dan's math@home - problem of the week - archives
Problem Archives page 9

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+ . prob index

81 Change 4A Dollar
82 - - Solve Or Die !
83:The Tree Amigos
84 Crazy Ma'ticians?
85 - - Total-ly Equal!
86- Number of Clues
87: 2 Balls and a Box
89-Gear's The Thing
90 - Cross Numbers!

Problem #81 - Posted Friday, May 12, 2000
I went into a store to get change for a dollar.
The cashier asked, "How do you want that, 100 pennies, or what?
There are 292 ways involving pennies, nickels, dimes, quarters, or halves."
"Don't give me any pennies," I said. "That'll cut down the number of ways, eh?"
a) Just how many ways does this leave? . b) What are all the ways (with no pennies)?
c) What's the probability I'll get exactly one half-dollar coin in my change?
List them clearly, like "10 n + 5 d," etc. Assume all ways equally likely. Note: penny:1c, nick:5c, dime:10c, quar:25c, half:50c.

Solution: a) There are a total of 40 ways to make change for a dollar without using pennies.
b) Here they are. pasted from Filipe:
20 n , 1 d + 18 n , 2 d + 16 n , 3 d + 14 n , 4 d + 12 n , 5 d + 10 n , 6 d + 8 n , 7 d + 6 n , 8 d + 4 n ,
9 d + 2 n , 10 d , 1 q + 15 n , 1 q + 1 d + 13 n , 1 q + 2 d + 11 n , 1 q + 3 d + 9 n , 1 q + 4 d + 7 n ,
1 q + 5 d + 5 n , 1 q + 6 d + 3 n , 1 q + 7 d + 1 n , 2 q + 10 n , 2 q + 1 d + 8 n , 2 q + 2 d + 6 n ,
2 q + 3 d + 4 n , 2 q + 4 d + 2 n , 2 q + 5 d , 2 q + 5 n , 3 q + 1 d + 3 n , 3 q + 2 d + 1 n , 4 q ,
1 h + 10 n , 1 h + 1 d + 8 n , 1 h + 2 d + 6 n , 1 h + 3 d + 4 n , 1 h + 4 d + 2 n , 1 h + 5 d ,
1 h + 1 q + 5 n , 1 h + 1 q + 1 d + 3 n , 1 h + 1 q + 2 d + 1 n , 1 h + 2 q , 2 h:
c) 29 of the 40 ways use no half-dollars, 1 uses 2 halves (halfs?); 10 (red in list) use one half.
So the chance is 10/40 = 1/4 = 25%. Filipe claims there's a prob of 49/292 = 16.78% if pennies are allowed.

Bonus pt for Tim Nelson for being first to point out my mistake!)

Another note: The number of ways of change for a dollar is the coefficient of x^100 in the expansion of
(1/(1 - x))(1/(1 - x^5))(1/(1 - x^10))(1/(1 - x^25))(1/(1 - x^50)); can you say why?

+1 pt for Phil, who noticed that 1/(1 - x^k) = 1 + x^k + x^2k + ... , so the ways to make x^100 are the
ways to make change for 100 cents, because the exponents add to 100 by using various multiples of the
numbers 1, 5, 10, 25, and 50! (not factorial).

Filipe Gonçalves. . . 10 pts - Thanks for the list- you also got 292 like me.
Arthur Morris. . . . . . 7 pts - That's right; I grade these individually by 'hand'.
Chen Sin Tak . . . . . . 5 pts - You didn't miss any of them; thanks for worrying!
Phil Sayre . . . . . . . . . 4 pts - I like your 'ordered quadruples' notation. and +1 pt bonus ^
Tim Nelson. . . . . . . . 4 pts - Only 12 minutes behind Phil. +1 pt for fixing my error.
Allen Druze . . . . . . . 3 pts - Nice equation-based attack, and it paid off!
Sue B . . . . . . . . . . . . . 3 pts - Organized approach by a fathful entrant! ;-}
Shinsuke Kuzuya. . . 2 pts - Close: 9/39, but you missed one (of the 10). Merci!
Ji-Hong Min . . . . . . . 1 pt - You might be multiple-counting things; good try.

Problem #82 - Posted Thursday, May 25, 2000
. A man has two medications that he has to take: Med A and Med B. He must take one of each every morning and every evening; any deviation from the prescribed amount will result in death. Each pill costs \$1000. The two meds look exactly alike: neither have any writing on them and they weigh exactly the same amount. Once the pills are outside of the bottle, there's no way anyone can tell them apart.
. One morning the man shook one pill out of the Med A bottle, into his hand. He recapped the bottle, and picked up the Med B bottle. When trying to shake one pill out of the Med B bottle, two fell into his hand. The man now has one Med A and two Med B pills in the same hand and can't tell them apart.
. What can the man do? He is too poor to throw away 3 pills at \$1000 each, and if he takes them incorrectly, or not at all, he will die.

Solution: Most of you assumed the pills could be cut in half, ground into a powder, or liquified. Here's how he
can survive: He takes one more pill from bottle A, then cuts all four pills in half; 4 'lefts' and 4 'rights'.
He can now eat the four right halves in the morning and then take what's 'left' at night, so he gets the proper
dose and lives to see another day! This is a simple answer, but many answers are simple after you know them!

Dennis Borris . . . . . .10 pts - Welcome to my contest; you've got a great start!
Arthur Morris. . . . . . 7 pts - Thanks for your frequent entries; right again!
Lisa Schechner. . . . . 5 pts - Hey yo! Another Newyawker; welcome to dnasmath!
Filipe Gonçalves. . . . 4 pts - Yes, I'm fine, now that I've been correctly medicated!
Alison Muñoz. . . . . . 3 pts - Welcome; your first answer is fine but the second is too B.
Allen Druze. . . . . . . . 3 pts - Yes, two separate but equal piles of pills!
Sue B . . . . . . . . . . . . . 3 pts - Right about the 2/3 chance if he takes 2 pills at random.
Marcin Mika. . . . . . . 2 pts - Another correct answer from a new entrant; good job!
Shinsuke Kuzuya. . . 2 pts - Right answer but you'll lose the suit with the drug company.
Tim Nelson. . . . . . . . 2 pts - Your answer was fine, but let's not knock the black market!
Monique S. . . . . . . . .1 pt - Good try but you can't identify the first 3 pills no matter what.
Adam Levy . . . . . . . .1 pt - If he died last night from too much A then how's he alive now?
Phil Sayre. . . . . . . . . 1 pt - Answer was a bit late but I'll give you a point for a good effort!

 Problem #83 - Posted Sunday, June 4, 2000 The Tree Amigos! (back to top) I like fruit trees; I have ten in my big flat backyard. The trees form five straight lines of four trees each! Is this actually possible? If so, how? If not, why not? If yes, give (x,y) coords of your trees, or describe exactly how they are placed. If it's not possible, try to prove or at least explain why not.

Solution: The answer lies in the stars! Well, star. One popular solution was to make a five-pointed star; the 5 outer points and the
5 intersection points represent the ten trees; each tree is used in two different rows, so 5 * 4 / 2 = 10 trees. Phil noticed that drawing
lines in succession that intersect all of the previous lines will do it. A few of you gave coordinates; Arthur found all integers and
wondered what the smallest integers that work might be; good question!
Bonus pt (+1) to Arthur for lowering his own coords: divide all y-coords by 3! (Not factorial)

Chen Sin Tak . . . . . 10 pts - Good answer and well-illustrated in ASCII !
Arthur Morris. . . . . . 7 pts (now 8)- Yep: (6,0), (0,6), (4,6), (8,6), (12,6), (3,9), (9,9), (6,12), (0,18), (12,18).
Al B. . . . . . . . . . . . . . . 5 pts - Geometry and trigonometry - a powerful combination!
Shinsuke Kuzuya. . . 4 pts - Good pair of pictures - did you splurge on "Mac Paint"?
Sue B . . . . . . . . . . . . . 3 pts - Wow, simultaneous linear equations to find coords! ;-}
Lisa Schechner. . . . . 3 pts - Good; the lines can be watering ditches for the trees!
Phil Sayre . . . . . . . . . 1 pt - Answer again late but I'll give you a point 4 being quoted.
Greg Hoggarth . . . . . 1 pt - Also late; ten trees in a row is 'at lest 4' not '4', but herezapoint.

Problem #84 - Posted Sunday, June 11, 2000
All mathematicians are either pure or applied. All mathematicians ('mats') are either sane or insane.
Pure mats always tell the truth about their beliefs; applied mats always lie about their beliefs.
Sane mathematicians' beliefs are correct; insane mathematicians' beliefs are incorrect.
Alice says: "I am insane" and "Charlie is pure." Bob says: "I am pure" and "Dorothy is insane."
Charlie says: "I am applied" and "Bob is applied." Dorothy says: "I am sane" and "Charlie is sane."
Describe each mathematician as Pure or Applied, Sane or Insane; explain your reasoning!

Solution: What I did was to make a table of what each person could be based on each comment, and halve the
possibilities each time. But I'll paste in Tim's solution, althought I also liked Arthur's, Al's and Sue's!

A says: A is insane, C is pure. . . . . . . . B says: B is pure, D is insane.
C says: C is applied, B is applied. . . . . .D says: D is sane, C is sane.

If someone is insane and applied, they: Believe they are sane, and lie about it, and say: I am insane.
. . . . . . . . . . . . . . . . . . . Believe they are pure, and lie about it, and say: I am applied.
If someone is insane and pure, they: . . Believe they are sane, and tell the truth, and say: I am sane.
. . . . . . . . . . . . . . . . . . Believe they are applied, and tell the truth, and say: I am applied.
If someone is sane and applied, they: .. Believe they are sane, and lie about it, and say: I am insane.
. . . . . . . . . . . . . . . . . . Believe they are applied, and lie about it, and say: I am pure.
If someone is sane and pure, they: . . . Believe they are sane, and tell the truth, and say: I am sane.
. . . . . . . . . . . . . . . . . . Believe they are pure, and tell the truth, and say: I am pure.

This leads us to: 1) Only Insane people say they are applied. 2) Only Applied people say they are insane.
So we have a revised table: A is applied, says C is pure. . . . B is pure, says D is insane.
. . . . . . . . . . . . . . . . C is insane, says B is applied. . . . D is pure, says C is sane.

We know C is insane, and D says he is sane. If D is pure, and tells
. . the truth, he must be insane himself. This makes Dorothy insane/pure.
B is telling the truth about D being insane, and since he is pure, Bob is sane/pure.
C says that B is applied. Since we know C is already insane, he must be
. . telling the truth about his delusions, and so Charlie is insane/pure.
A says that C is pure. He's telling the truth, which forces Alice to be insane/applied.

Dan's note: Notice that C and D come out the same, so you can't use 'process of elimination.'

Arthur Morris. . . . .10 pts - You got it; nice and concise way of displaying the info.
Tim Nelson . . . . . . . 7 pts - You've been quoted! (and slightly edited for space)
Sue B. . . . . . . . . . . . 5 pts - Welcome back from the insane! (sanity clause ;-} )
Lisa Schechner. . . . 3 pts - Correcto! Also see above for what 'show reasoning' means.
Filipe Gonçalves. . . 2 pts - That's the spirit; try your best and learn something!
Al B. . . . . . . . . . . . . . 2 pts - Good answer but I wanted both categ's for each one.
Phil Sayre . . . . . . . . 1 pt - I feel for you; going purely insane is worth a point.

Problem #85 - Posted Friday, June 23, 2000
Each icon in the table represents a real number.
Some of the sums are given on the outside.
The total of all five icons is a 2-digit factorial.
It's up to you to find the value of each icon,
and figure out the totals x, y, z, and w.
Show your steps and reasoning; be clear!
Solution: This time I stole (and edited) Marcin's solution, being basically lazy myself...

Replace the icons with a = apple, e = eye, f = folder, d = document, p = play button;
new matrix is:
a e f d 20
p d a f 17
a a f e 35
f p d d x?
y? 19 18 z? w?
 = 20 = 17 = 35 = x || || || || \\ y 19 18 z w
1. a+e+f+d+p=4!=24 since 24 is the only 2-digit factorial... 2. a+e+f+d=20, so p=4, by subtraction.
Taking two other equations we can find the value of e:
a+e+f+d=20 ; a+f+d=13 (subtract p=4 from sides of row 2) ; e must equal 7.
Take a row and a column, a+e+f+d-1 = 20 - 1 = e+d+a+p = 19.
a+7+f+d -1 = p+7+a+4 ; f-1 = 4 (destroy some variables from both sides) ; f=5.
Row 3: 2a+f+e=35 ; 2a+5+7=35 ; 2a=23 ; a=11.5
Now we have 4 variables out of the 5, finding d is trivial, its value is -3.5. Final answer:
a = apple = 11.5 ; e = eye-con = 7 ; f = folder = 5 ; d = document = -3.5 ; p = play button = 4 ;
and the missing totals are : x = 2 ; y = 32 ; z = 5 ; w = 9.5 .
Arthur mentioned he was 'suspicious of me' and wasn't surprised the eqns were dependent! ;+=}

(Nobody got the 10 pts this time; but a couple of you were close!)
Arthur Morris. . . . . 9 pts - That's it,, good! I'd still like to see details in 'addition' to verification
Marcin Mika. . . . . . 7 pts - You were first and got the icons right, but no x, y, z, w.
Sue B . . . . . . . . . . . . 6 pts - You're easily impressed by my pilfered graphics!
Al B. . . . . . . . . . . . . . 5 pts - Everything checks out! (except your inflated value of w.)
Filipe Gonçalves. . . 4 pts - You win the prize for the most equations (10) in 9 variables.
Phil Sayre . . . . . . . . 3 pts - I'm not offended you think my system of equations is 'ugly'. ; *}
Shinsuke Kuzuya. . 2 pts - Thanks for i.d.-ing the Photoshop and QuickTime icons!

Problem #86 - Posted Saturday, July 1, 2000
(Ok, I admit it -- I stole this problem from one of my many e-mail fans. Thanks!)
I have eight integers in my list. Use the clues to find my numbers.
1. The range (statistical sense) of the numbers in my list is 93.
2. One of my numbers is an abundant number between 21 and 25.
3. One of the numbers in my list is a perfect cube less than 10.
4. The arithmetic mean of the numbers in my list is 40.
5. The mode of the list is the sum of the first two perfect numbers.
6. The median of the numbers in my list is 29.5.
7. The largest number in my list is the smallest 3-digit palindrome.
8. The sum of the digits of one of the numbers is its square root.
What are the eight numbers? . . Show your steps and reasoning.

Solution: Stolen from Phil Sayre -- thanks Phil!
Clue 3. A cube less than 10: that would have to be 8=2^3.
Clue 7. The largest number is the smallest 3 digit palindrome: 101
Clue 1. The range, maximum - minimum, is 101 - 8 =93, so that checks out.
Clue 4. The arithmetic mean of 8 numbers is 40, so the sum of the numbers is 8*40=320.
Clue 2. The first few abundant numbers (numbers such that the sum of the divisors less
than the number is greater than the number) are 12, 18, 20, 24, 30, ... hence the abundant
number between 21 and 25 it would have to be 24.
Clue 5. The mode of the list is the sum of the first two perfect numbers, numbers such
that the sum of the proper divisors equal the ,number. The first two are 6 and 28, hence
the mode is 34. Since a mode would not be defined unless at least two numbers in the
list are the same, we assume two number 34s.
Clue 6. The median is 29.5=mean of two successive entries, one of which is the max-
imum of the lower four numbers and one of which is the minimum of the upper four
numbers, e.g., 29 and 30, or 28 and 31, or 27 and 32, or 26 and 33, or 25 and 34. Aha!
We've got it: 25 and 34.
Clue 8. Looking at a table of squares, the only one under 121 such that the sum of its
digits is the root is 81. . . "Summing up"; from least to greatest we have:
8, x, 24, 25, 34, 34, 81, 101. x is a slack variable,
chosen so that the sum of the numbers is 320, that is x=13. There they are!

Arthur Morris. . . . . 10 pts - There you go- followed all the clues, and found the useless one (#3)!
Phil Sayre . . . . . . . . 7 pts - You got the soapbox this week, also second answer in!
Sue B . . . . . . . . . . . . 5 pts - There were 8 clues for 8 numbers and one wasn't needed; hmm!
Al B. . . . . . . . . . . . . . 4 pts - Here's another detective that wasn't baffled by these clues.
Filipe Gonçalves. . 3 pts - Another friendly, and correct, answer; thanks!

Problem #87 - Posted Thursday, July 13, 2000
A perfect cube is inscribed in a big sphere, and then another (smaller) sphere is inscribed in the cube.
a. What is the ratio of the radii of the two spheres?* b. What's the ratio of the volumes of the two spheres?*
c. Which is bigger, the ratio of the volume of the big ball to the vol of the box, or the vol of box to the vol of small ball?
.* Answers to a and b should be more than 1. Show your steps and reasoning; be clear!

Solution: Let R be the radius of the big sphere, and r be the radius of the small sphere. If you look at a main diagonal
of the cube, that's 2 R, while a side of the cube is 2 r. Drawing a cross-section of the sphere and cube, we see that there's
a right triangle with sides r , r -/2 , and R ; using the trusty Pythagorean Theorem we find that R = r -/3 ; so
a) R / r = -/3 ~=~ 1.73205 , and b) V / v = [(4/3) Pi R^3] / [(4/3) Pi r^3] = (R / r)^3 = (-/3)^3 = 3 -/3 ~=~ 5.19615
c) if the cube has side s , then s = 2 r , then V / B = [(4/3) Pi R^3] / [s^3] = [(4/3) Pi R^3] / [(2 r)^3] =
= (4/3) Pi [(R / r)^3] / 8 = (4/3) Pi [3 -/3] / 8 = Pi -/3 / 2 ~=~ 2.72070 , where B = vol of the cube, while
B / v = [s^3] / [(4/3) Pi r^3] = [(2 r)^3] / [(4/3) Pi r^3] = 8 / [(4/3) Pi] = 6 / Pi ~=~ 1.90986 ; so the answer is
the volume of the big ball to the volume of the box. (Checking; 1.90986 * 2.72070 ~=~ 5.19615.)

Arthur Morris. . . . . 9 pts - Good job, some values/details omitted but not enough to lose first place
Sue B. . . . . . . . . . . . . 7 pts - Nice explanation and exact values; watch the order of your ratios!
Phil Sayre. . . . . . . . . 5 pts - Those are right, but throwing numerical evidence in woulda been nice!
Al B. . . . . . . . . . . . . . . 4 pts - Ideal answer including a 3D right triangle.
Lisa Schechner . . . . 3 pts - Good answer; but I never think details are boring; send em in!
Chen Sin Tak . . . . . .3 pts - You do a good job there; keep it up!
Filipe Gonçalves . . 2 pts - Good job but an early mistake sorta mounted up.
Patrik Petersson . . . 2 pts - Welcome back! Slight error there in number c)

Problem #88 - Posted Monday, July 24, 2000
Tenny the Elephant starts along a 10-mile long road and walks one mile per day. Each night the road lengthens by 10
miles, proportionally on each side, so beginning the second day there are 2 miles behind Tenny and 18 miles in front.
At the end of that day he'll have 3 miles behind and 17 miles in front, and that night the road will (proportionally) get
another 10 miles longer. If he walks one mile each day and the road gets 10 miles longer each night, will he ever reach
the end of the road? If not, why not? If so, how many days will it take? Show your steps and reasoning; be clear!

Solution: Well, I'll be a fly on this elephant's trunk, but Tenny DOES reach the end of the road! I thought it might
take some cleverly converging geometric series; one way to think of it is that Tenny goes 1/10 of the way the first day,
then 1/20 of the way the second day, 1/30 the third, and ignore all this 'lengthening' stuff if you're looking at fractions.
So since 1/10 + 1/20 + 1/30 + . . . is a divegent harmonic series, we use enough terms so that the total exceeds 1.

Mult thru by 10 to get that Tenny reaches the end in n days if 1/1 + 1/2 + 1/3 + . . . + 1/n > 10. The left is approx.
equal to the integral (area) of 1/x from 1 to n, so that means ln(n) ~ 10 ; expect n ~ e^10 ~ 22000, an overestimate.

The true n is
12367 days because the sum up to 1/12366 is 9.9999621 and up to 1/12367 it's 10.000043.
So that works out to 33 years, 314 days, depending on when the leap years come. Long time for 10 miles; actually by
this time the road has grown to 123,670 miles long, which is about 5 times around the earth. But that's a problem for the engineers!

Sue B. . . . . . . . . . . . 10 pts - Yep, 12367, and if the road meets at the back there's no end!
Lisa Schechner . . . . 7 pts - Good; Euler's approx 1+1/2+......+1/n ~ log(n) + .5772157
Sam Wilson . . . . . . .5 pts - Hey welcome; a family tradition of good answers continues!
Arthur Morris. . . . . 3 pts - Nice try, but Tenny is just his name, he gets farther than ten percent along
Filipe Gonçalves . . 2 pts - Good job but again, an early mistake sorta mounted up.

Problem #89 - Posted Thursday, August 3, 2000
 Lants Headstrong pedals at the rate of 105 revolutions per minute. His bike chain connects a 53-tooth sprocket on the front (pedal) ring and a 12-tooth sprocket on the back wheel. If the diameter of his (back) wheel is 27 inches, how many times will his left leg push down in a 43-km time trial, and what will his (winning) time be (in minutes and seconds)? Show your steps and reasoning; be clear!

Solution: What about all those different units? From Lisa S:
Answer: It takes about 4519 revolutions to cover 43 km, for a total time of appx. 43 minutes.
Reasoning: A single revolution covers (53/12)xPIx27 inch.(1)
43 km = 43x.6214x63380 inch (2)
Divide 2nd quantity by 1st quantity results in
4519 revolutions.(= left foot pushing)
[ Dan's note: 5280 * 12 = 63360 , but the answer of 4519 was right ]
At a rate of 105 r.p.m total time required is 4519/105= 43 min.
[ Dan's note: More exactly it's about 43 min 2 sec. Thanks Lisa!]

Lisa Schechner . . . . 10 pts - You showed most of the details; time was within 2 seconds!
Filipe Gonçalves. . . 7 pts - Right number of pedal pushes; very close to 43 min.
Sue B . . . . . . . . . . . . . 5 pts - '1 km = .621 mi' was your roundoff error; 2nd answer perfect.
Sam Wilson . . . . . . . 4 pts - Exactly right; 0.999151 km/min. is not the same as 1!
Al B. . . . . . . . . . . . . . . 3 pts - Well-phrased answer, and correct to 1/100 sec!
Phil Sayre. . . . . . . . . 3 pts - Yes, that's 37 mph all right; hope it's not a school zone!
Arthur Morris . . . . . 2 pts - You were first, but some of your conversions were off.
Patrik Petersson . . . 2 pts - Again, your reasonimg looks good but numbers went off

Problem #90 - Posted Sunday, August 13, 2000
 Fit a bunch of the words ONE, TWO, ... , TEN, all connected crossword-style, at most once each, into a 6x6 grid. The sample shown has 'sum 6' because 1 + 2 + 3 = 6. The winner was the one with the largest sum before the deadline; ties will be broken by largest number of numbers, then most total letters in grid, then entry date. Answers, showing sum, and 'row form' using '*' for blank. Row form for entry shown is {******,*TWO**,*H*N**,*R*E**,*E****,*E****}.
Solution: After a rough executive session I've decided to allow all entries, even those with
illegally crossed words and uncomputed totals. I gave each person the maximum total I could
for the connected words. The highest total submitted was 41 by our season winner Sue B.
Al B went for 45 but the 9 and 5 were illegally connected. Good tries by all of you! %;-}

Sue B . . . . . . . . . . . . .10 pts - Total 41: Good strategy; go for the big numbers.
{ **FOUR, **I***, SEVEN*, I*E*I*, X***N*, ***TEN }
Arthur Morris . . . . . 7 pts - Total 40: The top five numbers add up to 5 * 8.
{ *N****, EIGHT*, *N**E*, SEVEN*, I*****, X***** }
Phil Sayre. . . . . . . . . 5 pts - Total 40: Another member of the 6-7-8-9-10 club.
{ *E*SIX, NINE**, *G*V**, *H*E**, *TEN**, ****** }
Lisa Schechner . . . . 4 pts - Total 40; Your 2 was disconnected but I counted the rest.
{ SEVEN*, I**I**, X**G*N, ***H*I, ***TEN, TWO**E }
Filipe Gonçalves. . . 3 pts - Total 32; Your 41 was tied but the 9 was not part of another word.
{ T**T**, EIGHT*, N**R*N, *S*E*I, FIVE*N, *X***E }
Al B. . . . . . . . . . . . . . . 3 pts - Total 31; The best I could do for you was take off the 9 and 5.
{ EIGHT*, ****E*, SEVEN*, I***I*, X***N*, *FIVE* }
Pat Gremban . . . . . . 2 pts - Total 17 after amputating the offending 4. Welcome to the contest!
{ ***TWO, ***H*N, FOUR*E, *S*E**, FIVE**, *X****}

 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
Problems only . . . 1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90

Probs & answers . . . 91-100 . 101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151+
Problems only . . . . . 91-100 . 101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151+

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