answer:First, we identify all two-digit primes less than 50. These primes are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47. This gives us a total of 11 choices for the two-digit prime WX. Since WX and YZ must be distinct primes, after choosing WX, there are 10 remaining choices for YZ. The total number of different combinations of WX and YZ that form a valid four-digit number WXYZ is calculated by multiplying the number of choices for WX by the number of choices for YZ. This gives: [ 11 times 10 = 110 ] Thus, the total number of possible house numbers is boxed{110}.

answer:First, we find the magnitude of overrightarrow{a}: |overrightarrow{a}| = sqrt{2^2 + 0^2} = 2. Next, we use the dot product formula to find the dot product of overrightarrow{a} and overrightarrow{b}: overrightarrow{a} cdot overrightarrow{b} = |overrightarrow{a}| |overrightarrow{b}| cos frac{pi}{3} = 2 cdot 1 cdot frac{1}{2} = 1 Now, we can find the magnitude of overrightarrow{a} - 2overrightarrow{b} using the dot product formula and the fact that |overrightarrow{b}| = 1: begin{align*} |overrightarrow{a} - 2overrightarrow{b}|^2 &= (overrightarrow{a} - 2overrightarrow{b}) cdot (overrightarrow{a} - 2overrightarrow{b}) &= overrightarrow{a} cdot overrightarrow{a} - 4overrightarrow{a} cdot overrightarrow{b} + 4overrightarrow{b} cdot overrightarrow{b} &= |overrightarrow{a}|^2 - 4(overrightarrow{a} cdot overrightarrow{b}) + 4|overrightarrow{b}|^2 &= 2^2 - 4(1) + 4(1^2) &= 4 end{align*} Therefore, |overrightarrow{a} - 2overrightarrow{b}| = boxed{2}.

answer:Since y=2^x and y=log_2(x+1) are both increasing functions on the interval [0,1], it follows that y=2^x+log_2(x+1) is an increasing function on the interval [0,1]. Therefore, the sum of the maximum and minimum values is: 2+log_2(0+1)+2^1+log_2(1+1)=4. Hence, the answer is boxed{4}.

answer:1. Consider the sequence ( {a_n} ) which satisfies the condition that any natural number can be uniquely expressed in the form ( a_i + 2a_j + 4a_k ), where ( i, j, k ) are not necessarily distinct. 2. To solve the problem, we introduce the generating function: [ f(x) = sum_{n geq 0} x^{a_n} ] By hypothesis, we know that: [ f(x) f(x^2) f(x^4) = sum_{n geq 0} x^n = frac{1}{1 - x} ] This is because each term on the left corresponds to ( x^{a_i + 2a_j + 4a_k} ), and due to the uniqueness of the representation, there is exactly one term for each degree. 3. Now, replacing ( x ) by ( x^2 ) in the above equation, we get: [ f(x^2) f(x^4) f(x^8) = frac{1}{1 - x^2} ] 4. Combining both equations, we have: [ f(x) = frac{1 - x^2}{1 - x} f(x^8) = (1 + x) f(x^8) ] 5. Iterating this, we find: [ f(x) = prod_{k geq 0} (1 + x^{8^k}) ] 6. The only nonzero terms of ( f(x) ) are those where the exponents have digits 0 or 1 in base 8. Thus, the ( n )-th such number is obtained by writing ( n ) in base 2 and reading the result in base 8. 7. To find ( a_{1998} ), write 1998 in binary: [ 1998_{10} = 11111001110_2 ] 8. Reading this binary number as a base 8 number, where each group of three binary digits corresponds to one digit in base 8: [ 11111001110_2 = 17716_8 ] This can be further broken down in terms of powers of 8: [ a_{1998} = 8^{10} + 8^{9} + 8^{8} + 8^{7} + 8^{6} + 8^{3} + 8^{2} + 8 ] 9. Therefore, calculating the resultant value: [ 8^{10} + 8^9 + 8^8 + 8^7 + 8^6 + 8^3 + 8^2 + 8 = 1227096648 ] 10. Hence, the ( 1998 )-th term of the sequence is: [ a_{1998} = 1227096648 ] Conclusion: [ boxed{1227096648} ]