+0  
 
+2
132
4
avatar

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=?

Guest Dec 9, 2017
Sort: 

3+0 Answers

 #2
avatar
0

[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.

Guest Dec 9, 2017
 #3
avatar
+1

[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.

Guest Dec 10, 2017
edited by Guest  Dec 10, 2017
 #4
avatar+18827 
+2

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:

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

 

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

 

laugh

heureka  Dec 11, 2017

16 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details