Math Team Problems

KH1031

United States862 Posts

March 19 2008 02:36 GMT

1. (2^1)*(2^2)*(2^3)*...*(2^100)
If the above expression is multiplied out as a number, what is the product of the ones digit and the tens digit of that number?

2. If a random group of teenager is chosen (A Teenager is defined as someone whose age is 13, 14, 15, 16, 17, 18, or 19), and the product of their ages is 705,600. What is the average of their age?

I saw these two questions while working in my school today. Took me a bit to solve. Thought it was interesting and thus sharing with the TL community.

micronesia

United States24752 Posts

March 19 2008 02:37 GMT

Why do you need these solved? Is your team supposed to do it? :o

KH1031

United States862 Posts

March 19 2008 02:39 GMT

Oh...I don't need to have it solved. I solved it and thought it was a good question to share.

I work at a school as a math tutor to the students there.

Muirhead

United States556 Posts

March 19 2008 02:51 GMT

For the first one, we just need to compute 2^5050 modulo 100, which by the Chinese Remainder Theorem is a matter of finding mod 25. By Euler's Theorem, 2^20 is 1 mod 25, so we just need to compute 2^10 modulo 25, which is 24. Using the Chinese Remainder Theorem we conclude that the last two digits are 2 and 4, so the answer is 8.

For the second one, I'm too lazy to factor 705,600...

mikeymoo

Canada7170 Posts

March 19 2008 02:53 GMT

Problem one:
+ Show Spoiler +

.. = 2^(101*50) = 2^5050
If you look at two's, the pattern repeats.
2
4
6
8
1*6 = 6
3*2 = 6
6 *4 = 24
(1) 2*8 = 16
(2) 5*6 = 30
(5) 1*2 = 2
(10) 2*4 = 8
(20) 4*8 = 32
(40) 9*6 = 54
(81) 9*2 = 18
(163) 8*4 = 32
(327) 6*8 = 48
(655) 3*6 = 18
...72
...44
...88
...76
...52
...04
...08
...16

last two digits of 2^22 line up with 2^2
So 2^nmod20 = 2^n (digit wise)
2^5050 = 2^10
2*4 = 8

Not elegant, maybe not right, but brute force wins some of the time when you're tired.

EDIT: lol beat to it.

Anyway. Q2.

+ Show Spoiler +

705600 = 2*352800
= 2*2*176400
= 2*2*2*5*2*5*1764
= 2*2*2*5*2*5*2*882
= 2^5*5^2*2*21^2
= (2^6)(5^2)(3^2)(7^2)

14*14*16*15*15

Avg = 74/5

No calculators allowed, right?

infinity21

Canada6683 Posts

March 19 2008 21:05 GMT

+ Show Spoiler +

1. okay so it's 2^5050 iirc
knowing that 2^10 = 1024, we can reduce it to 24^505 => 4^505 => 2^1010 => 24^101 => 4^101 => 2^202 => 4^21 => 2^42 => 2^8*4 => 256*4 => 24? So 8? ~_~ Too lazy to check lol

2. factoring gives us (hope I factored correctly >_<

705600 = 2^6 * 3^2 * 5^2 * 7^2 which implies that there are two 14 year-old teenagers. Which leaves us with 2^4 * 3^2 * 5^2. We know that there is no 17 or 19 year-olds by default. We also know that there can't be an 18 year-old since that would leave us with a couple of five lying around with no 3 to be matched up with (or is it the other way around? ~_~). So therre are two 15 year-olds and that leaves us with a 16 year-old. Taking the average of their age is
(2*14 + 2*15 + 16) / 5 = 74/5 = 14.8?

KH1031

United States862 Posts

March 19 2008 22:05 GMT

Answers:

1. 2*4=8. 8 is the answer.
I found Muirhead's solution most intriguing. I have never taken number theory classes

2. 14, 14, 15, 15, 16

74/5=14.8

14.8 is the answer.

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