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# The sum of 18 consecutive odd integers is a perfect fifth power of n.

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The sum of 18 consecutive odd integers is a perfect fifth power of n. If p is the smallest first number of the series, then the product np=?

Dec 9, 2017

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[F + L] / 2 x 18 =n^5

[415 + 449] /2 x 18 =6^5

n=6, p=415. So, we have:

6 * 415 =2,490.

Dec 9, 2017
#3
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[F + L]/2 x 18 =n^5

[p + p + 34]/2 x 18 =n^5

18[p + 17] =n^5

18p + 306 =n^5

Iterate: [3^5 - 306] / 18 =No integer solution

[4^5 - 306] / 18 =No integer solution

[5^5 - 306] / 18 = No integer solution

[6^5 -306] / 18 =415 as the first term. So, we have:

n = 6 and p =415. Therefore:

6 x 415 =2,490.

Dec 10, 2017
edited by Guest  Dec 10, 2017
#4
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The sum of 18 consecutive odd integers is a perfect fifth power of n.

If p is the smallest first number of the series,

then the product np=?

arithmetic progression:

$$\begin{array}{|lcll|} \hline p + 0 \cdot 2,~ p + 1 \cdot 2,~ p+ 2 \cdot 2,~ p+ 3 \cdot 2,~\ldots,~ p+ 17 \cdot 2 \\ p,~ p +2,~ p+ 4,~ p+ 6,~ \ldots,~ p+ 34 \\ \hline \end{array}$$

sum of a arithmetic progression:

$$\begin{array}{|rcll|} \hline && a_1 = p \qquad a_m = p+34 \qquad d = 2 \\\\ s_m &=& \left( \frac{a_1+a_m}{2} \right) \cdot m \\ \hline \end{array}$$

$$\begin{array}{rcll} s_{18} &=& \left( \frac{p+(p+34)}{2} \right) \cdot 18 \\ s_{18} &=& \left( \frac{2p+34}{2} \right) \cdot 18 \\ s_{18} &=& (p+17) \cdot 18 \\ s_{18} &=& 18p+17\cdot 18 \quad & | \quad s_{18} = n^5 \\ n^5 &=& 18p+17\cdot 18 \\ n^5 - 17\cdot 18 &=& 18p \quad & | \quad : 18 \\ \frac{n^5}{18} - 17 &=& p \\ p&=& \dfrac{n^5}{18} - 17 \quad & | \quad 18 = 2^1 \cdot 3^2 \\\\ p&=& \dfrac{n^5}{2^1 \cdot 3^2} - 17 \quad & | \quad n = 2^i\cdot 3^j \\\\ p&=& \dfrac{(2^i\cdot 3^j)^5}{2^1 \cdot 3^2} - 17 \\\\ p&=& \dfrac{2^{5i}\cdot 3^{5j}}{2^1 \cdot 3^2} - 17 \\\\ p&=& 2^{5i-1}\cdot 3^{5j-2} - 17 \\\\ p_{\text{min}}&=& 2^{5i-1}\cdot 3^{5j-2} - 17 \quad & | \quad p_{\text{min}} \text{ if }i = 1 \text{ and } j = 1 \\\\ p_{\text{min}}&=& 2^{4}\cdot 3^{3} - 17 \\ p_{\text{min}}&=& 16\cdot 27 - 17 \\ \mathbf{p_{\text{min}}}& \mathbf{=}& \mathbf{415} \\\\ n &=& 2^i\cdot 3^j \quad & | \quad i = 1 \text{ and } j = 1 \\ n &=& 2^1\cdot 3^1 \\ \mathbf{n}& \mathbf{=}& \mathbf{6} \\\\ n\cdot p &=& 6\cdot 415 \\ &=& 2490 \end{array}$$

Dec 11, 2017