Find out n∈N such that n^2 +n is divisible by 30.


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Question Number 166475 by Rasheed.Sindhi last updated on 20/Feb/22

$F i n d o u t n \in N$ $s u c h t h a t$ $n^{2} + n i s d i v i s i b l e b y 30 .$
Commented by mr W last updated on 20/Feb/22

$n^{2} + n = n (n + 1) = 30 m$ $5 \times 6 \Rightarrow n = 5$ $9 \times 10 \Rightarrow n = 9$ $(30 k - 1) (30 k) \Rightarrow n = 30 k - 1$ $(30 k) (30 k + 1) \Rightarrow n = 30 k$
Commented by mathsmine last updated on 20/Feb/22

$n = 15 w o r c k a l s o$
Commented by Rasheed.Sindhi last updated on 21/Feb/22

$A l s o n = 14, 20, 24$
Commented by Rasheed.Sindhi last updated on 21/Feb/22

$T h a n X S i r s!$
Commented by SANOGO last updated on 21/Feb/22

$m e r c i b i e n$
	Answered by Rasheed.Sindhi last updated on 21/Feb/22

	$n (n + 1) = 30 k$ $∙ 2 ∣ n \land 15 ∣ (n + 1)$ $n = 2 k \Rightarrow 15 ∣ (2 k + 1)$ $2 k + 1 = 15 m$ $n = 2 k = 15 m - 1$ $n = 2 k = 15 m - 1$ $\begin{matrix} n = 2 k = 15 m - 1 \end{matrix}$ $k = \frac{15 m - 1}{2} \in N$ $n = 2 (\frac{15 m - 1}{2} \in N)$ $\begin{array}{cc} m & 1 & 3 & 5 & . . . & 2 a + 1 & . . . \\ n & 14 & 44 & 74 & . . . & \underset{a = 0, 1, 2, . . .}{30 a + 14} & . . . \end{array}$ $∙ 3 ∣ n \land 10 ∣ (n + 1)$ $n = 3 k \Rightarrow 10 ∣ (3 k + 1)$ $3 k + 1 = 10 m$ $n = 3 k = 10 m - 1$ $\begin{matrix} n = 3 k = 10 m - 1 \end{matrix}$ $k = \frac{10 m - 1}{3} \in N$ $n = 3 (\frac{10 m - 1}{3} \in N)$ $\begin{array}{cc} m & 1 & 4 & 7 & 10 & . . . & 1 + 3 a & . . . \\ n & 9 & 39 & 69 & 99 & . . . & \underset{a = 0, 1, 2, . . .}{30 a + 9} & . . . \end{array}$ $∙ 5 ∣ n \land 6 ∣ (n + 1)$ $n = 5 k \Rightarrow 6 ∣ (5 k + 1)$ $5 k + 1 = 6 m$ $n = 5 k = 6 m - 1$ $\begin{matrix} n = 5 k = 6 m - 1 \end{matrix}$ $k = \frac{6 m - 1}{5} \in N$ $n = 5 (\frac{6 m - 1}{5} \in N)$ $\begin{array}{cc} m & 1 & 6 & 11 & . . . & 5 a + 1 & . . . \\ n & 5 & 35 & 65 & . . . & \underset{a = 0, 1, 2, . . .}{30 a + 5} & . . . \end{array}$ $∙ 6 ∣ n \land 5 ∣ (n + 1)$ $n = 6 k \Rightarrow 5 ∣ (6 k + 1)$ $6 k + 1 = 5 m$ $n = 6 k = 5 m - 1$ $\begin{matrix} n = 6 k = 5 m - 1 \end{matrix}$ $k = \frac{5 m - 1}{6} \in N$ $n = 6 (\frac{5 m - 1}{6} \in N)$ $\begin{array}{cc} m & 5 & 11 & 17 & . . . & 6 a + 5 & . . . \\ n & 24 & 54 & 84 & . . . & 30 a + 24 & . . . \end{array}$ $∙ 10 ∣ n \land 3 ∣ (n + 1)$ $n = 10 k \Rightarrow 3 ∣ (10 k + 1)$ $10 k + 1 = 3 m$ $n = 10 k = 3 m - 1$ $\begin{matrix} n = 10 k = 3 m - 1 \end{matrix}$ $k = \frac{3 m - 1}{10} \in N$ $n = 10 (\frac{3 m - 1}{10} \in N)$ $\begin{array}{cc} m & 7 & 17 & 27 & . . . & 10 a + 7 & . . . \\ n & 20 & 50 & 80 & . . . & \underset{a = 0, 1, 2, . . .}{30 a + 20} & . . . \end{array}$ $∙ 15 ∣ n \land 2 ∣ (n + 1)$ $n = 15 k \Rightarrow 2 ∣ (15 k + 1)$ $15 k + 1 = 2 m$ $n = 15 k = 2 m - 1$ $\begin{matrix} n = 15 k = 2 m - 1 \end{matrix}$ $k = \frac{2 m - 1}{15} \in N$ $n = 15 (\frac{2 m - 1}{15} \in N)$ $\begin{array}{cc} m & 8 & 23 & 38 & . . . & 15 a + 8 & . . . \\ n & 15 & 45 & 75 & . . . & \underset{a = 0, 1, 2, . . .}{30 a + 15} & . . . \end{array}$ $∙ 30 ∣ n$ $n = 30 a$ $\begin{matrix} 0 & 30 & 60 & 90 & . . . & 30 a & . . . \end{matrix}$ $∙ 30 ∣ (n + 1)$ $n + 1 = 30 a \Rightarrow n = 30 a - 1 \Rightarrow n = 30 a + 29$ $\begin{matrix} 29 & 59 & 89 & . . . & 3 a + 29 & . . . \end{matrix}$ $n = 30 a, 30 a + 5, 30 a + 9, 30 a + 14,$ $30 a + 15, 30 a + 20, 30 a + 24, 30 a + 29$ $W h e r e a = 0, 1, 2, . . .$
	Commented by Rasheed.Sindhi last updated on 21/Feb/22

	${Y o u}^{'} r e w e l c o m e!$
	Commented by SANOGO last updated on 21/Feb/22

	$m e r c i b i e n$
	Answered by Rasheed.Sindhi last updated on 21/Feb/22

	$n^{2} + n \equiv 0 [30] \Rightarrow n^{2} \equiv - n [30]$ $\Rightarrow n^{2} \equiv 30 - n [30]$ $\begin{array}{ccccccccccccccccccccccccccccccc} n & n^{2} [30] & (30 - n) [30] \\ \underset{―}{0} & 0 & 0 \\ 1 & 1 & 29 \\ 2 & 4 & 28 \\ 3 & 9 & 27 \\ 4 & 16 & 26 \\ \underset{―}{5} & 25 & 25 \\ 6 & 6 & 24 \\ 7 & 19 & 23 \\ 8 & 4 & 22 \\ \underset{―}{9} & 21 & 21 \\ 10 & 10 & 20 \\ 11 & 1 & 19 \\ 12 & 24 & 18 \\ 13 & 19 & 17 \\ \underset{―}{14} & 16 & 16 \\ \underset{―}{15} & 15 & 15 \\ 16 & 19 & 14 \\ 17 & 19 & 13 \\ 18 & 24 & 12 \\ 19 & 1 & 11 \\ \underset{―}{20} & 10 & 10 \\ 21 & 21 & 9 \\ 22 & 4 & 8 \\ 23 & 19 & 7 \\ \underset{―}{24} & 6 & 6 \\ 25 & 25 & 5 \\ 26 & 16 & 4 \\ 27 & 9 & 3 \\ 28 & 4 & 2 \\ \underset{―}{29} & 1 & 1 \end{array}$ $n = 0, 5, 9, 14, 15, 20, 24, 29$ $I n g e n e r a l :$ $30 k, 30 k + 5, 30 k + 9, 30 k + 14,$ $30 k + 15, 30 k + 20, 30 k + 24, 30 k + 29$ $f o r k = 0, 1, 2, . . .$
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