dan's math@home - problem of the week - archives
 
 
Problem Archives page 7
with Answers and Winners. For Problems Only, click here.
 
1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90 . 91-100
101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151-160 . 161-170 . 171-180
181-190 . 191-200 . 201-210 . 211-220 . 221-230 . 231+ . prob index
 
61 - The Right Sticks
62 - Eat Lots of Fruit
63 - Sqrs into Primes
64 A Prime Complex
65 -- Goat or Sheep ?
66 - Holiday Cookies
67 - Into 2 Thousand
68 -- Powerful Digits
69 -- The Do-It-Alls!
70 -- "Can" Be Done!
 
 
Problem #61 - Posted Sunday, October 24, 1999 (back to top)
The Right Sticks!
Stella sees six sticks, of lengths 2, 3, 4, 5, 6, and 7 inches.
She arranges them into two separate triangles (each with positive area).
a) How can she maximize the sum of the two areas?
b) How can she minimize the sum of the two areas?
c) What are these extreme area sums (to 3 decimal places)?
Winner is first with the best answers. Supply steps and any formulas used.
Solution: (Thanks to Patrik Petersson)
There are 10 different combinations of two triangles. (Some of these triangles
cannot be constructed) I used Herons formula to calculate the area sums. *
Triangle 1 Triangle 2 Sum of areas
(2,3,4) (5,6,7) 17.602 (= 2.9047 + 14.6969)
(2,3,5) (4,6,7) -
(2,3,6) (4,5,7) -
(2,3,7) (4,5,6) -
(2,4,5) (3,6,7) 12.744 (= 3.7997 + 8.9443)
(2,4,6) (3,5,7) -
(2,4,7) (3,5,6) -
(2,5,6) (3,4,7) -
(2,5,7) (3,4,6) -
(2,6,7) (3,4,5) 11.562 (= 5.562 + 6.000)
"-" means that at least one of the triangles cannot have a positive area.
Maximum sum: 17.602 square inches. Triangles: (2,3,4) and (5,6,7)
Minimum sum: 11.562 square inches. Triangles: (2,6,7) and (3,4,5)
* Herons formula: Area of a triangle = sqrt((p(p-a)(p-b)(p-c))
where a,b and c are the sides and p = half the circumference (perimeter) of the triangle.
WINNERS - Problem 61 . Keep 'em coming! (back to top) . leader board
Patrik Petersson. . . . 10 pts- (nice, concise, and complete - so I used it!)
Rebecca Wooten. . . . 7 pts - Good - also gave the lowest overall sum 2,5,6+3,4,7 = 4.684
Sue B. . . . . . . . . . . . . . 5 pts - Good expl'n - did you check 2,4,5 + 3,6,7 ?
Al B. . . . . . . . . . . . . . . 4 pts- Clear description of derivation of area formula.
Danielle Janney . . . . 3 pts - Didja use Heron's formula or another construction?
Chris Pentacoff. . . . . 3 pts - You got a hold of a bad formula from somewhere.
Shinsuke Kuzuya . . . 3 pts - Nice feeling for what shape max should be.
Dave Seaman. . . . . . . 2 pts - Nice proof and ASCII picture but I wanted total (sum) of areas.
Big Dave. . . . . . . . . . . 1 pt - Thanks 4 entering & giving max and min triangles
 
 
Problem #62 - Posted Thursday, Novenber 4, 1999 (back to top)
Eat Lots of Fruit!
Cafe de la Peche offers three fruit bowls:
Bowl A has 2 apples and one banana;
Bowl B has 4 apples, 2 bananas, and 3 pears;
Bowl C has 2 apples, 1 banana, and 3 pears.
Your doctor tells you to eat exactly:16 apples, 8 bananas, and 6 pears per day.
How many of each type of bowl can you buy so there's no fruit left over?
Find all possible answers. The numbers of bowls must be integers; show equations and steps.
Solution (Answer from Al B.) Since Pears only exist in Bowls B & C, to have 6 Pears
. . . requires 2 Bowl B, or 2 Bowl C, or 1Bowl B and 1 Bowl C.
By subtracting the no. of Apples and Bananas in the above from 16 Apples and 8 Bananas,
. . . the no. of Bowl A`s to make up the difference is obtained.
1. 4 Bowl A, 2 Bowl B 2. 6 Bowl A, 2 Bowl C 3. 5 Bowl A, 1 Bowl B, 1 Bowl C
WINNERS - Problem 62 . Thanks for waiting; lotsa correct answers! (back to top) . leader board
Big Dave. . . . . . . . . . . 10 pts - That's good healthy reasoning. And first!
Danielle Janney . . . . . 6 pts - A little glitch in your notation but good job
Patrik Petersson. . . . . 6 pts - Another good answer - wait until next week!
Dave Seaman . . . . . . . 5 pts - You want harder problems? Take Calculus!
Sue B. . . . . . . . . . . . . . . 4 pts - You got it. You want honey on that fruit?
Al B. . . . . . . . . . . . . . . . 3 pts - Some pears might exist elsewhere in the world?
Carl Main. . . . . . . . . . . 3 pts - Best answer - linear algebra + logical soundness
Alix Erie . . . . . . . . . . . 2 pts - Thanks for joining my contest! Well-explained!
Rebecca Wooten. . . . .2 pts - I guess bags of fruit is as good as bowls to go!
Kavita Aggarwal . . . . 1 pt - 4A+2B ok, I think you meant 6A+2C not 2B.
Shinsuke Kuzuya. . . . 1 pt - 2A + 2B leaves you with a vitamin deficiency.
T. Jackowski. . . . . . . . 1 pt - Thanks 4 entering; your answer was 1 of em!
 
 
Problem #63 - Posted Saturday, November 13, 1999 (back to top)
Squares Into Primes
Some primes can be expressed as the sum of two squares, as in 13 = 2^2 + 3^2;
some can't: 7 =/= a^2 + b^2. Other primes can be c^2 + 2 e^2, like 11 = 3^2 + 2 * 1^2;
still others are f^2 + 3 g^2 , like 31 = 2^2 + 3 * 3^2. Find all primes less than 100
that can each be written in all three ways: p = a^2 + b^2 = c^2 + 2 e^2 = f^2 + 3 g^2,
and show how it's done with each one (as above with 13, 11, and 31).
The numbers you're squaring must be integers; show your thinking!
Solution: There are only two such primes! (This one leads to algebraic number theory! - Dan)
"I was rather surprised that there were only two," says programmer Tim Nelson,
"I have to admit - I was bored and wrote a program to solve it. My results:
73 is prime and can be constructed using all 3 methods.
73 = 8^2 + 3^2 = 1^2 + 2 * 6^2 = 5^2 + 3 * 4^2
97 is prime and can be constructed using all 3 methods.
97 = 9^2 + 4^2 = 5^2 + 2 * 6^2 = 7^2 + 3 * 4^2
WINNERS - Problem 63 .(back to top) . leader board
Sue B. . . . . . . . . . . . . . . 9 pts - I had to dock you for thinking 1 was prime.
Patrik Petersson. . . . . 6 pts - Saw 12 primes are a^2 + b^2, only two all 3 ways
Al B. . . . . . . . . . . . . . . . 6 pts - Your answer was earlier but also cast 1 as a prime
Tim Nelson . . . . . . . . . 5 pts - Welcome back. So you got a C++ in programming, eh?
Colleen Frye . . . . . . . . 4 pts - You were right, you just didn't believe it. Go Niners!
Joseph Hendricks . . . 3 pts - Neat you found those that can't be done any of the 3 ways.
Rebecca Wooten. . . . .2 pts - You classified each prime but didn't pick these two out...
(Dan's note: One additional bonus point to Sue B, the first person that e-mailed
me with the primes that aren't writable any of our 3 ways: 23, 47, and 71.)
 
 
Problem #64 - Posted Wednesday, November 24, 1999 (back to top)
A Prime Complex
One of our contestants asked me, "What is it with you and primes?" Well, I admit to a certain
'infinity' towards primes, and not just real ones. Let's call m + ni an 'iprime' if it can't be factored
as (a+bi)(c+di), with a,b,c,d integers (not counting "units" 1, -1, i, -i as factors; the symbol i represents the
imaginary square root of -1, so that i^2 = -1.) For example, 2 is not an iprime: (1+i)(1-i) = 1^2-i^2 = 1+1 = 2.
Neither is 7+6i, which is (2+i)(4+i). However, 2+i itself is an iprime, and so is 3, because they can't be factored
into 'smaller' parts. What are all iprimes with positive (integer) real part at most 7 and (integer)
imaginary part between -7 and 7 (inclusive)?
 
Solution: One system is to draw a grid of integer (m,n) points representing m + n i for 0 <= m <= 7 and -7 <= n <= 7.
Then cross out all multiples of small primes like 1 + i, 2 + i, etc. up to the "square root" of the biggest size, 7 + 7 i.
We can measure size by the "norm" N(m + n i) = m^2 + n^2 ; you can prove that N(zw) = N(z) N(w) , and this limits
your crossouts to complex multiples of primes a + b i where N(a + b i) <= Sqrt(7^2 + 7^2) ; meaning a^2 + b^2 < 10;
so after losing all multiples of 1+i (N=2) , 2+i (N=5) , 2-i (N=5) , and 3 (N=9), we're left with these iprimes:
m + n i = 3, 7, 1+i , 2+i , 4+i , 6+i , 3+2i , 5+2i , 5+4i , 6+5i , 7+2i ,
(and related ones like 2-i , 1+2i , and 1-2i ; see brown comment).
You should draw a graph of these using lattice points (a, b) to represent a+bi; there's some nice geometry here.
(Dan's note: Dave saw that, if m+ni is prime, so is m-ni, and Dan says then so are n+mi and n-mi.)
 
WINNERS - Problem 64 .(back to top) . leader board (sorry, nobody really nailed this one)
Sue B. . . . . . . . . . . . . . . 3 pts - Good to try a program but some of your leftovers factored.
Dave Seaman. . . . . . . . 3 pts - Glad you noticed the conjugate rule, but 3+i factors, etc.
Miao . . . =^o^= . . . . . . 2 pts - Some real iprimes are 3, 7, 11, 19, etc. (1 less than 4k).
 
 
Problem #65 - Posted Monday, December 6, 1999 (back to top)
Goat or Sheep?
My friend Charlie sometimes gets confused; 80% of the time he identifies a sheep as
a sheep, and 80% of the time he calls a goat a goat. (Otherwise he thinks it's the other animal.)
In his area, 85% of the animals are sheep and the rest are goats.
Charlie sees a random animal and says it's a goat. What's the probability that he's right?
 
Solution: You might think his chance of being right about anything is 80%, but that's baa-aa-aa-d reasoning.
There are two ways Charlie could claim he saw a goat:
(a) It was a goat and he was right, or (b) it was a sheep and he was wrong.
In every random 100 animals there will be 15 goats, so he'll say 80% of 15, or 12, are goats.
But in the 100 animals there will be 85 sheep, and he'll say 20% of those, or 17, are goats.
So he's called 29 of the 100 animals "goats," and as he's correct in 12 of the 29 calls, his accuracy,
or probability of being right, is just 12/29 or only about 41.4%. Fooled most of you! A popular
answer was 80% of 15% = 12% correct; that leaves out the 17% of incorrectly called sheep.
 
WINNERS - Problem 65 .(back to top) . leader board
Patrik Petersson. . . . 10 pts - Nice explanation and use of Bayes' Law!
Dave Seaman . . . . . . . 7 pts - That's right; you have 12 correct out of 29 called goats.
Sue B . . . . . . . . . . . . . . 4 pts - Table of four events was good, but you need the 17% too!
Big Dave . . . . . . . . . . . 3 pts - Your product makes sense but not the sheep.
Rajwant Singh. . . . . . .3 pts - There are 17 more sheep he called goats.
Phil Sayre . . . . . . . . . . 3 pts - Welcome to the animal farm- I mean contest!
Al B . . . . . . . . . . . . . . . 3 pts - Another one in the 12% trap.
Greg Silin. . . . . . . . . . 3 pts - Nice to hear from you again!
Shinsuke Kuzuya . . . 2 pts - You submitted two answers, and I couldn't find 12/29.
Charles Stanton. . . . . 2 pts - I'm not sure of the subtraction, but there is addition with the mult!
Tom DeSanctis . . . . . 2 pts - Welcome to the dansmath (dysfunctional) family!
 
 
Problem #66 - Posted Thursday, December 16, 1999 (back to top)
 Holiday Cookies?
You decide to bake just two cookies for your holiday party. One is circular, the other is square. The diameter D of the circle and the side S of the square (in inches) are both (positive) integers less than 100. a) What choice of D and S give cookies of closest area, which cookie has more area, and by how much? b) If these closest areas were equal, what would be the implied value of Pi? Include a brief summary of your method and/or steps involved.

 
Solution: Sh--ucks, I meant that the diameter and side were unequal integers, or integers bigger than 1.
But I couldn't change the question once the problem had "leaked" onto my site.
 
So you were right if you said D = 1 and S = 1 was the best answer;
Acirc = Pi * (.5)^2 = Pi/4 :=: 0.7854 , Asqu = 1^2 = 1.000 ; diff :=: 0.2146, but the square has 27.3% more area.
The implied value of Pi is that "Pi / 4 = 1" gives a lousy "Pi = 4"; see below for Egyptian answer.
 
The one I meant was D = 9, S = 8: this gives Acirc = Pi * (4.5)^2 :=: 3.14159 * 20.25 :=: 63.617 , and
Asqu = 8^2 = 64 ; diff :=: 0.383 ; here the square has only 0.6% more area than the circle.
Here, "Pi * (4.5)^2 = 64" gives the "Egyptian" approx. Pi :=: 256/81 :=: 3.1605.
Some other near misses are: (D, S) = (2,2), (8,7), (18,16), (26,23), (35,31), (44,39), (79,70), (88,78).
The bold one, D = 44, S = 39, gives an area diff of only 1.88 sq.in and a Pi estimate of (39/22)^2 :=: 3.14256; not bad!
 
WINNERS - Problem 66 .(back to top) . leader board
Sue B . . . . . . . . . . . . . .10 pts - Good program & answer, with all the runners-up!
Phil Sayre . . . . . . . . . . 7 pts - Nice list of runners up! What's the Python language?
Vvalker Kellogg . . . . 5 pts - Vvelcome; thanks for knowing what I meant to ask too!
Al B . . . . . . . . . . . . . . . 4 pts - Thanks for spelling out your approach; you got it.
Big Dave . . . . . . . . . . . 2 pts - You were a bit off in the circles but D=S=3 is close...
B. M. W. . . . . . . . . . . . 1 pt - There are lots of answers to check; careful with formulas.
 
 
Problem #67 - Posted Tuesday, December 28, 1999 (back to top)

Into Two Thousand!

The number 2000 has 20 divisors, as shown in "Dan's Divisor Chart" at the right. a) What are all (natural) numbers less than 2000 that have at least that many (20) divisors? b) How many divisors does each of those have? c) Which of them has (have) the most divisors? Find as many as you can; partial answers get partial credit. Explain your thinking or give a procedure.

Solution: There are 109 such numbers! The best one of the bunch is 1680 (the number of hours in 10 weeks), with 40 divisors. You can arrange prime powers and divisors as hinted at in the 5 x 4 chart. Here 2000 = 2^4 * 5^3 ; the exponents are 4 and 3, so there are (4+1)*(3+1) = 20 divisors. One way to look for numbers with at least 20 divisors is to make sure the product of 'one more than each exponent' is at least 20. This is a long but do-able process!
 
Another way is to write a program... here are the results according to Sue B (and confirmed by Phil S).
 Number
# of
divisors
  Number
# of
divisors
  Number
# of
divisors
  Number
# of
divisors
  Number
# of
divisors
240
336
360
420
432
480
504
528
540
560
576
600
624
630
648
660
672
720
756
780
792
810
20
20
24
24
20
24
24
20
22
20
21
24
20
24
20
24
24
30
24
24
24
20
816
840
864
880
900
912
924
936
960
990
1008
1020
1040
1050
1056
1080
1092
1104
1120
1134
1140
1152
20
32
24
20
27
20
24
24
28
24
30
24
20
24
24
32
24
20
24
20
24
24
1170
1176
1188
1200
1224
1232
1248
1260
1296
1320
1344
1350
1360
1368
1380
1386
1392
1400
1404
1428
1440
1456
24
24
24
30
24
20
24
36
25
32
28
24
20
24
24
24
20
24
24
24
36
20
1470
1488
1500
1512
1520
1530
1536
1540
1560
1584
1596
1600
1620
1632
1638
1650
1656
1680
1710
1716
1728
1740
24
20
24
30
20
24
20
24
32
30
24
21
30
24
24
24
24
40
24
24
28
24
1760
1764
1776
1782
1800
1820
1824
1836
1840
1848
1860
1872
1890
1904
1920
1932
1944
1950
1960
1968
1980
2000
24
27
20
20
36
24
24
24
20
32
24
30
32
20
32
24
24
24
24
20
36
20
The number with the most divisors (40) is 1680.
See also Problem 10 and Problem 43 for more on divisors.
 
WINNERS - Problem 67 .(back to top) . leader board
Sue B . . . . . . . . . . . . . .10 pts - That's a great discussion about 3D divisor charts!
Phil Sayre . . . . . . . . . . 7 pts - 'Python' squeezes out another solution!
Al B . . . . . . . . . . . . . . . 3 pts - You were correct, there were s'more, but good job!
 
 
Problem #68 - Posted Thursday, January 6, 2000 (we lived!) (back to top)
Powerful Digits The "powers of 2" are 1, 2, 4, 8, ...
a) Which digits (0-9) can be the last digit of a power of 2? Which occur infinitely often?
b) Which pairs of digits (00-99) can be the last two digits of a power of 2?
. .Which ones occur infinitely often? (Consider 1 as 01, etc.)
c) What's the last digit of 2^34 ? What are the last two digits of 2^345 ?
Explain your thinking and give a procedure.
 
Solution: a) There's a pattern to this; the last digits are 1, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, . . . with a repeating
period of 4 digits; 2, 4, 6, 8 repeat infinitely, 1 does not.
b) The last 2 digits according to my TI-85 (2, Enter, Ans*2, Enter, Enter, Enter, . . . ) are:
01, 02, 04, 08, 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52, 04, 08, . . .
There are two non-repeaters: 01 and 02, and the rest repeat in a cycle of 20.
Gee, there is a pattern of how the last 3 (or n) digits repeat; with more non-repeats and longer cycles!
c) Now 2^34 occurs 32 spots after 2^2, so the last digit is the same, namely 4.
Also, 2^345 is 340 spots after 2^5. Since 340 is a mult of 20, the last digits of 2^345 are 32.
 
WINNERS - Problem 68 .(back to top) . leader board
Sue B . . . . . . . . . . . . . .10 pts - Good discussion of cyclical patterns; new 99/00 leader!
Al B . . . . . . . . . . . . . . . 6 pts - You were correct, there were s'more, but good job!
Phil Sayre. . . . . . . . . . 5 pts - Your answer's a year late; arriving January 2001!
Samuel McArthur. . . 4 pts - Welcome to the contest! You want unsolved problems next time?
 
 
Problem #69 - Posted Sunday, January 16, 2000 (back to top)
The Do-It-Alls!
There's a job opening at dansmath.com for a Webmaster. There are three desirable skills:
Writing, Design, Programming. I received 45 applications; 80% have at least one of these skills.
Twenty know design, 25 can write well, and 21 have programming ability. There are 7 with
writing and design skills only, 9 with just writing and programming, and six who can design
and program but can't write. I will only interview those with all three skills.
How many interviews will there be?
 
Solution: I'll post the reasons later this month (Ok, it's 2003 and I still haven't), but the answer is 4 interviews.
Draw a Venn Diagram and fill it in with variables to solve for the inner amount.
 
WINNERS - Problem 69 .(back to top) . leader board
Sue B . . . . . . . . . . . . . 10 pts - Good job once again; interesting 'grid' method
Phil Sayre . . . . . . . . . .7 pts - Sorry about the delay; I didn't look at next year.
Patrik Petersson . . . . 6 pts - Good use of systems of equations (sorry, Phil was first)
Al B . . . . . . . . . . . . . . . 4 pts - Nice concise answer, could use more details
Dave Seaman. . . . . . . 4 pts - Very persistent at algebra, arencha?
Richard Miller. . . . . . 3 pts - Yes, variables rule! (Constants just sit there.)
TripleVCourtois . . . . 3 pts - Lots of equations, 4 interviews
Shinsuke Kuzuya . . . 3 pts - I'm pretty sure those calculations are right.
Kevin Williams . . . . .3 pts - Nice idea, counting up 'skills' instead of people!
Melina Hidayat. . . . . .3 pts - I'd say this problem is screaming for 3 circles
Max Mojab. . . . . . . . . 2 pts - Get answers in early by e-mail for more pts!
Lorenzo Grespan. . . . 2 pts - Good explanation but too many interviews
Andy Harmon . . . . . . 2 pts - Ok, now I know who you are, do you?
Brian Lin. . . . . . . . . . . 2 pts - Right answer, steps could be clearer
Ray&Dina Leduc. . . . 2 pts - Good work but original percentage was off
Truman Lee . . . . . . . . 2 pts - The answer is 4 - what was the question? ;-}
Jason M. McCargar . 2 pts - Brevity is the soul of wit, but here show steps
Gene Oldfield . . . . . . 1 pt - Yes, it's a venn-er, but what about the numbers?
 
 
Problem #70 - Posted Wednesday, January 26, 2000 (back to top)
"Can" Be Done!
At a carnival game, you see nine paint cans stacked and numbered as shown. You get three throws, and you must knock down one (and only one) can per throw. Your first throw scores the number on that can, the second one counts twice the number on the can, and the third shot counts triple its number. To win a prize you must score exactly 50 points; no more, no less. Describe precisely how you can win the prize.

Solution: You can't knock down a trapped can and expect the can(s) above to drop down magically.

Since the total is 50, which is even, you need either all even cans, or odd cans on 1st and 3rd because 1*odd and 3*odd are odd, and you need zero or two odds to make 50. But knocking down all evens means 10, 8, 10 (10 + 16 + 30 = 56) or 8, 10, 10 (8 + 20 + 30 = 58), which don't work. So try odd, odd, odd: (you can't); maybe odd, even, odd: only 7, 10, 7 (7 + 20 + 14 = 41) or 7, 10, 9 (7 + 20 + 18 = 45).

None of these work, but finally there's 7, 8, 9 (7 + 16 + 27 = 50) -- That's it! %;-}

8 10 7
10 7 9
7 9 8
WINNERS - Problem 70 .(back to top) . leader board . . Ok, I gotta make the problems harder.
Jason M. McCargar .10 pts - You got it; 7-8-9. McCongrats for being first!
Sue B . . . . . . . . . . . . . . 7 pts - Thorough look at all possibilities!
Patrik Petersson . . . . 5 pts - Yes, transatlantic throws knock down cans
Al B . . . . . . . . . . . . . . . 4 pts - Lemme throw em, Al B. Sure to hit em
Joya Guha. . . . . . . . . . 3 pts - Welcome to the contest!
Esther Li . . . . . . . . . . . 2 pts - You too, good to say only one can falls at a time.
Lorenzo Grespan. . . . 2 pts - 50 pts, and a good weekend!
Max Mojab. . . . . . . . . 2 pts - Thanks for spelling out your thinking!
Kevin Williams . . . . .2 pts - Good reasoning, process of elimination
A. Anaya. . . . . . . . . . . 2 pts - You got it; double A.
Melina Hidayat. . . . . .2 pts - Thanks for coming back!
Dave Seaman. . . . . . . 2 pts - Dramatic buildup to the total!
Ileen Huh . . . . . . . . . . 2 pts - Absolutely, 7-8-9.
Andy Harmon . . . . . . 2 pts - Yes, forcing math on carnival customers!
Reyy. . . . . . . . . . . . . . . 2 pts - Good problem-solving skill, nice expl'n.
Cheryl Fredricks . . . . 2 pts - Persistence pays off! How you been?
XCute. . . . . . . . . . . . . . 1 pt - Your 9 would knock over the 7.
L Mac. . . . . . . . . . . . . . 1 pt - Is that your final answer? Wanna throw at a friend?
Kimy Leung. . . . . . . . 1 pt - That 9 would knock over the 7 and 10
 
 THANKS to all of you who have entered, or even just clicked and looked.
My site is now in its fifth season - OVER 25,000 HITS so far! (Not factorial.)
Help it grow by telling your friends, teachers, and family about it.
YOU CAN ALWAYS FIND ME AT dansmath.com - Dan the Man Bach - 3*23*29 A.D.
 
Problem Archives Index
 
Probs & answers . 1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90 . 91-100
Problems only . . . 1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90 . 91-100
Probs & answers . 101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151-160 . 161-170 . 171-180
Problems only . . . 101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151-160 . 161-170 . 171-180
Probs & answers . 181-190 . 191-200 . 201-210 . 211-220 . 221-230 . 231+
Problems only . . . . 181-190 . 191-200 . 201-210 . 211-220 . 221-230 . 231+
 
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