BuilderBoi

2) No number can be squared and have a ones digit of 2 or 3. 

The only way it can have a last digit of 4 is if the number ends in 2 (2, 12, and 22). 

So there are \(\color{brown}\boxed{3}\) numbers that work. 

1) If it is divisible by 7 and 11, it must be divisible by \(\text{lcm}(7, 11) = 77\). 

Can you take it from here?

you forgot to post them...

5) If a 2-digit number is prime, it must end in 1, 3, 7, or 9. 

This means that the only numbers that satisfy the condition will contain 2 of the 4 numbers listed above. 

Try each one out and find the ones that work.

4) Track the last digit: 3, 9, 7, 1, 3, 9, 7, 1, 3 ...

Notice a pattern?

3) If a number has an odd number of divisors, it must be a square number. 

Now, just try each square number until you find the smallest one with 5 factors. 

2) If it is divisible by 12, it must be divisible by 4 and 3. The only cases where it is divisible by 4 are 2, 4, 6, and 8. 

If it is divisible by 3, d must be 1 more than a multiple of 3. The only case that works is 4, so the sum is \(\color{brown}\boxed{4}\)

1) Maximize the x coordinate of the midpoint, and go from there. 

4) First, maximize the quarters, then maximize the dimes with what's left, then repeat for the nickels and pennies. 

So, use 3 quarters for 75 cents

Then, use 1 dime to get to 85

Then, use 1 nickel to get to 90

Then, use 3 pennies to get to 93

So \(3 + 3 + 1 + 1 = \color{brown}\boxed{8}\) coins.

3) \(12 = 2^2 \times 3\), \(15 = 3 \times 5\)

This tells us that the number will have 3 prime factors (2, 3, and 5). 

The key here is that \(2^x \times 3^y \times 5^z\) has \((x+1)(y+1)(z+1)\)

The only 3 numbers that multiply to 12 are 2, 2, and 3. 

So, z = 1, y = 1, and x = 2 meaning the number is \(2^2 \times 3^2 \times 5 = \color{brown}\boxed{60}\)