Puzzle for March 21, 2019  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 30 eq.2) D + F = C eq.3) C + D + E - B = A + B + F eq.4) A + F = B eq.5) C - B = D - F eq.6)* CD - AB = DE - A - F

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

Scratchpad

 

Help Area

Hint #1


In eq.5, replace C with D + F (from eq.2): D + F - B = D - F Subtract D from both sides of the above equation: D + F - B - D = D - F - D which becomes F - B = -F Add B + F to both sides: F - B + B + F = -F + B + F which makes 2×F = B


  

Hint #2


In eq.4, substitute 2×F for B: A + F = 2×F Subtract F from both sides of the above equation: A + F - F = 2×F - F which means A = F


  

Hint #3


Subtract F from both sides of eq.2: D + F - F = C - F which becomes eq.2a) D = C - F


  

Hint #4


eq.6 may be written as: 10×C + D - (10×A + B) = 10×D + E - A - F which is the same as 10×C + D - 10×A - B = 10×D + E - A - F Add (A - D) to both sides of the above equation: 10×C + D - 10×A - B + (A - D) = 10×D + E - A - F + (A - D) which becomes eq.6a) 10×C - 9×A - B = 9×D + E - F


  

Hint #5


Substitute F for A, and 2×F for B in eq.6a: 10×C - 9×F - 2×F = 9×D + E - F Add F to both sides of the above equation: 10×C - 9×F - 2×F + F = 9×D + E - F + F which becomes 10×C - 10×F = 9×D + E which may be written as 10×(C - F) = 9×D + E Substitute D for C - F (from eq.2a): 10×(D) = 9×D + E Subtract 9×D from both sides: 10×(D) - 9×D = 9×D + E - 9×D which makes D = E


  

Hint #6


Substitute D for E, 2×F for B, and F for A in eq.3: C + D + D - 2×F = F + 2×F + F which becomes C + 2×D - 2×F = 4×F Substitute (C - F) for D (from eq.2a) in the above equation: C + 2×(C - F) - 2×F = 4×F which is the same as C + 2×C - 2×F - 2×F = 4×F which becomes 3×C - 4×F = 4×F Add 4×F to both sides: 3×C - 4×F + 4×F = 4×F + 4×F which simplifies to 3×C = 8×F Divide both sides by 3: 3×C ÷ 3 = 8×F ÷ 3 which makes C = 2⅔×F


  

Hint #7


Substitute 2⅔×F for C in eq.2a: 2⅔×F - F = D which makes 1⅔×F = D which also makes E = D = 1⅔×F


  

Solution

Substitute F for A, 2×F for B, 2⅔×F for C, and 1⅔×F for D and E in eq.1: F + 2×F + 2⅔×F + 1⅔×F + 1⅔×F + F = 30 which simplifies to 10×F = 30 Divide both sides by 10: 10×F ÷ 10 = 30 ÷ 10 which makes F = 3 making A = F = 3 B = 2×F = 2×3 = 6 C = 2⅔×F = 2⅔ × 3 = 8 D = E = 1⅔×F = 1⅔ × 3 = 5 and ABCDEF = 368553