Puzzle for March 5, 2021 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit non-negative integer.
Scratchpad
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Hint #1
In eq.2, replace B + C with D – E (from eq.4): D + E = A + D – E In the above equation, subtract D from each side, and add E to both sides: D + E – D + E = A + D – E – D + E which makes eq.2a) 2×E = A
Hint #2
Add C to both sides of eq.5: A + F + C = D – C + E + C which becomes A + F + C = D + E In the above equation, replace D + E with A + B + C (from eq.2): A + F + C = A + B + C Subtract A and C from each side: A + F + C – A – C = A + B + C – A – C which simplifies to F = B
Hint #3
In eq.5, substitute 2×E for A: 2×E + F = D – C + E In the equation above, add C to both sides, and subtract E from both sides: 2×E + F + C – E = D – C + E + C – E which becomes E + F + C = D which may be written as eq.5a) F + C + E = D
Hint #4
Substitute F for C + E (from eq.3) in eq.5a: F + F = D which makes eq.5b) 2×F = D
Hint #5
Divide both sides of eq.5b by 2: 2×F ÷ 2 = D ÷ 2 which makes F = ½×D and also makes B = F = ½×D Substitute B for F in eq.5a: eq.5c) B + C + E = D
Hint #6
eq.1 may be written as: A + B + C + E + D + F = 22 Substitute D for B + C + E (from eq.5c), and ½×D for F in the above equation: A + D + D + ½×D = 22 which becomes A + 2½×D = 22 Subtract 2½×D from both sides of the equation above: A + 2½×D – 2½×D = 22 – 2½×D which becomes eq.1a) A = 22 – 2½×D
Hint #7
To make eq.1a true, check several possible one-digit non-negative integer values for D and A. Do not check odd integers for D, as 2½×D would be non-integer if D is odd, and would make A non-integer in eq.1a. If D = 8, then A = 22 – 2½×8 = 22 – 20 = 2 If D = 6, then A = 22 – 2½×6 = 22 – 15 = 7 If D = 4, then A = 22 – 2½×4 = 22 – 10 = 12 If D < 4, then A > 12 Since A and D must be one-digit non-negative integers, then A = 2 and D = 8 or A = 7 D = 6
Hint #8
Check: A = 7 ... Substituting 7 for A in eq.2a would yield: 2×E = 7 Dividing both sides of the equation above by 2 would yield: 2×E ÷ 2 = 7 ÷ 2 which would make E = 3½ Since E must be an integer, then E ≠ 3½ which means A ≠ 7 and, therefore makes A = 2 and D = 8 making B = F = ½×D = ½×8 = 4
Hint #9
Substitute 2 for A in eq.2a: 2×E = 2 Divide both sides of the above equation by 2: 2×E ÷ 2 = 2 ÷ 2 which makes E = 1
Solution
Substitute 1 for E, and 4 for F in eq.3: C + 1 = 4 Subtract 1 from both sides of the above equation: C + 1 – 1 = 4 – 1 which makes C = 3 and makes ABCDEF = 243814