In eq.5, replace D with 0: C ^ 0 = E - A which becomes 1 = E - A (assumes C > 0) In the equation above, add A to both sides, and subtract 1 from each side: 1 + A - 1 = E - A + A - 1 which means eq.5a) A = E - 1


  

Hint #3


In eq.3, substitute (E - 1) for A (from eq.5a): E - (E - 1) = B - E which is equivalent to E - E + 1 = B - E which makes 1 = B - E Add E to each side: 1 + E = B - E + E which makes eq.3a) 1 + E = B


  

Hint #4


Substitute 1 + E for B (from eq.3a) in eq.2: 1 + E + E = C which makes eq.2a) 1 + 2×E = C


  

Solution

Substitute E - 1 for A (from eq.5a), 1 + E for B (from eq.3a), 1 + 2×E for C (from eq.2a), and 0 for D in eq.1: E - 1 + 1 + E + 1 + 2×E + 0 + E = 21 which simplifies to 5×E + 1 = 21 Subtract 1 from each side of the above equation: 5×E + 1 - 1 = 21 - 1 5×E = 20 Divide both sides by 5: 5×E ÷ 5 = 20 ÷ 5 which means E = 4 making A = E - 1 = 4 - 1 = 3 (from eq.5a) B = 1 + E = 1 + 4 = 5 (from eq.3a) C = 1 + 2×E = 1 + 2×4 = 1 + 8 = 9 (from eq.2a) and ABCDE = 35904