By starting with a million and alternatively dividing by 2 and multiplying by 5, Anisha created a sequence of integers that starts 1000000, 500000, 2500000, 1250000, and so on. What is the last integer in her sequence? Express your answer in the form \(a^b\), where a and b are positive integers and a is as small as possible.

Thanks :)

Guest Sep 28, 2019

#1**0 **

I think your sequence DIVERGES and does not converge !!

1,000,000, 500,000, 2,500,000, 1,250,000, 6,250,000, 3,125,000, 15,625,000.....and blows up!!!

Guest Sep 28, 2019

#3**0 **

Make sure that you have stated the question accurately. If you were to start with a 1,000,000 and DIVIDE by 5, then MULTIPLY by 2, then it would CONVERGE to something. But, NOT the way the question is stated !!.

Guest Sep 28, 2019

#4**+1 **

I think what he/she means is: What is the last WHOLE INTEGER ??? If that is what is meant, then:

1,000,000, 500,000, 2,500,000, 1,250,000, 6,250,000, 3,125,000,15,625,000, 7,812,500, 39,062,500, 19,531,250, 97,656,250, 48,828,125, 244,140,625.

**The last WHOLE INTEGER is =244,140,625 =5^12**

Guest Sep 28, 2019

#6**+1 **

By starting with a million and alternatively dividing by 2 and multiplying by 5, Anisha created a sequence of integers that starts 1000000, 500000, 2500000, 1250000, and so on. What is the last integer in her sequence? Express your answer in the form , where a and b are positive integers and a is as small as possible.

You start with \(10^6\) which is \( 5^6 * 2^6 \)

After you have divided this by 2 six times you will not be able to divide it by 2 any more, not if you want an integer answer.

So the last integer term will be \(\frac{5^6 * 2^6}{2^6} * 5^6 = 5^{12}\)

.Melody Oct 5, 2019