Puzzle for January 31, 2019 ( )
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.
* "E ^ B" means "E raised to the power of B".
Scratchpad
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Hint #1
Add (D - F) to both sides of eq.5: C - D + (D - F) = D + F + (D - F) which becomes eq.5a) C - F = 2×D
Hint #2
Adding D to each side of eq.3: C - D + F + D = A + D + D which becomes eq.3a) C + F = A + 2×D
Hint #3
In eq.3a, replace 2×D with C - F (from eq.5a): C + F = A + C - F Add (F - C) to each side of the above equation: C + F + (F - C) = A + C - F + (F - C) which simplifies to 2×F = A
Hint #4
In eq.4, substitute 2×F for A: 2×F - C = C - F Add F + C to each side of the equation above: 2×F - C + F + C = C - F + F + C which makes 3×F = 2×C Divide both sides by 2: 3×F ÷ 2 = 2×C ÷ 2 which makes 1½×F = C
Hint #5
In eq.4, replace A with 2×F: 1½×F - F = 2×D which becomes ½×F = 2×D Divide both sides of the above equation by 2: ½×F ÷ 2 = 2×D ÷ 2 which makes eq.3b) ¼×F = D
Hint #6
Subtract both A and E from each side of eq.1: D + E - A - E = A + B - A - E which becomes D - A = B - E Substitute 2×F for A, and ¼×F for D in the above equation: ¼×F - 2×F = B - E which becomes eq.1a) -1¾×F = B - E
Hint #7
Substitute 1½×F for C, and ¼×F for D in eq.2: 1½×F + ¼×F = B + E which makes eq.2a) 1¾×F = B + E
Hint #8
Add the left and right sides of eq.1a to the left and right sides of eq.2a, respectively: -1¾×F + 1¾×F = B - E + B + E which becomes 0 = 2×B which means 0 = B
Hint #9
Substitute 0 for B in eq.6: E ^ 0 = D which means E ^ 0 = 1 (assumes E > 0) which makes D = 1
Hint #10
Substitute 1 for D in eq.3b: ¼×F = 1 Divide both sides by ¼: ¼×F ÷ ¼ = 1 ÷ ¼ which means F = 4
Solution
Substitute 4 for F, and 0 for B in eq.2a: 1¾×4 = 0 + E which means 7 = E making A = 2×F = 2 × 4 = 8 C = 1½×F = 1½ × 4 = 6 and ABCDEF = 806174