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1. Since we know that x,y,z are all positive, we can tell that 2x<15, 3y<15,5z<15, or else some other term wouldn’t be a positive integer. So, x<7.5,y<5,z<3. We can try out the values of z, since there are only two options, 1 and 2. If z = 1, 2x+3y+5(1) = 15. 2x+3y =10. Two values that work for this are x=2, y=2. So, a possible solution is (2,2,1). However, there are more than one possible solution, such as (1,1,2).
If we were asked to solve over reals, there would be an infinite number of solutions. We can isolate each variable to find the values.
x = 7/2 – 3y/2 – 5z/2
y = 5- 2x/3 – 5z/3
z = 3- 2x/5 – 3y/5
2. 3 distinct factors: 4. 1,2,4
4 distinct factors: 6. 1,2,3,6
3.We can look at the pattern for remainders when power of 2 are divided by 3.
So, half of the powers of 2 have a remainder of 1 when divided by 3. If k<=100, then and it has a remainder of 1 for all even exponents, then there must be 51, including 0.
4. We can use the sum of numbers formula to get S = 25(24)/2 = 300, which is divisible by 25. So, the remainder is 0.
Another way to do this is to group the numbers: (1+24)+(2+23)+(3+22)….(12+13) = 25+25+25…+25. All of the terms are 25, so it must be divisible by 25.
5.Since at least half of them answered too low, we know that the age must be above half, which is 36. Since two of them are off by 1, that means that it has to be directly in between two integers, since no number is repeated. The only two pairs that have numbers directly in between them are 36 and 38, and 47 and 49. So, the possible numbers are 37 and 48. However, 48 is not a prime number, so he must be 37 years old.