**dan's math@home - problem of the week - archives****Problem Archives**page 1**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 index1- The Most Divisors 2 - Trains versus Fly! 3 - Unit fractions 1/n 4 - The Census Taker 5 - Square root probs 6- Buying house nos. 7 - Name that pattern! 8 - Clink wine glasses 9 - Strange Powers !! 10- Odd & Abundant? - Problem #1
- The Most Divisors (back to top)
- A "divisor" of a positive whole number n is a whole number that divides evenly into n
- (with no remainder).
**What integer from 1 through 1000 has the most divisors?** - To win you must prove it and explain your method.

**Solution:**- A number with a lot of small prime factors will have lots of divisors.
- For example, 12 = 2 * 2 * 3 and 12 has 6 divisors: 1, 2, 3, 4, 6, and 12.
- Since 12 has 6 divisors, we say d(12) = 6, where d is the "divisor function."
- Restricting to 2's and 3's, 2^5 * 3^3 = 864 and d(864) = 24
- Testing out numbers like n = 2^a * 3^b * 5^c we find that
- 2^4 * 3^2 * 5 = 720 and d(720) = 30, also
- 2^6 * 3 * 5 = 960 and d(960) = 28, but if we allow 7's then
- 2^3 * 3 * 5 * 7 = 840 and d(840) = 32.
- This is the answer:
**840 has 32 divisors.** - I call it "
**supercomposite**" because it has more divisors than any smaller number. - The first few supercomposites are:
**2**(2 divisors),**4**(3),**6**(4),**12**(6),**24**(8),**36**(9),**48**(10), and**60**(12).- There is a way to find d(n) from the prime factorization of n; it has to do with the exponents, not the primes, so it pays to use higher powers of smaller primes for lots of divisors. See also problem 10. - Dan "the Man" Bach %;-}

**WINNERS - Problem 1 . . .**(back to top)**. . .**leader board**Jonathan Ediyanto**- 7 pts (correct number, 840; flawed solution/explanation)**Joao Sousa**- 3 pts (answered 960, only 28 divisors)**Sarah DeSanctis**- 1 pt - honorable mention- Problem #2 - Posted Friday, November 28, 1997
- Trains versus Fly (back to top)
**Two trains are 2 miles apart**and are traveling towards each other on the same- track, each train going 30
mph.
**A fly going 60 mph**starts at the nose of one train, - flies toward the other train, and upon reaching the second train immediately turns
- around and flies back towards the first train. The fly buzzes back and forth until all
- three collide.
**How far did the fly fly?** **Solution:**By Joao Sousa, DVC Calculus Student (first correct answer)- The trains will each go 1 mile in 1/30 hr = 2 min, so they collide in 2 min.
- The fly flies 60 mi/hr = 1 mi / min, so in 2 min it goes 2 miles.
- The fly flew 2 miles before is was frog food. - Dan %;-}

**WINNERS - Problem 2 . . .**(back to top)**. . .**leader board**Joao Sousa**- 10 pts (first correct answer, good explanation)**Peter Williams**- 7 pts (Dec.1)**Sarah DeSanctis**- 5 pts - (later that day)**Hoodin Hamidi**- 3 pts - (just over the wire)- Problem #3 - Posted Sunday, December 7, 1997
- Unit Fraction Problems (back to top)
**A unit fraction is of the form 1 / n**(where n = whole no. >= 1.)**#3 a)**Starting with 1 / 1 , then 1 / 2 , 1 / 3 , etc., how many unit fractions- does it take to add up to more than pi ?
**#3 b)**Express 3 / 23 as the sum of two unit fractions; 3 / 23 = 1 / a + 1 / b .**#3 c)**Write 5 / 4 as the sum of distinct unit fractions. Fine print . . .- The winner in part c) will be the one using the smallest number of fractions.
- In case of a tie, the winner is the one with the "smallest biggest denominator."
**Solution:**(By Sarah DeSanctis)- a) 1/1 + 1/2 + . . . + 1/13 = about 3.18,
so it takes
**13 unit fractions**to exceed pi. - b) Try 3/23 - 1/1, 3/23 - 1/2 , etc. until
positive: 3/23 - 1/8 = 1/184 so
**3/23 = 1/8 + 1/184.** - c) Is this a trick question?
**5/4 = 1/1 + 1/4**. That seems too easy. - No, Sarah, it wasn't a trick question, just a mistake in my statement; I meant to say,
- "How do you make 5/4 with unit fractions starting with 1/2 ?"
- Then the best answer is 5/4 = 1/2 + 1/3 + 1/4 + 1/6. - - Dan - %;-}

**WINNERS - Problem 3 . . .**(back to top)**. . .**leader board**Sarah DeSanctis**- 10 pts (first correct answer, good explanation)**Peter Williams**- 6 pts (1 pt penalty for too-brief expl and wrote 3/24 not 3/23))**Joao Sousa**- 2 pts - (partly late, no part c)- Problem #4 - Posted Thursday, December 18, 1997
- The Census Taker Problem (back to top)
- A census-taker rings Mr. Simpson's bell and asks how many children he has.
- "Three daughters," he replies.
- "And how old are they, in whole numbers?" asks the census-taker.
- "Well, I'll tell you
this: the
**product of their ages is 72**, and the**sum of their ages is my house number**." - "But that isn't enough information!" complains the census taker.
- "Okay, my oldest daughter (in years) likes chocolate milk," replies Mr. Simpson.
- With that, the census-taker nods and writes down the three ages.
**How old are the Simpson girls, and how did the census-taker figure it out?**- Include a full explanation!

**Solution:**. . . . . . . . . . . . . . . . . . . . . . . .- Don't assume that only kids like chocolate milk, or that a "daughter" can't be 24 or 36 or even 72.
- To avoid repetition we'll go youngest to oldest. (Think primes and factorizations!)
Youngest 1 1 1 1 1 1 2 2 2 **2****3**3

Next 1 2 3 4 6 8 2 3 4 **6****3**4

Oldest 72 36 24 18 12 9 18 12 9 **6****8**6

Sum 74 39 28 23 19 18 22 17 15 **14****14**13

- The key lies in the statement "But that isn't enough information!".
- Since the census taker can see the house number, if it was anything but 14,
- the issue would be settled. With 2, 6, 6 there would be no "oldest" daughter,
- so the answer has to be :
**The daughters are 3, 3, and 8.**

**WINNERS - Problem 4 . . .**(back to top)**. . .**leader board**Sarah DeSanctis**- 10 pts (nice job, have some chocolate milk on me!)**Joao Sousa**- 2 pts - (too many assumptions made)**Peter Williams**- 1 pt (see, you don't need the house number)- Problem #5 - Posted Monday, December 29, 1997
- A Tree-o of Square Root Problems (back to top)
- Let
**Sqrt(x) or -/x**denote the (positive) square root of x, - as in
**Sqrt(100) = -/100 = 10**. - Also
**x ^ 2**will mean x squared, as in**10 ^ 2 =**10 * 10 =**100**.

**1)**. . If:**Sqrt(m) + Sqrt(n) = 13**, and**m and n differ by 65**,- what is the
**largest possible value of m**? - Solution
**:**m - n = (-/m + -/n)(-/m - -/n) = 65 ; (-/m + -/n) = 13 ; - (13)(-/m - -/n) = 65 ; -/m - -/n = 5 ; 2
-/m = 18 ; -/m = 9 ;
**m = 81**.

**2)**. . Notice that the equation**x^2**-**3 = 0**has a**solution x = Sqrt(3)**.- Find a polynomial equation
in
**x**, with integer coefficients, having **x = Sqrt(3) + Sqrt(5)**as a**solution**.- Solution
**:**x - -/3 = -/5 ; (x - -/3)^2 = (-/5)^2 ; x^2 - 2-/3 x + 3 = 5 ; - x^2 - 2 = 2-/3 x ; (x^2 - 2)^2 = (2-/3 x)^2 ; x^4 - 4 x^2 + 4 = 12 x^2 ;
**x^4 - 16 x^2 + 4 = 0 is our equation.****Number Theory Note: This is called the "minimal polynomial"****of the "degree 4 algebraic number" -/3 + -/5 .**

**3)**. . What is a really good**fraction**approximation for**Sqrt(17)**,- and why? Generalize your answer if possible to Sqrt(n^2 + 1).
- Solution
**:**-/16 = 4 ; -/17 = 4.1231056... :=: 4.125 = 4 + 1/8 =**4 1/8 or 33 / 8**. - This is good beacuse it's really close with a small denominator.
- 4.123 = 4123/1000 is less impressive since it's "inefficient."
- (
**Got Calculus?**Notice that on the graph y = f(x) = -/x, - the slope of the tangent line at x = 16 is m = f ' (16) = 1/(2-/16) = 1/8 = rise * 1 ,
- hence the approx
**-/17 :=: 4 1/8**.) - So, -/17 = -/(4^2 + 1) :=: 4 + 1/(2*4)
**In general, -/(n^2 + 1) :=: n + 1/(2 n).**- For example -/101 :=: 10 + 1/20 :=: 10.05. (Actual -/101 :=: 10.0498756...)

**Number Theory Note: (33/8)^2 :=: 17 --> 33^2 :=: 17 * 8^2 --> 1089 :=: 1088****So {x = 33, y = 8} is a solution to "Pell's Equation" x^2 - 17 y^2 = 1.**

**WINNERS - Problem 5 . . .**back to top**. . .**leader board**Sarah DeSanctis**- 5 pts (1: good job, 2:..., 3: ok but generalize)**Joao Sousa**- 4 pts - (good job on #2)- Problem #6 - Posted Wednesday, January 7, 1998
- New Year, More House Numbers! (back to top)
- The people living on Sesame Street all decide to buy new house numbers,
- so they line up at the store in order of their addresses: 1, 2, 3, . . . .
- If the store has 100 of each digit, what is the first address that
- won't be able to buy its house numbers?

- Solution: The digits are fairly balanced from house #1 to #99, but
- after that the 1's are used more, so the 1,s will rum out first.
- From 1-99: 1, 10, 11, . . . , 19, 21, 31, . . . , 91 (21 1's).
- From 100 - 109 (11 1's). From 110 - 119 (21 1's). So far: 53.
- Then 11 each for 120-129,
130-139, etc: 97 up to 159;
**100**up to**162**. - So #162 gets theirs but
**#163 is the first house not to get its number!**

**WINNERS - Problem 6 . . .**back to top**. . .**leader board**Sarah DeSanctis**- 10 pts (credit for "it's 163")**Joao Sousa**- 5 pts - (answered first, nice counting method, but off by 1. Missed it by that much!)- * The rules are 7 pts for the second
*correct*answer, so I'm being generous. *

- * The rules are 7 pts for the second
- Problem #7 - Posted Friday, January 23, 1998
- Patterns and Sequences (back to top)
**What number comes next? Give reasons!****a) 2, 3, 5, 8, 13, _?_****Answer: 21**(add two numbers to get the next - "Fibonacci Sequence")**b) 2, 3, 5, 7, 11, _?_****Answer: 13**(the next prime number; there are 25 primes under 100 and 168 under 1000.)**c) 3, 3, 5, 4, 4, 3, 5, 5, 4, _?_****Answer: 3**(the number of letters in**"one, two, . . . , ten"**)**d) 1, 3, 7, 15, 31, _?_****Answer: 63**(each one is 1 less than the next power of 2 ; an = 2^n - 1**e) 1, 4, 27, 256, 3125, _?_****Answer: 46,656**(1^1, 2^2, 3^3, . . . , 6^6)**f) 1, 2, 6, 24, 120, 720, _?_****Answer: 5040**(the factorials; 1!, 2!, . . . , 7!, where e.g. 4! = 4*3*2*1 = 24.- n! is the product of the first n natural numbers.
- Or, we could say "times 2, times 3, times 4, etc.")

**g) 1, 2, 6, 30, 210, _?_****Answer: 2310**(the product of the first n-1 primes,- or, "times 2, times 3, times 5, times 7, times 11")

**Answer key: a) 21 . b) 13 . c) 3 . d) 63 . e) 46656 . f) 5040 . g) 2310 .**

**WINNERS - Problem 7 . . .**back to top**. . .**leader board**Joao Sousa**- 6 pts - (Got 4 of 7 sequences perfect - good try on 2 of the other 3.)**Leo Bach**- 1 pt - (For at least asking a good question!)- Problem #8 - Posted Friday, January 30, 1998
- Clinking Wine Glasses
- When I have wine with a few people and we clink glasses and say "salud", I can always
- tell if everyone has "clinked" with everyone else, because I know math! Let's assume
- each person clinks each other person exactly once. If there are 2 people, there is one
- "clink." If there are 3 people, there are 3 clinks.
- 8a. How many clinks are there for 4, 5, 6, . . . 10 people?
**Answer:**Each person clinks with all others, so for 4 people it's 3 + 2 + 1 = 6 clinks.- For 5 it's 4 + 3 + 2 + 1
= 10, and we can use the
**binomial coefficient****nC2**to do each one. People:Clinks:**4:6, 5:10, 6:15, 7:21, 8:28, 9:36, 10:45**. - 8b. How many people were there if I heard 903 clinks?
**Answer:**We want nC2 = 903 clinks, let's answer (c) first and come back!- 8c. What is the formula for the quantity : c(n) = number of clinks for a group of n people ?
**Answer:****c(n)**= nC2 = n(n-1)/2**=****(n^2**-**n) / 2**is "n choose 2",- which can be gotten from Pascal's Triangle or from the arithmetic series
- n-1 + n-2 + . . . + 3 + 2 + 1 = n(n-1)/2 [ there are n/2 pairs that add to n-1 ]. Now to finish #8b...
- nC2 = n(n-1)/2 = 903, (n^2 - n) = 1806, n^2 - n - 1806 = 0, (n - 43)(n + 42) = 0 ,
- n = 43 or -42, so there were
**43 people**at the clinkfest.

**WINNER - Problem 8 . . .**(back to top)**. . .**leader board**Joao Sousa**- 5 pts - (Be more thorough - and 42 was close but no cigar with your wine.)- Problem #9 - Posted Sunday, February 15, 1998
- Strange Powers (back
to top)
- 9a) If 3 ^ a = 4 and 4 ^ b = 8, what is 9 ^ (a - b) ?
**Answer:**9^a = (3^2)^a = 3^(2a) = (3^a)^2 = (4)^2 = 16;- Let's find b: 4^b = (2^2)^b = 2^(2b) = 8 = 2^3 , so 2b = 3 ; b = 3/2.
- So
**9 ^ (a - b)**= (9^a) / (9^b) = 16 / (9^b) = 16 / (9^(3/2)) = 16 / (3^3) =**16/27**. - 9b) Can you find numbers a =/= b such that a ^ b = b ^ a ?
**Answer:**Sure,**4^2 = 2^4 = 16,**but did you know there is a whole curve of (a,b) in the ab-plane with a^b = b^a ?! Use an implicit plot to graph x^y - y^x = 0 ; it goes through any (a,a); also (2,4), (4,2), and a host of other points such as (2.478,3) approx.- Check this :
**2.478^3 :=: 3^2.478**. (The :=: means "approx equal") - An answer of "no" might be considered correct, if you couldn't find the numbers!
- 9c) If a $ b means a ^ b - b ^ a , what is 4 $ 6 ?
**Answer:**4^6 - 6^4 = 4096 - 1296 =**2800**. Note a $ b is often positive if a < b. Is it always?- Is there a simple condition for when a $ b > 0 ?
- 9d) Is 2 $ (3 $ 4) the same as (2 $ 3) $ 4 ?
**Answer: No.**Here's a proof that $ is "not associative" :**2 $ (3 $ 4)**= 2 $ (3^4 - 4^3) = 2 $ (81 - 64) = 2 $ 17 = . . .- . . . = 2^17 - 17^2 = 16384
- 289 =
**130783**, while **(2 $ 3) $ 4**= (2^3 - 3^2) $ 4 = (8 - 9) $ 4 = (-1) $ 4 = . . .- . . . = (-1)^4 - 4^(-1) =
1 - 1/4 =
**3/4**. That's very**different**!

**WINNERS - Problem 9 . . .**(back to top)**. . .**leader board**Andy Murdock**- 8 pts - First answer was 16/27 , not your approx value of 0.593.**Joao Sousa**- 4 pts - Your 9a) worked out to be 0.4103, and 9c) wasn't finished.- Problem #10 - Posted Saturday, February 28, 1998
- Odd and Abundant?
- A natural number is
**abundant**if its proper divisors (not including itself) add up to more - than the number. (For example: 12 is abundant because the divisors of 12 are 1, 2, 3, 4, 6, and 12,
- and 1 + 2 + 3 + 4 + 6 = 16 > 12.)
Notice 6 = 1 + 2 + 3 ;
6 is called
**perfect**. - Also 15 is
**deficient**(not abundant or perfect) because 1 + 3 + 5 = 9 < 15. **What is the smallest odd abundant number?**Prove your answer and spell out your thinking.- Hint: Numbers like 12 = 2 * 2 * 3 . that have lots of small prime factors, tend to be abundant.

**Solution:**The hint means that powers of smaller primes give smaller numbers with lots of divisors that might add up to more than the original number.- Divisors of powers of 3 never add up: 1 + 3 + 9 = 14 < 27, etc. So 3^b isn't enough.
- (Note 14/27 :=: 1/2, the limit. We want this ratio to exceed 1. Good job, Andy.)
- Try 3^b * 5^c: n = 3*5 = 15: 1 + 3 + 5 = 9 < 15; 9/15 = 0.6
- n = 45: 1 + 3 + 5 + 9 + 15 = 33 < 45; 33/45 :=: 0.733
- (For n = 3^6 * 5^4 = 455,625: sum of proper div = 398,008 ; ratio :=: 0.874)
- This seems to fall short of our goal of 1, so we bring in powers of 7.
- n = 3^2 * 5 * 7 = 315; sum of all div = (1+3+9)(1+5)(1+7) = 624,
- sum of proper div = 624 - 315 = 309, just missed!
- n = 3^3 * 5 * 7 = 945; sum of all div = (1+3+9+27)(1+5)(1+7) = 1920,
- sum of proper div = 1920 - 945 = 975 , it's abundant!
- The answer is:
**The smallest odd abundant number is 945.**

**WINNERS - Problem 10 . . .**(back to top)**. . .**leader board**Andy Murdock**- 9 pts - You got it and explained how, but didn't prove it had to be smallest.**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.

**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