##### MATHCOUNTS Problems 207 and 208: SOLVE EITHER ONE, SOLVE BOTH AND I'LL BEST IT!

label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5

207) The product of three consecutive odd integers is between 64,000,000 and 65,000,000. What is the greatest of the three integers?

208) What is the 2015th digit after the decimal point in the decimal representation of 1/13 ?
May 3rd, 2015

207) 403

208) 2

Best me please !!!

May 3rd, 2015

can you show me how you solved it please?

May 3rd, 2015

of course. I'm going to provide you the answer on my next post

May 3rd, 2015

okay thanks

May 3rd, 2015

Let's call x the smallest integer, you can write x*(x+1)*(x+2) < 65 000 000 or x < x+1 and x < x+2 and x, x+1 and x+2 are positive numbers so you can write that x^3 < x(x+1)(x+2)<65 000 000

Also x < 65 000 000 ^(1/3) (or cube root of 65 000 000)

x < 402,07. Let's try the smallest odd integer inferior to 402,07 :

401 * 403 * 405 = 65 449 215 (it doesn't work)

Let try with 399 * 401 * 403 = 64 479 597 (it works)

You can also see that with 377 we have a result inferior to 64 000 000, the only good combination is 399 x 401 x 403 and the number you want is 403.

Is this ok for this one ?

May 3rd, 2015

thanks

May 3rd, 2015

Do you know how to solve the second problem ?

May 3rd, 2015

no can you tell me?

May 3rd, 2015

Yes, let's go.

May 3rd, 2015

okay

May 3rd, 2015

1/13 = 0.0769230769230769230 ...

The number 769230 is recurring and there are 6 digits in this number.

2015/6=335,83333

so we have the first 0 after the decimal point, then we could put 335 times 769230. Right now we have 1 + 335 * 6 = 2011 digits after the decimal point.

Then you can write 7 (2012 digit) 6 (2013 digit) 9 (2014 digit) 2 (2015 digit)

Your digit is 2.

Is it enough clear ?

May 3rd, 2015

yes and thanks

May 3rd, 2015

You are welcome.

May 3rd, 2015

do you mind solving 2 or 3 more problems for me?

May 3rd, 2015

I have just solved for you the 205. We have solved 205, 207 and 208. What remains ?

May 3rd, 2015

can you solve 201 and show me the work with the answer please?

May 3rd, 2015

I don't see right now how to solve this one, I'm sorry. For 202, it's 4/17.

May 3rd, 2015

It's okay. Can you show me how you solved 202 please?

May 3rd, 2015

you are seeking for 2 integers x and y as x/y= around 0,236

So 0,236 * y = x. Thus you should multiply 0,236 by an integer (1 then 2 then 3, etc.) until you find another integer. That happens when you multiply 0,236 by 17, you find about 4 so the fraction is 4/17.

I don't believe there is a "pure" method to find this result. You have to do multiple tests.

May 3rd, 2015

okay thanks

May 3rd, 2015

...
May 3rd, 2015
...
May 3rd, 2015
Sep 22nd, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle