Puzzle for June 14, 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 non-negative integer.
* "C ^ E" means "C raised to the power of E".
Scratchpad
Help Area
Hint #1
Add D and F to both sides of eq.2: C – D + D + F = A + D – F + D + F which becomes eq.2a) C + F = A + 2×D Add C to both sides of eq.3: B – C + C = A + E + F + C which may be written as: eq.3a) B = A + E + C + F
Hint #2
In eq.3a, replace C + F with A + 2×D (from eq.2a): B = A + E + A + 2×D which becomes eq.3b) B = 2×A + E + 2×D
Hint #3
In eq.4, replace B with 2×A + E + 2×D (from eq.3b): C + D = 2×A + E + 2×D – D – E which becomes C + D = 2×A + D Subtract D from both sides of the equation above: C + D – D = 2×A + D – D which makes eq.4a) C = 2×A
Hint #4
In eq.1, substitute 2×A for C: A + E = 2×A Subtract A from each side of the above equation: A + E – A = 2×A – A which becomes E = A
Hint #5
Substitute 2×A for C, and A for E in eq.6: D = A ÷ (2×A ÷ A) which becomes D = A ÷ 2 Multiply both sides of the above equation by 2: 2 × D = 2 × A ÷ 2 which makes 2×D = A and also makes E = A = 2×D
Hint #6
Substitute (2×D) for A in eq.4a: C = 2×(2×D) which makes C = 4×D
Hint #7
Substitute (2×D) for A and E in eq.3b: B = 2×(2×D) + (2×D) + 2×D which becomes B = 4×D + 4×D which makes B = 8×D
Hint #8
Substitute 4×D for C, and 2×D for A in eq.2a: 4×D + F = 2×D + 2×D which becomes 4×D + F = 4×D Subtract 4×D from both sides of the equation above: 4×D + F – 4×D = 4×D – 4×D which makes F = 0
Hint #9
Substitute 2×D for A and E, 8×D for B, and 4×D for C in eq.5: 2×D × 8×D = 4×D ^ 2×D which becomes 16×D² = 4×D ^ 2×D which may be written as eq.5a) 16×D² = 4×D ^ (D + D) (implies D ≠ 0)
Hint #10
When multiplying two equal numbers (or terms) that have exponents, the result is the number (or term) with the two exponents added together. Mathematically, this would be expressed as: X^(m+n) = (X^m) × (X^n) This rule can be applied to eq.5a: eq.5b) 16×D² = (4×D ^ D) × (4×D ^ D)
Hint #11
When multiplying any two numbers or terms that have the same exponent, the result is the two numbers or terms multiplied together, along with the exponent. Mathematically, this would be expressed as: (X^n)×(Y^n) = (X×Y)^n This rule can be applied to eq.5b: 16×D² = (4×D × 4×D) ^ D which becomes eq.5c) 16×D² = (16×D²) ^ D
Hint #12
Divide both sides of eq.5c by 16×D²: 16×D² ÷ 16×D² = ((16×D²) ^ D) ÷ (16×D²) which becomes 1 = ((16×D²) ^ D) ÷ (16×D²) which may be written as eq.5d) 1 = ((16×D²) ^ D) ÷ ((16×D²) ^ 1) When dividing two equal numbers or terms that have exponents, the result is the number or term with the two exponents subtracted from each other. Mathematically, this would be expressed as: (X^m)/(X^n) = X^(m–n) This rule can be applied to eq.5d: eq.5e) 1 = (16×D²) ^ (D – 1)
Solution
To make eq.5e true, then: 1 = (16×D²) ^ 0 which means 0 = D – 1 which makes D = 1 making A = E = 2×D = 2×1 = 2 B = 8×D = 8×1 = 8 C = 4×D = 4×1 = 4 and ABCDEF = 284120