Puzzle for November 15, 2020 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit positive integer.
* BC is a 2-digit number (not B×C).
Scratchpad
Help Area
Hint #1
eq.5 may be written as: eq.5a) 10×B + C = D + F Since D and F must be one-digit positive integers, then: D ≤ 9 and F ≤ 9 which means D + F ≤ 9 + 9 which makes D + F ≤ 18 In the above inequality, replace D + F with 10×B + C (from eq.5a): eq.5b) 10×B + C ≤ 18
Hint #2
To make the inequality eq.5b true, check several possible values for B and C: If B = 1, then 10×1 + C ≤ 18 which makes 10 + C ≤ 18 and makes C ≤ 8 If B = 2, then 10×2 + C ≤ 18 which makes 20 + C ≤ 18 and makes C ≤ –2 If B > 2, then C < –2 Since B and C must be positive integers, the only value for B that makes eq.5b true is: B = 1
Hint #3
Logarithmic equation eq.4 may be re-written as the following equivalent exponential equation: C ^ B = E ("C ^ B" means "C raised to the power of B") In the above equation, replace B with 1: C ^ 1 = E which makes C = E
Hint #4
Inverting (i.e, taking the reciprocal of) both sides of eq.3 yields: 1 / (A ÷ D) = 1 / (E ÷ (B + C)) which becomes D ÷ A = (B + C) ÷ E In the above equation, replace D ÷ A with B + E (from eq.1): eq.3a) B + E = (B + C) ÷ E
Hint #5
In eq.3a, substitute 1 for B, and C for E: 1 + C = (1 + C) ÷ C Multiply both sides of the equation above by C: C × (1 + C) = C × (1 + C) ÷ C which makes C × (1 + C) = (1 + C) Divide both sides by (1 + C): C × (1 + C) ÷ (1 + C) = (1 + C) ÷ (1 + C) which becomes C × 1 = 1 which makes C = 1 and also makes E = C = 1
Hint #6
Substitute 1 for B and E in eq.2: 1 × D = 1 + F which makes eq.2a) D = 1 + F
Hint #7
Substitute 1 for B and C, and 1 + F for D (from eq.2a) in eq.5a: 10×1 + 1 = 1 + F + F which becomes 11 = 1 + 2×F Subtract 1 from both sides of the above equation: 11 – 1 = 1 + 2×F – 1 which makes 10 = 2×F Divide both sides by 2: 10 ÷ 2 = 2×F ÷ 2 which makes 5 = F
Hint #8
Substitute 5 for F in eq.2a: D = 1 + 5 which makes D = 6
Solution
Substitute 6 for D, and 1 for B and C and E in eq.3: A ÷ 6 = 1 ÷ (1 + 1) which becomes A ÷ 6 = 1 ÷ 2 Multiply both sides of the equation above by 6: 6 × A ÷ 6 = 6 × 1 ÷ 2 which makes A = 3 and makes ABCDEF = 311615