dan's math@home - problem of the week - archives
Problem Archives page 11

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

101 -- 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
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.
b)
The mean of the original set is 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!)

Arthur Morris . . . . 10 pts - Thanks for providing this week's solution!
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 . . . . . . . . . 2 pts - Right on the Fab Five, part c) is indeed doubtful usually.
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
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 p = 19
(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.)

Arthur Morris . . . . 10 pts - Good sys of eqns using k = i/j ; right: 1, 2, 4 not a triangle.
Lisa Schechner . . . . 7 pts - Also if m^3 - n^3 = (m-n)(m^2 + mn + n^2) = prime then m-n = 1.
Phil Sayre. . . . . . . . . 5 pts - Oh, my, yes, a cubic equation! (Devoured by the Python - language.)
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
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,...

Ludwig Deruyck . . . 10 pts - Good answer, thanks for the mathematica code!
Lisa Schechner. . . . . 7 pts - Thanks! I had dinner with someone with an 'Erdos number' of 1!
Irina Kolesnik . . . . . 5 pts - You got 'em! How did your program prove there was no subset?
Phil Sayre . . . . . . . . . 4 pts - Python strikes again, or at least paralyzes the problem.
Arthur Morris . . . . . 3 pts - Good try - 70 works, but 102 = 17 + 34 + 51.
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
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 . . . . . 10 pts - COLIC - a good thing, in this case! (Used MATLAB)
Sue B . . . . . . . . . . . . . 8 pts - My dictionary has COLIC too - should I operate on it?
Phil Sayre . . . . . . . . . 6 pts - 'Python' says: JUNE, MODAL and BUOY are good 'n' close!
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
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 . . . . . 10 pts - I worked through your solution, others can too! (Hints in brown)
Lisa Schechner. . . . . 5 pts - You were a 20th century entry but not exact pennies at each hand.
Phil Sayre . . . . . . . . . 4 pts - Good use of exponentials but I think you rounded to nearest 1c.
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
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. But let's try a numerical argument instead, for maximum style.
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. . . . . . . 10 pts - First in again, good use of equations and MATLAB search for y = 2192.
Ludwig Deruyck . . . . . 7 pts - Got y^2 / 2 - x = 1000x + y ; used BASIC program to search for whole # x.
Lisa Schechner. . . . . . . 6 pts - Good answer involving 137 k where 20002 = 2*73*137. (Bonus pt!)
Sue B . . . . . . . . . . . . . . . 5 pts - Did a great 'guess/check/adjust' approach, backed up by algebra!
Phil Sayre . . . . . . . . . . . 4 pts - Nice approach; b = 1 + sqrt(1 + 20002 a), let PYTHON do the search!
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
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... This was a three-part problem, so let's bring in three experts!
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. . . . . . . 10 pts - The Art Morris juggernaut plows on! First and fastest yet thourough!
Ludwig Deruyck . . . . . 7 pts - You got the (600,169, 1440,169) intersec pt like I did. Good taste!
Phil Sayre . . . . . . . . . . . 5 pts - I'll let our new president check your equations for validiosity.
Lisa Schechner. . . . . . . 4 pts - Correct on all answers, even though rather light on the reasons...
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}

Arthur Morris. . . . . . . 10 pts - Thanks again, allow me to copy and paste your list to save typing!
Sue B . . . . . . . . . . . . . . . 6 pts - Got all 10, but one was an illegal L and another was a repeat.
Lisa Schechner. . . . . . . 5 pts - All yours were good; you just overlooked the staircase one.
Phil Sayre . . . . . . . . . . . 4 pts - Nice names, 9 shapes, original visual system, like chess positions.
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.

Arthur Morris. . . . . . 10 pts - Yes, it's 'doubly weird' all right, no 9 in base 8, just 'nine' = 11.
Lisa Schechner. . . . . . 8 pts - You got the base 8 answer & were first (bonus pt) to show me 175.
Ludwig Deruyck . . . . 5 pts - I appreciate how you explicitly solved for the base B. 163 rules!
Phil Sayre . . . . . . . . . . 4 pts - Your rhyme is self-explanatory; repeated below for posterity!
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
"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.)

Lisa Schechner. . . . . 10 pts - Yep, answer the question that's asked! Two examples prove it's 'no'.
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 . . . . . . . . . . 4 pts - Found the most (13) answers with 27<=F<=75, M>=S+18, 10<=S<=25.
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 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+

Browse the complete problem list, check out the weekly leader board,