## Monday, January 1, 1990

This post is backdated intentionally.

Problem 01: 54
Problem 02: $\frac{13}{60}$
Problem 03: 9
Problem 04: 119. Thank you Jonathan. Looking back at my notes, I did the exact same reasoning but, for some reason, put 720 - 1 = 719. I really have to pay closer attention. Maybe bad handwriting? Somehow I got an extra 6.
Problem 05: 150
Problem 06: Have not looked at this yet.
Problem 07: 4
Problem 08: $\pi$
Problem 09: 38
Problem 10: 6
Problem 11: 11, 19, 43
Problem 12: 100 or 1
Problem 13: $\frac{19}{90}$
Problem 14: 18 square units
Problem 15: a = -2, b = 4, c = 1
Problem 16: 126/17
Problem 17: 6 square units
Problem 18: 7*6*5*4 = 840; Thanks to Marc and Eric !
Problem 19: 8/95
PeggyU: I assumed there were 20 X 19 or 380 ways of picking two different numbers. There were 20 pairs where one number was the double of the other, and 12 pairs where one was the triple of the other, and 20 + 12 = 32, so 32/380 = 8/95 ...?
Problem 20: 5, 6, & 7
Problem 21: $3^3 + 4^3 + 5^3 = 6^3$
Problem 22: I got $\frac{\pi}{6}$ and $\frac{-\pi}{2}$
Problem 23: 24 ~ each face will have four edges, making a new cube. I think.
Problem 24: 13,440
Problem 25: 27 -- look at the equivalent lengths. PM is PQ + QM ...
Problem 26: 1
Problem 27: 2310
Problem 28: LCM of 1414 and 1000 = 707000. But this feels wrong.
Problem 30: 40
Problem 31: 1990.  Found it! There is always a problem that contains the year.
Problem 32:
Problem 33: I got 27$\pi$. A commenter got 3$\pi$. Confirmation?
Problem 34: 44.
Consider the function $x^{17} - 45x^{16}$.
This function is "Overall S-Shaped" and $\lim_{x\rightarrow -\infty}f(x)=-\infty$. It has 16 roots at 0, making a very flat broad shelf, and one root at 45, through which the function is passing at a positive slope (m>20). If the function is translated vertically 27 units, the roots will shift ever so slightly to the left, 44<r<45.
Problem 35:
Problem 36:
Problem 37:
Problem 38: 2, 3, 4
Problem 39: 9/4
Problem 40:
Problem 41:

Any thoughts on any of these answers?.

1. Answer for #4, I believe, is 119.

There are no duplicate values possible, so no subtraction is needed.

There are four choices for 100s (0, 1, 2, or 3), times five choices for 10s (0, 1, 2, 3, or 4), times two for 5s (none or one), times 3 for ones (0, 1, or 2).

4 x 5 x 2 x 3 (minus the 0, 0, 0, 0 non-positive choice).

It is equivalent to asking how many factors greater than 1 126,000 has.

Jonathan

2. For #18, I think it is a Counting Principle problem. 7*6*5*4

3. 18 is simple counting. Each of the 4 rows must have exactly one "1", slotted in a different column than the others. So should be 7*6*5*4.

4. For #31, y = 1990

5. #38 caught my eye because I thought you could graph it somehow at first. I then realized if the three absolute values have to add to 5, then each one can be at most 5. So each absolute value only allows for a window of 10 possibilities. I tried the union of these sets (ie. checking all values) and realized quickly it should have been the intersection of these sets, allowing now for only the numbers from 0 to 6 as possibilities. I checked each and found the answer to be x=2,3,4.

I thought it was interesting that no values are strictly less than 5. There may be a bit more thinking that can further reduce the possibilities to check here, but I didn't see it at first and 6 wasn't too many (especially when I had started checking at 14 possibilities already).

6. Just took a stab at #33 and got 3(pi) minutes. Can anyone confirm?