The N-series (edited)

Number Theory Level pending

Let $\displaystyle N_{k} (a) = \sum_{i=0}^k a^{i}$ for any positive odd integer $a$ .

For example, $N_{5} (a) = 1 + a + a^{2} + a^{3} + a^{4} + a^{5}$ .

What is the largest power of 2 that can completely divide all numbers of the form

$\large (N_{95}+ N_{63} - N_{31})(a) ?$

Clarifications:

$a$ has to be an odd integer.

The answer is 128.

1 solution

Efren Medallo
May 9, 2015

The expansion of $N_{95} (a)$ is as follows:

$N_{95} (a) = 1 + a + a^{2} + a^{3} + a^{4} + a^{5} ... +a^{94} + a^{95}$

This can be factored out as $(1 + a)(1+a^{2})(1+a^{4})(1+a^{8})(1+a^{16})(1+a^{32}+a^{64})$

$N_{63} (a) - N_{31} (a)$ is equal to $a^{32} (1 + a + a^{2} + a^{3} + a^{4} + ... + a^{31})$ this can be factored out as

$(1+a)(1+a^{2})(1+a^{4})(1+a^{8})(1+a^{16})(a^{32})$

So, in totality, the complete factorization is $(1 + a)(1+a^{2})(1+a^{4})(1+a^{8})(1+a^{16})(1+2a^{32}+a^{64})$ $(1 + a)(1+a^{2})(1+a^{4})(1+a^{8})(1+a^{16})(1+a^{32})^{2}$

There are 7 factors. Since a is known to be odd, then each of these factors are assured to be even. Thus the largest power of two that can divide all numbers of this form is

$2^{7} = \boxed{128}$

There may be possibilities that one factor may contribute two or more factors of 2 (say a = 3, or a= 15), so these numbers can be divided by powers of two greater than 128. However, the number needed is the power of two that can divide all of these numbers, or the greatest common power-of-two factor that can divide all of these numbers.

The last line needs to be justified. You have shown that 128 divides these numbers, but have not clearly shown that 128 is the largest power of 2 that can divide all of these numbers. For example, if $a = 3$ , then the term $(1+a)$ would contribute 2 powers of 2.

Calvin Lin Staff - 6 years, 1 month ago

The N-series (edited)

The answer is 128.

1 solution

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