**dansmath contest problems 1-10 - with answers**

**dansmath contest problems 1-10 - with answers**

### problem #1** **- posted friday, november 21, 1997

*the most divisors*

*the most divisors*

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), **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 my supercomposites page; also problem 10. - Dan "the Man" Bach %;-}*

### problem #2 - posted friday, november 28, 1997

**trains versus fly**

**trains versus fly**

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 minutes it goes 2 miles.

The fly flew 2 miles before it was frog food. - Dan %;-}

### problem #3 - posted sunday, december 7, 1997

**unit fraction problems **

**unit fraction problems**

A **unit fraction** is of the form** 1/n** (where n = whole no. ≥ 1.)

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 ?

b) Express 3/23 as the sum of two unit fractions; 3/23 = 1/a + 1/b .

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 %;-}*

### problem #4 - posted thursday, december 18, 1997

**the census taker problem**

**the census taker problem**

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.

Let's make a list. 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**68**

6

**Sum**

74

39

28

23

19

18

22

17

15**1414**

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.**

### problem #5 - posted monday, december 29, 1997

**a tree-o of square root problems**

**a tree-o of square root problems**

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.

a ) If: Sqrt(m) + Sqrt(n) = 13 , and m and n differ by 65, what is the largest possible value of m ?

b ) 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.

c ) What is a really good fraction approximation for Sqrt(17), and why? Generalize your answer if possible to Sqrt(n^2 + 1).

**Solution: **

**a) **m – n = (-/m + -/n)(-/m – -/n) = 65 ; (-/m + -/n) = 13 ;

(13)(-/m – -/n) = 65 ; -/m – -/n = 5 ; 2 -/m = 18 ; -/m = 9 ; **m = 81**.

**b)** 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 .

**c)** -/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, whose derivative is f'(x) = 1/(2-/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/(2n).**

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.

### problem #6 - posted wednesday, january 7, 1998

**new year, more house numbers!**

**new year, more house numbers!**

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!**

### problem #7 - posted friday, january 23, 1998

**patterns and sequences **

**patterns and sequences**

**What number comes next in each sequence? 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; a_n = 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**

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

### problem #8 - posted friday, january 30, 1998

**clinking wine glasses**** **

**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.

a) ** 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 ****C(n,2)** to do each one.

People:Clinks: **4:6, 5:10, 6:15, 7:21, 8:28, 9:36, 10:45**.

b) ** How many people** were there if I heard 903 clinks?

**answer:** We want nC2 = 903 clinks, let's answer (c) first and come back!

c) **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.

problem #9 - posted sunday, february 15, 1998

**strange powers**

**strange powers**

The expression **b ^ n** means b to the power n.

**a) 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**

**b) 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!*

**c) 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 ?

**d) 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 = 131072 – 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**!

### problem #10 - **posted saturday, february 28, 1998**

**odd and abundant? **

**odd and abundant?**

A natural number is **abundant** if its proper divisors (not including itself) add up to more than the number. (12 is abundant: the divisors of 12 are 1, 2, 3, 4, 6, and 12, and 1 + 2 + 3 + 4 + 6 = 16 > 12.)

Note: 6 = 1 + 2 + 3 ; 6 is called **perfect**; 15 is **deficient** (not abundant or perfect): 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

n = 3^6 * 5^4 = 455,625:

sum of proper div = (1+3+...+729)(1+5+...+625) = 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.**