Odd Ones Out

Algebra Level 3

Given the positive integers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \ldots,$ Jeremy--a lover of odd numbers--does the following:

$(*)$ Choose only the numbers at odd positions, i.e, those at the first, third, fifth, seventh... positions, and remove the rest.

So, now he has $1, 3, 5, 7, 9, 11, 13, 15, 17, 19, \ldots.$ But he does $(*)$ again, and this time he has $1, 5, 9, 13, 17, 21, 25, 29, 33, 37, \ldots.$ If Jeremy does $(*)$ 8 more times (thus 10 times in total), what will be the next number after 1 in the sequence?

2^{10}

1 + 2^{10}

1 + 2 \times {10}

1 + 2 + {10}

1 solution

Hana Wehbi
Jun 24, 2017

Every time the numbers are increasing by $2^1, 2^2, 2^3, 2^4, ...$ , doing it ten times, the numbers are gonna increase by $2^{10}$ , so the number after $1$ is going to be $1+2^{10}.$

How do you know that the numbers are always increasing by a power of 2? Does this pattern still hold true after 10 times? Where's your proof? Did you first write all the 1000s of numbers to make this conjecture?

Pi Han Goh - 3 years, 11 months ago

The author's intention is to increase the numbers by power of 2 in each trial, so if he does $10$ times, the numbers are going to increase by $2^{10}$ , thus the number after $1$ in the tenth trial is $1+2^{10}$

Hana Wehbi - 3 years, 11 months ago

Odd Ones Out

1 solution

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