**dan's math@home - problem of the week - archives****Problem Archives**page 11**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+ .**prob index101 -- The Fab Five ! 102 - Similar but Diff 103 - Weird Numbers 104 Dan's Prime Code 105- Ez Come, Ez Go 106- New Year Phone 107- Geometry Gems 108 - Cover The Cube 109- Based On What? 110 - Family Of Four - Problem #101 - Posted Sunday, November 19, 2000
- The Fab Five ! (back to top)
- A set contains five integers. When distinct elements of this set are added together,
- two at a time, the complete list of different possible sums that result is:
- 637, 669, 794, 915, 919, 951, 1040, 1072, 1197.
- a) Figure out (if possible) the original five integers in the set.
- b) What are the means; of the original set, and the set of sums ?
- c) Given a set of 'sums', do five such numbers always exist ?

**Solution**: From Arthur Morris : Let the integers be x1, x2, x3, x4 and x5 in ascending order.

**a)**Assume the three smallest sums are { x1 + x2 = 637 , x1 + x3 = 669 , x2 + x3 = 794 } .

Solve the 3 simultaneous equations ; { x1 = 256 , x2 = 381 , x3 = 413 } .

There are only 9 different sums given, and the 5 elements in the set produce ten sums so there- are two summations with the same value. These two sums must involve four different elements
- with the sum of the smallest and largest
equal two the sum of the middle terms.

x1 + x4 = x2 + x3 and so x4 = 538. The largest sum must be the sum of the two largest elements, - x5 = 1197 - 538 = 659. The five elements
are
**256, 381, 413, 538 and 659.**The mean of the original set is

b)**449.4**. The mean of the sum set is**910.44**. (some of you put 898.8)

**c)**Five such numbers**do not always exist**, because I would have come up the sum set without - using values 4 through 8 in the original set. (Dan's note: 9 equations in 5 unknowns aren't always consistent!)

**WINNERS - Problem 101 .**(back to top)**.**leader board**Arthur Morris**.**Irina Kolesnik**. . . . . . 7 pts - Nice use of remainders (794 = 2 mod 4 )**Ludwig Deruyck**. . .5 pts - Good answer; I meant the actual 9 sums average.**Sudipta Das**. . . . . . . 4 pts - Clever, let S = repeated sum; S + 8194 = 0 mod 4.**Slanky**. . . . . . . . . . . . 3 pts - Start with the means, get info on the nos, nice strategy!**Lisa Schechner**. . . . 3 pts - Nice idea on c), impossible if all 'sums' were odd!**Phil Sayre**.**Al B**. . . . . . . . . . . . . . 2 pts - c) is true if the sums come from actual numbers, yes.**Montta**. . . . . . . . . . . . 2 pts - I like how you started with estimating a ~ 250, etc.**Kia Davidson**. . . . . . 1 pt - Thanks for trying it; here they are, the mystery numbers!- Problem #102 - Posted Monday, November 27, 2000
- Similar but Different (back to top)
- Drew drew two similar triangles, both with integer sides. Two sides of one triangle
- were the same as two in the other.The other (unmatching) sides differed by a prime.
- a) What are the smallest triangles making this possible? (Give corresp. prime difference.)
- b) What is the smallest prime for which this is possible? (Give corresponding triangles.)
- Answers for a) and b) may or may not be different.

**Solution**(by frequent responder Lisa Schechner):- I'm not sure what the formal definition for "the smallest triangles" would be, so I can't answer exactly.
- Let one triangle be given by its edges: a<b<c. Then there exists k>1 such that the second triangle's edges
- are given by: ka<kb<kc, with ka=b, kb=c. Since k must be rational, we can assume that k=m/n, with
- m, n are relatively prime, m < [(1+sqrt(5))/2]
n and n>1. (Dan's note
: this is the Golden Ratio.)

This means that the two triangles should have the form of {n^3,mn^2,nm^2} and {mn^2,nm^2,m^3} - and the prime number should have the
form: m^3 - n^3.

The smallest prime p having the form p= m^3 - n^3, such that n>1 and n, m are rel. prime is - (m=3, n=2). So my conclusion is that
the
**smallest prime = 19**. - This corresponds to the triangles:
**{8,12,18}**and**{12,18,27}**. (Dan's note: 27 - 8 = 19 = 3^3 - 2^3.)

**WINNERS - Problem 102 .**(back to top)**.**leader board**Arthur Morris**.**Lisa Schechner**. . . . 7 pts - Also if m^3 - n^3 = (m-n)(m^2 + mn + n^2) = prime then m-n = 1.**Phil Sayre**.**Jesse Goldlink**. . . . . 4 pts - Not 234, 468; not 124, 248, but yes! 8.12.18, 12.18.27 !**Al B**. . . . . . . . . . . . . . 3 pts - I bumped you up a point because 3 < 19 and 2,3,4 ~ 3,4,5 almost.**Sudipta Das**. . . . . . . 2 pts - Pretty good but 1, 2, 4 can't form a triangle because 1 + 2 < 4.**Ludwig Deruyck**. . . 2 pts - Yes, if 2 _corresp_ sides are the same the triangles are congruent.**Irina Kolesnik**. . . . . 2 pts - if a1 = p/(k^3 - 1) then no solution if k = integer, but k = 1.5.**Slanky**. . . . . . . . . . . . 2 pts - 2,2,3 isn't similar to 2,3,4 , one being isosceles, but p=2 is small.**Les Billig**(new) . . . . . 2 pts - Welcome to the contest! (Good but same prob with the 1,2,4 'triangle'.)**Montta**. . . . . . . . . . . . 2 pts - For the triangles to be similar they must have the same shape.**Matt Albrecht**(new) . . 2 pts - Welcome! Right, you can answer as long as the problem is still up!- Problem #103 - Posted Friday, December 8, 2000
- Weird Numbers! (back to top)
- You may know a 'perfect number' is one whose proper divisors add up to the number,
- such as 6: 1+2+3 = 6. In an 'abundant number' the divisors add up to more than the
- number, like 12: 1+2+3+4+6 = 16 > 12. A 'weird number' is an abundant number with
- no subset of divisors adding to the number. The number 12 isn't weird 'cause 1+2+3+6 = 12.
- What are the first two weird numbers?

**Solution**: The first 2 weird numbers are**70**and**836**. You can try to find a slightly abundant- number with a divisor sum that exceeds the number by a non-divisor, then look more closely.
- Thanks to Stanley Benkoski of West Valley College in Calif. for his 'weird' concept; see Lisa's score comment.
- A few of you (including me) wrote programs or did a search; but I think Phil put it best when
- describing his program that spit out abundant candidates to be submitted to manual recounts:
- "This is a prize example of the difference between machine and human intelligence."
- According to Lisa, the late great Paul Erdos once offered a (deficient) 10 dollars to anyone
- finding any odd weird numbers, their existence is still in question! (Is the $10 still on the table?)
- (Dan's note: Same with odd perfect numbers; for odd abundant numbers look in my Problem Archives.)
- Lisa also gave the longest list of weird nos (for us weirdos that appreciate them, she says!)
- 70, 836, 4030, 5830, 7192, 7912, 9272, 10430, 10570, 10792, 10990, 11410, 11690, 12110,
- 12530, 12670, 13370, 13510, 13790, 13930, 14770, 15610, 15890, 16030, 16310, 16730,...

**WINNERS - Problem 103 .**(back to top)**.**leader board**Ludwig Deruyck**. . . 10 pts - Good answer, thanks for the mathematica code!**Lisa Schechner**.**Irina Kolesnik**. . . . . 5 pts - You got 'em! How did your program prove there was no subset?**Phil Sayre**.**Arthur Morris**.**Al B**. . . . . . . . . . . . . . 2 pts - Right; 70 is first; the next one is after quite a gap.**Aimee Jacot**(new) . . . 1 pt - Hi, thanks for entering; 2 and 3 are weird-esque but not abundant.- Problem #104 - Posted Sunday, December 17, 2000
- Dan's Prime CodeTM (back to top)
- The first 26 primes (2, 3, 5, ...) can be put in correspondence with the letters A through
- Z, so the 'Prime Code' for the word CAB would be the product 5 * 2 * 3 = 30.
- a) What's the Prime Code for the word BIKER ?
- b) Decode this message... 913 1511191 110618.
- c) What (English) word comes closest to a million?

- Please submit all parts in one message. No proper nouns. Ranked by order received AND best c); results avgd.
- One point penalty for each resubmission of 'improved' answers to c).

**Solution**: The 'alphabet primes' are 2, 3, 5, 7, 11, . . . 97, 101. (There are 25 primes under 100.)- a)
**BIKER**makes a product of 3 * 23 * 31 * 11 * 61 =**1,435,269**(not all that close to a million). - b)
**913**= 11 * 83 = E W ;**1511191**= 11 * 37 * 47 * 79 = E L O V ;**110618**= 2 * 19 * 41 * 71 = A H M T - The message decodes as
**WE LOVE MATH**, which is pretty obvious by now, after over 100 problems! - c) This was a popular activity, thanks for the 'positive' feedback; you've 'primed' me for more hard 'code' facts.
- The best entry (so far!) is
**COLIC**, an acute abdominal pain: 5 * 47 * 37 * 23 * 5 =**999925**(within 75; ouch!) - FFFRY = 999973 (27) is more onomotopaeic than a real word. Next closest (thanks mostly to Phil) include:
**JUNE = 1001341**(1341),**BUOY = 998421**(1579),**MODAL = 998186**(1814),- Also near a million are: SELL = 1008953, POLE = 1013837, AVOID = 1195586, PASTA = 1008484, DON'T = 1004437;
- ERASE, WARY, CHIME, FINED; less English are ADIOS = 1013978, JOSE = 1004531, KOPF = 1003873.
- You can try to make sentences of just these: (Neon door: "June & Jose sell pasta - adios, colic! Don't erase.)

**WINNERS - Problem 104 .**keep sending in good words!**.**(back to top)**.**leader board**Arthur Morris**.**Sue B**. . . . . . . . . . . . . 8 pts - My dictionary has COLIC too - should I operate on it?**Phil Sayre**.**Ludwig Deruyck**. . . 5 pts - Good words: DOOR = 943923 , NEON = 955923; first entry.**Irina Kolesnik**. . . . . 5 pts - Yes, JUNE and BUOY; A-ZINC = 998890 is a vitamin slogan!**Slanky**. . . . . . . . . . . . 4 pts - Got COLIC? (f.y.i. 9131511191110618 = 2*17*7643*60631*579569)**Martin Gritch**(new) . .3 pts - Welcome to a Mathematica user! (FFFRY sounds like something frying!)**Al B**. . . . . . . . . . . . . . 3 pts - Hey Al, I enjoy eating PASTA but I don't SELL or KILL it.**Robert Hussey**. . . . . 2 pts - Hello again! AVOID was good; you slightly missed 'E' in BIKER.- Problem #105 - Posted Monday, December 25, 2000
- Easy Come, Easy Go! (back to top)
- At their traditional end-of-the-millenium poker game, Clifton and Lawrence agree on
- the stakes for each hand: The loser pays 1/3 of the money he has remaining, to the winner.
- After a while, Lawrence gives up: "You now have exactly three times the cash I have,
- you've won the last few hands, and I've lost just about four bucks!"
- "But you won every hand before that," Clifton replied, "in fact you've won the same
- number of hands I have!" How many hands were played, how much money did they
- each start with, and how much do they have now?

**Solution**: From Arthur Morris - Thanks again!!- Let Lawrence's holding at the end be
x pennies, Clifton's at 3x and the length of each streak at N.

At the start of Lawrence's losing streak, he had (1.5 x)^N. (1.5 = reciprocal of 2/3 = 1 - 1/3 ; Dan) - N<4 or else Lawrence would have had more than their combined holdings.
- If "few" >1. then N=2 or N=3. For N=2, Lawrence has 2.25 x at the start of his losing streak,
- and has 0.0625 at the start of the
game and is, in fact a winner. Thus N=3.

Working backward, Larry had 3.375 x (= (27/8) x ) at the start of his losing streak, and 1.890625 x - at the start of the game. He lost 0.890625 x ( = (399/448) x ). We must find x such that the loss is an
- integer and about 400 pennies ( 448 x = 399 ). In working backward, I used a factor or 1/2 a total of
- 6 times, so will try some multiple of 2^6. I find x = (7 times 64) = 448 to solve the problem.
- They played
**6 hands**.**Clifton**started with**$9.45**and ended with**$13.44**.

**Lawrence**(don't call me Larry) started with**$8.47**and ended with**$4.48**. **Dan's Note:**I made a table to show how much each player had at each stage, after 0, 1, . . . , 6 hands.- Their
**total**always equals**$17.92**, and Lawrence lost $3.99, about four bucks. Check that 14.48 = 3 * 4.48. - Some of you sent in solutions in which L. had lost exactly 4 dollars, but the 1/3 payoff isn't exact then.
- Some close answers: L: 8.49 -> 4.49, C: 9.47 -> 13.47 (Lisa and Phil); Lo: 8.50 (Sue), 8.55 (Al).
After -->

0 hands

1 hand

2 hands

3 hands

4 hands

5 hands

6 hands

**Clifton****9.45**6.30

4.20

2.80

7.84

11.20

**13.44****Lawrence****8.47**11.62

13.72

15.12

10.08

6.72

**4.48**

**WINNERS - Problem 105 .**(besides Clifton!)**.**(back to top)**.**leader board**Arthur Morris**.**Lisa Schechner**.**Phil Sayre**.**Ludwig Deruyck**. . . 3 pts - I like your use of tables but Lawrence's loss is over last 3 games.**Al B**. . . . . . . . . . . . . . 3 pts - Your total of $18.20 is too rich for this penny poker game!**Sue B**. . . . . . . . . . . . . 3 pts - Very good use of fractions to find divisors of x; it almost worked.**Slanky**. . . . . . . . . . . . 2 pts - Yep, some of those n's give those answers but are they all $4 losses?- Problem #106 - Posted Wednesday, January 3, 2001
- New Year, New Phone! (back to top)
- After my recent move, I got a new phone number. I forgot to write it down, but it had a
- 3-digit prefix and the rest was another 4 digits, like this: xxx-xxxx. I did remember that
- the prefix, subtracted from half the square of the rest of the number, gave me the whole
- phone number as a result ! What was my new number ? (Explain steps fully for best ranking!)

**Solution -**It's fine to write a program to square all the 4-digit numbers and test 'em out with an exhaustive- search, and better to set up an equation.
- Let's call the number N = abc-defg
; put
**x**= abc and**y**= defg; then N = 10000 x + y. - The condition is that y^2 - x = N =
10000 x + y , so y^2 - 2 y = 20002 x ;
**(y - 1)^2 = 20002 x + 1.** - Thus y - 1 = +1 or -1 mod 20002 ; since 20002 = 73 * 274 we have y - 1 = + 1 or -1 mod 73 and mod 274.
- The only interesting solution is y
= 2 mod 73 and y = 0 mod 274 ;
**y = 2192**by the Chinese Remainder Theorem. - Then (y-1)^2 = 2191^2 = 4800481 = 20002
x + 1 ;
**x = 240**; My phone number is**240-2192**(don't call!) - -- Nikita, Sue, and others noticed 000-0000 works; my sources say 000-0002 is OK2. -- ALL OF YOU GOT 240-2192 !!

**WINNERS - Problem 106 . .**(back to top)**.**leader board . . A few of you mentioned your 'language of choice'!**Arthur Morris**.**Ludwig Deruyck**. . . . . 7 pts - Got y^2 / 2 - x = 1000x + y ; used BASIC program to search for whole # x.**Lisa Schechner**.**Sue B**. . . . . . . . . . . . . . . 5 pts - Did a great 'guess/check/adjust' approach, backed up by algebra!**Phil Sayre**.**Sudipta Das**. . . . . . . . . 4 pts - Excellent answer, good use of primes 10001 = 73 * 137. (Bonus pt!)**Al B**. . . . . . . . . . . . . . . . 3 pts - Good to bound it : 1416 < xxxx < 4472 , cuts down the search!**Nikita Kuznetsov**(new) 2 pts - Welcome! 000-0000 ok too but you get an annoyed operator! Visual C++**Paul Langford**(new) . . . 1 pt - Got your partial answer as a YAHOO greeting card! Good unconventionality!- Problem #107 - Posted Friday, January 12, 2001
- Gee, Geometry Gems ! (back to top)
- a) Given that two of the three sides of a right triangle are 3 and 4, what's the shortest possible
- length for the third side? b) If (6,9) and (10,3) are the coordinates of two opposite vertices of
- a square, what are the coords of the other two? c) One circle has radius 5 and center at (0, 5).
- A second circle has radius 12, center (12,0). Find the radius and center of a third circle which
- passes through the center of the 2nd circle and both intersection pts of the first two circles.

**Solution:**I'm bach after a short absence...**a) Slanky sez:**What fun remembering my Geometry for this weeks problem! Since it's a right triangle, the- sum of the squares of the two shorter sides equals the square of the longest side. If I want the shortest possible
- third side, I will assume that 4 is the longest side and solve the equation a^2 + 3^2 = 4^2. a^2 = 7, so
- a =
**square root of 7 =**approx=**2.65**. (Dan's note: Sqrt[7] is 'the' answer, but just for fun, Mathematica gives us - 2.64575131106459059050161575363926042571025918308245 . . .)
**b) Sue B**(contest winner 1999/2000)**sez:**Find the midpoint of the two points (6, 9) and (10, 3) using the- midpoint formula. midpoint: x = (6 + 10) / 2 = 8 ; y = (9 + 3) / 2 = 6 ; midpoint = (8 , 6)
- Find the slope of the line using the two given points. m
= (9 - 3) / (6 - 10) = 6 / (-4) = - 3/2

The line perpendicular to this one has a slope of 2/3. Two points on this second line that are equidistant - from the midpoint are
**(5, 4) and (11, 8)**. 8 + 3 = 11 , 6 + 2 = 8 ; 8 - 3 = 5 , 6 - 2 = 4 . - Use the distance formula to check that the distances between all four outside points are the same, verifying
- that the object is indeed a square. (Dan's note: The distance from any corner to the center (8, 6) is Sqrt[13].)
**c) Phil Sayre sez:**c) Let f(x,y,r)=x*x+y*y-r*r and let (x1,y1) be the intersection of f(x,y-5,5)=0 and

f(x-12,y,12)=0. The difference of these two equations yields the relation y=12*x/5, from which we obtain- x1=24*25/13*13, y1=10*12*12/13*13. Three chords of the desired circle are formed by the three points
- (0,0), (12,0), and (x1,y1) . The perpendicular bisectors of each chord meet at the center of the circle.
- In particular, the x-coordinate, x0=12/2=6. Then from f(x0,y0,r)=0 we obtain y0=sqrt(r*r-36).
- From f(x0-x1,y0-y1,r)=0 we obtain, after substitutions for x0, y0,
- r^2-36={[x1*(x1-12)+y1*y1]/2*y1}^2=(5/2)^2, hence r^2=42.25, r=36. From this result,
- y0=sqrt(42.25-36)=2.5. Summary:
**radius = 6.5, center = (6, 2.5)**. (Dan's note: What can I add?!)

**WINNERS - Problem 107 .**(back to top) . leader board . The center and all intersec's were 'rational points' - Dan**Arthur Morris**.**Ludwig Deruyck**. . . . . 7 pts - You got the (600,169, 1440,169) intersec pt like I did. Good taste!**Phil Sayre**.**Lisa Schechner**.**Al B**. . . . . . . . . . . . . . . . 4 pts - Wow, a 12-step square process and arctangents of 5/12 and 3/2 !**Sudipta Das**. . . . . . . . . 3 pts - I liked your use of fractions and eqns of circles; part a) was a bit off.**Sue B**. . . . . . . . . . . . . . . 3 pts - Your certificate for 1999/2000 has been authenticated and is on its way!**Slanky**. . . . . . . . . . . . . . 3 pts - How's it feel to have your work quoted on a worldide basis!?**Conrad Tan**(new) . . . . . . 2 pts - Yep, go sqrt[52]/2 out from (8, 6) with slope 2/3. That's the spot!**Daria Eiteneer**(new) . . . 2 pts - Welcome to the contest, you two! The radius of the circle is smaller.- Problem #108 - Posted Sunday, January 21, 2001
- Cover The Cube ! (back to top)
- The T-shape at the right can cover the six faces of a cube. How many other shapes can you find that cover the cube? Please give your answers as lists of the six squares used. The T-shape in the picture would be called {a5, b2, b3, b4, b5, c5}. Rotations and reflections (flips) don't count as different. Shapes must be connected; squares must touch along a whole edge.

**Solution:**These are the**eleven**(including the T-shape) that I'm aware of: Arthur sent them all in; Phil provided some personal names; other names and graphics courtesy dansmath.com (me).- (Note: This is one of my
**all-time favorite**problems; I've been saving it for you!)

- 1. "Tee-shape" . . . {a4,
b1, b2, b3, b4, c4} (pictured
above right) . . .
**There are ten others:**2. "High F" . . . . . {a4, b1, b2, b3, b4, c3} . . . . 3. "Low F". . . . {a4, b1, b2, b3, b4, c2}

4. "Backwds S" . . {a4, b1, b2, b3, b4, c1} . . . . 5. "Cross" . . . . {a3, b1, b2, b3, b4, c3}

6. "Saguaro cactus"{a3, b1, b2, b3, b4, c2} . . . .7. "Skinny" . . . {a3, a4, a5, b1, b2, b3}

8. "Scary stairs" . . {a3, a4, b1, b2, b3, c1} . . . . 9. "Saguaro 2" .{a3, a4, b1, b2, b3, c2}

10."High five" . . . {a3, a4, b1, b2, b3, c3} . . . .11. "Staircase" . {a3, a4, b2, b3, c1, c2}

**WINNERS - Problem 108 .**(back to top) . leader board .**Arthur Morris**.**Sue B**. . . . . . . . . . . . . . . 6 pts - Got all 10, but one was an illegal L and another was a repeat.**Lisa Schechner**.**Phil Sayre**.**Al B**. . . . . . . . . . . . . . . . 4 pts - You found 8 of the extra 10; good going, no repeats.**Richard Clarke**(new). . . 3 pts - Six shapes, two repeats, welcome to the contest!**Ludwig Deruyck**. . . . . 3 pts - Four of your 5 were different; 2 were the same (hey, that makes 6 !?)**Bruce**(new) . . . . . . . . . . 2 pts - Three shapes, all good & different ; welcome to the contest!**Edwin Wijono**(new) . . . 2 pts - Welcome; also three (other) shapes; keep on entering, all you math-heads!- Problem #109 - Posted Tuesday, January 30, 2001
- Based On What?! Here's a fun rhyme I found; see if you can answer it! (back to top)
- The square of nine is 121 ; I know it looks quite weird.
- But still I say it's really true ; the way we figure here.
- And nine times ten is 132 ; the self-same rule, you see.
- So whatcha say I'd have to write for five times twenty-three?

**Solution**(s)**:**More than one of you said the title was a clue. I knew only method**(1)**, for which- Phil S. added his own rhyme (see below) but Lisa, Po, and others showed me another
**(2)**, and - Chris North and Sue B convinced me
**(3)**there was yet a third alternative! **(1)**Let's explore different bases: Say the base is n. Place values are 1, n, n^2, etc going right to left.- Then "The square of nine is
**121**" means 9^2 = 81 base 10 =**1***n^2 +**2***n +**1***1. Solving for n ; - n^2 + 2n + 1 = 81 ==> (n + 1)^2 = 81 ==> n + 1 = 9
==>
**n = 8.**(Ah, octal, from my home country!) - Checking , "nine times ten" is 90 base 10 , but
90 =
**1***64 +**3***8 +**2***1 , so it's**132**base 8. This - makes it clear: "five times twenty-three"
is 115 base 10 = 1*64 + 6*8 + 3*1 =
**163**written in octal. **(2)**Lisa suggested that in Joe the Contractor's math (2) you add 2 feet to every dimension, so if you- want a 9 x 9 ft room, it takes up (9+2)(9+2) = 121 sq ft of space. Also 9 x 10 would 'equal up' to
- (9+2)(10+2) = 11 * 12 = 132 ; therefore
5 x 23 would be done as 7 * 25 =
**175**in Joe's Math. **(3)**In (1) we multiplied decimal numbers, then converted to octal; let's convert first then view the- answer as decimal: 9 x 9 -> 11 *
11 = 121 ; 9 x 10 -> 11 * 12 = 132 ; 5 x 23 -> 5 * 27 =
**135**.

**WINNERS - Problem 109 .**(back to top) . leader board .**Arthur Morris**.**Lisa Schechner**.**Ludwig Deruyck**. . . . 5 pts - I appreciate how you explicitly solved for the base B. 163 rules!**Phil Sayre**.**Sue B**. . . . . . . . . . . . . . 4 pts - Thanks for makin' the octal/decimal vs decimal/octal distinction!**Chris North**. . . . . . . . 3 pts - I didn't support your 135 at first but I view it as a ful partner now!**Po-Nien Chen**. . . . . . 3 pts - Another supporter of the "add two and multiply" theory. Good!**Al B**. . . . . . . . . . . . . . . 3 pts - "Add two to everything" except the final answer; it works!**Sudipta Das**. . . . . . . . 3 pts - Good -175- you also solved for the base before proceeding.**Filipe Gonçalves**. . . . 2 pts - Nice to have you back; you must have added twos to get the 175.**Tat Tsui**(new). . . . . . . . 2 pts - That's the idea; add 2's , but 7 * 23 = 161 is only one two...**Slanky**. . . . . . . . . . . . . 1 pt - Sneaked the answer in under the bell ; see my version of (1).- Problem #110 - Posted Saturday, February 10, 2001
- Family Of Four (There may be more than one answer.) (back to top)
- "Hey, nice looking family!" said Fred, seeing the photo. "I met your younger boy today;
- he said he was nine, and your wife reminded me his brother is older."
- Frank agreed, saying, "It's odd about all our ages. If you total the squares of my age and
- the boys' ages, you get my wife's age times the total of my age and the boys' ages."
- If the ages are all whole numbers, can you figure them out?

**Solution:**From your entries I have thirteen answers, so Lisa's "No I can't and neither can you" is a winner.- Some of you did searches for ages (F, M, S) using Excel, Basic, Python, etc. I feel that mathematically the
- only restrictions should be F, M, S >= 9 ; how old do parents have to be? Older could mean a few days...
- Ludwig noticed the words 'odd about all our ages' and featured the all-odd solution (49, 37, 21).
- Here's a
**list**of submitted answers; Mo is mother's age at birth of oldest son. My source had only answer**j)**. - Red is biologically dubious, brown is outside some 'average age' assumptions of Mo, purp older bro = 9.
- a) (27, 21, 18) Mo=3 ; b) (29, 29, 14) Mo=15 ; c) (36, 27, 9) Mo=18 ; d) (36, 27, 18) Mo=9 ;
- e) (39, 29, 14) Mo=15 ; f) (39, 29, 15) Mo=14 ; g) (49, 37, 16) Mo=19 ; h) (49, 37, 21) Mo=16 ;
- i) (53, 41, 12) Mo=29 ;
**j)**(54, 41, 18) Mo=23 ; k) (54, 41, 23) Mo=18 ; l) (63, 51, 9) Mo=42 ; - m) (64, 49, 24) Mo=25 ; n) (64, 49, 25) Mo=25 ; o) (66, 53, 11) Mo=42 ; p) (74, 59, 15) Mo=44 ;
- q) (90, 75, 12) Mo = 63 (!)
**(Dan's note:**Some others I've found are (16, 13, 12), (18, 15, 15), (23, 17, 11), (25, 19, 15), (45, 39, 39),- (63, 51, 42), (72, 65, 65), (66, 57, 54), (81, 73, 72), (92, 83, 81), (99, 79, 54), etc.
- Don't gimme flack about a zero-year-old mother, please; it's a math problem... My 'AppleWorks' spreadsheet for
- M = (F^2+S^2+9^2)/(F+S+9)
gave whole numbers M at a whopping 49 places for the domain
**9****<= F,S<=****100**; - who's to say how old these
people are? We do assume F,M>=S>=9; this cuts it down to
**45 'legal' answers.)**

**WINNERS - Problem 110 .**(back to top) . leader board .**Lisa Schechner**.**Slanky**. . . . . . . . . . . . . 6 pts - Good use of algebra but soln i) not unique; your x < 60 missed (66,53,11)**Ludwig Deruyck**. . . . 5 pts - Found four solns, featuring the one with the 11-year old. Good 'odd one'!**Arthur Morris**. . . . . . 5 pts - Submitted ten answers assuming M,F>=S+18, M<60, F<75**Al B**. . . . . . . . . . . . . . . 4 pts - Four answers, including the "biologically difficult" (90, 75, 12).**Phil Sayre**.**Sudipta Das**. . . . . . . . 3 pts - The answer you got, (39, 29, 14), was a correct (and popular) one.**Sue B**. . . . . . . . . . . . . . 3 pts - Nine ways of solving your Diophantine equation, 'illogicals' are ok!**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 **- Dan the Man Bach**- 3*23*29 A.D.

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