Here are my answers, done on a 48gii. Corrections, comments and improvements are gratefully accepted.

1. Plugging in the numbers leads to [[4 -2 1] [9 3 1] [196 14 1]] * [a b c] = [6 3.5 29]. Since the calculator lets you "divide" matrices, I did this:

[6 3.5 29] [[4 -2 1] [9 3 1] [196 14 1] / OBJ-> DROP + + EVAL

Answer: 3.443181

2. 20931601446 FACTOR

Result is 347*227*37*19*7*3^3*2

3. 21456326 ENTER 3241157 GCD yields 1

So no common factors.

If the GCD were greater than 1, then applying FACTORS to it would yield the common prime factors.

4. Of course this depends on the current date, but the TIME menu makes quick work of it.

DATE 2.162097 DDAYS 24 * 3600 *

5. Graph the the function y=x+cos(30x)-1 from .9 to 2.1. There are 8 solutions that are clearly within the region and 2 (or is it 3?) that are near the boundaries.

Closer investigation with ROOT and ZOOM shows that the root near 1 is too small. Using ZOOM, BOX2 shows that there are two roots just slightly less than 2.

Answer is 10.

6. 37* (sum of digits) + 8 * (product of digits) > 45742.

The sum of the digits given is 24 and the product is 960, so if you have a digit x, then the formula become 37(24+2x+1) + 8(960x(x+1)) > 45742.

I programmed the formula and ran through the digits 0-8. Smallest result that's less than 45742 occurs with 2.

Answer: 2

7. The formula for each term is (-1^(x+1)) / x. So program it as

0. 1. 1000. FOR X -1 x 1 + ^ X / + NEXT

8. << DUP INV SWAP 2 / + >>

'P8' STO

1. ENTER P8 P8 P8 P8

Result is 1.4142135... So answer to problem is 3.

9. If I remember my math right, the sum is approximately ln(n-0.5), so n is about 148. Armed with this I wrote a program to compute the sum, flubbed it with an off-by-one error and tried again. I ended up with this:

%%HP: T(3)A(R)F(.);

\<< 0. 1. \-> S X

\<<

DO X INV 'S' STO+ 1. 'X' STO+

UNTIL S 5.5 >

END X 1. -

\>>

\>>

Plugging numbers in yielded 137 as the answer.

10. 11201 ENTER NEXTPRIME NEXTPRIME NEXTPRIME etc.

11. Solve 2x-sin(x)-10 = 0. Answer 4.5102

12. Starting with $5, each time she talks to someone, she adds $4, so the sums are 5, 9, 13, etc. Each value mod 4 = 1. So the first value that she'll have in the range is 2,134,565 and they continue to 2,134/585. Now when the problem says "strictly between" do they mean that the end points are not included? That determines whether the last one is included or not.

13. f(x) = cos(1/x)

g(x) = f(f(f(f(x))))

I graphed 10*g(x)+9 and found 4 roots in the range. From the graph window, From the graph window, Press FCN, move the cursor near a root and press ROOT. When it gets the value, press ENTER to copy the root the stack. Repeat for the other roots. Exit he graph window, press "+" 3 times and you have the sum of the roots. I got 0.3183757

14. Ratio of cicumference to area is 4/D.

To compute the sum of the ratios I wrote this program:

%%HP: T(3)A(R)F(.);

\<< { 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. } ADD INV \GSLIST 4. *

\>>

Plugging in values until I narrowed it down to the one closest to 1/100, I got a program input of 5992, which means the max circumference was 6007.

15. Enter the polynomial and press FACTOR. The hardest part of this one was avoiding typo's.

16. If everyone turned in their money then 43 + 81X + 22Y + 630 = 2887. Where X > 20 and Y < 5. Simplifying gives 81X + 22Y = 2214. There are no integer solutions to this.

The total money in the room is the money collected plus the money m still in their pockets, so 81X+22Y = 2214+m. Recast this as 2214-22Y = 81X-m. Modulo 81 gives (I think) (2214-22Y) mod 81 = 81-m.

I wrote this program to generate the m for each value of Y (1-5)>

%%HP: T(3)A(R)F(.);

\<< 1 5

FOR X 2214 X 22 * - 81 MOD 81 SWAP -

NEXT

\>>

Running it showed m=2 when Y=5.

17. Divide by 19 to get 7140766 factors of 19. Divide it by 31 to get 4376598 factors of 31. Now you've double counted the factors of 31*19 = 589. There are 230347 of those. So there are 11287017 unique factors of 19 and 31 between 1 and 135674561. 135674561-11287017 = 124387544.

But that isn't the right answer, because they want value of x between 1 and 135674561 EXCLUSIVE, so you have to remove 1 and 135674561. Therefore the answer is 124387542.

18. Here is a brute force program that works, but it's ugly. It tries every combination of adding and subtracting the numbers:

%%HP: T(3)A(R)F(.);

\<< -54 54

FOR X54 -74 74

FOR X74 -112 112

FOR X112 -131 131

FOR X131 -172 172

FOR X172 -191 191

FOR X191 -213 213

FOR X213 -261 261

FOR X261 -436 436

FOR X436

IF X54 X74 + X112 + X131 + X172 + X191 + X213 + X261 + X436 + 406 ==

THEN X54 X74 X112 X131 X172 X191 X213 X261 X436 9 \->LIST

END

872 STEP

522 STEP

426 STEP

382 STEP

344 STEP

262 STEP

224 STEP

148 STEP

108 STEP

\>>

There is one solution: 54-74-112+131-172+191+213-261+436

19. The general question of how many ways to form an amount from specific coins was the subject of a previous thread. Running my program, which took forever even on an emulator, gave an answer of 28050. There must be an easier way for to have been given on a test for high school kids.

20. Three stacks to the moon would stand 384402*3 km high, which is 1.153206E12 mm. At 0.11mm per dollar bill, that's 1.04836909E13 bills. The debt would have to grow by a factor of 1.13337199 to reach that value.

1.00016^n = 1.1333719. n = 782.55, so the smallest number of full days is 783.