Find-out-n-N-such-that-n-2-n-is-divisible-by-30-

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 & \dots & 2 a + 1 & \dots \\ n & 14 & 44 & 74 & \dots & \underset{a = 0, 1, 2, \dots}{30 a + 14} & \dots \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 & \dots & 1 + 3 a & \dots \\ n & 9 & 39 & 69 & 99 & \dots & \underset{a = 0, 1, 2, \dots}{30 a + 9} & \dots \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 & \dots & 5 a + 1 & \dots \\ n & 5 & 35 & 65 & \dots & \underset{a = 0, 1, 2, \dots}{30 a + 5} & \dots \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 & \dots & 6 a + 5 & \dots \\ n & 24 & 54 & 84 & \dots & 30 a + 24 & \dots \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 & \dots & 10 a + 7 & \dots \\ n & 20 & 50 & 80 & \dots & \underset{a = 0, 1, 2, \dots}{30 a + 20} & \dots \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 & \dots & 15 a + 8 & \dots \\ n & 15 & 45 & 75 & \dots & \underset{a = 0, 1, 2, \dots}{30 a + 15} & \dots \end{array}

∙ 30 ∣ n

n = 30 a

\begin{matrix} 0 & 30 & 60 & 90 & \dots & 30 a & \dots \end{matrix}

∙ 30 ∣ (n + 1)

n + 1 = 30 a \Rightarrow n = 30 a - 1 \Rightarrow n = 30 a + 29

\begin{matrix} 29 & 59 & 89 & \dots & 3 a + 29 & \dots \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, \dots

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, \dots