Question

als and tools- . . . ving Vehicle Educational learning method Medicine- Electrical products- Communications- Food Habits- Musical Instrument • Space - Agriculture - Periodic cognition- Entertainment devices/things- Fertilizer preparation - 2 Discuss with​

Answer 1

Solution 1:-

We see that,

$\small\longrightarrow 6\equiv-1\pmod{7}$

$\small\longrightarrow 6^{203}\equiv(-1)^{203}=-1\pmod{7}\quad\quad\dots(1)$

and,

$\small\longrightarrow 8\equiv1\pmod{7}$

$\small\longrightarrow 8^{203}\equiv1^{203}=1\pmod{7}\quad\quad\dots(2)$

Adding (1) and (2),

$\small\longrightarrow 6^{203}+8^{203}\equiv-1+1=\mathbf{0}\pmod{7}$

Hence the remainder is 0.

Solution 2:-

Given,

$\small\longrightarrow a=6^{203}+8^{203}$

We need to find the remainder obtained on dividing this number by 49.

Since 6 = 7 - 1 and 8 = 7 + 1,

$\small\longrightarrow a=(7-1)^{203}+(7+1)^{203}$

By binomial expansion,

$\small\displaystyle\longrightarrow a=\sum_{r=0}^{203}\,^{203}C_r\,7^{203-r}\,(-1)^r+\sum_{r=0}^{203}\,^{203}C_r\,7^{203-r}$

$\small\displaystyle\longrightarrow a=\sum_{r=0}^{203}\,^{203}C_r\,7^{203-r}\,\left((-1)^r+1\right)$

We see that for every whole number k,

$\small\text{ \displaystyle\longrightarrow(-1)^r+1=\left\{\begin{array}{ll}0,&r=2k-1\\2,&r=2k\end{array}\right. }$

So we can avoid all terms having odd value of r (since they are zero each) and only consider the terms having even value of r, then the sum becomes,

$\small\displaystyle\longrightarrow a=\sum_{k=0}^{101}\,^{203}C_{2k}\,7^{203-2k}\cdot2$

[The limit changes such that r = 0 for k = 0 and r = 202 for k = 0. Since r = 203 is odd, the term having this value of r is zero.]

$\small\displaystyle\longrightarrow a=2\sum_{k=0}^{101}\,^{203}C_{2k}\,7^2\cdot7^{201-2k}$

$\small\displaystyle\longrightarrow a=2\cdot49\sum_{k=0}^{101}\,^{203}C_{2k}\,7^{201-2k}$

Since the exponent of 7, 201 - 2k ≥ 0 ⇒ k ≤ 100, we take the term having k = 101 out of the sum.

$\small\displaystyle\longrightarrow a=2\cdot49\left[\sum_{k=0}^{100}\,^{203}C_{2k}\,7^{201-2k}+\,^{203}C_{202}\,7^{201-202}\right]$

$\small\displaystyle\longrightarrow a=2\cdot49\left[\sum_{k=0}^{100}\,^{203}C_{2k}\,7^{201-2k}+203\cdot\dfrac{1}{7}\right]$

$\small\displaystyle\longrightarrow a=2\cdot49\left[\sum_{k=0}^{100}\,^{203}C_{2k}\,7^{201-2k}+29\right]$

Taking $\small\displaystyle2\left[\sum_{k=0}^{100}\,^{203}C_{2k}\,7^{201-2k}+29\right]=m,$

$\small\displaystyle\longrightarrow a=49m$

Hence the remainder is 0.

Answer 2

Solution 1:-

We see that,

$\small\longrightarrow 6\equiv-1\pmod{7}$

$\small\longrightarrow 6^{203}\equiv(-1)^{203}=-1\pmod{7}\quad\quad\dots(1)$

and,

$\small\longrightarrow 8\equiv1\pmod{7}$

$\small\longrightarrow 8^{203}\equiv1^{203}=1\pmod{7}\quad\quad\dots(2)$

Adding (1) and (2),

$\small\longrightarrow 6^{203}+8^{203}\equiv-1+1=\mathbf{0}\pmod{7}$

Hence the remainder is 0.

Solution 2:-

Given,

$\small\longrightarrow a=6^{203}+8^{203}$

We need to find the remainder obtained on dividing this number by 49.

Since 6 = 7 - 1 and 8 = 7 + 1,

$\small\longrightarrow a=(7-1)^{203}+(7+1)^{203}$

By binomial expansion,

$\small\displaystyle\longrightarrow a=\sum_{r=0}^{203}\,^{203}C_r\,7^{203-r}\,(-1)^r+\sum_{r=0}^{203}\,^{203}C_r\,7^{203-r}$

$\small\displaystyle\longrightarrow a=\sum_{r=0}^{203}\,^{203}C_r\,7^{203-r}\,\left((-1)^r+1\right)$

We see that for every whole number k,

$\small\text{ \displaystyle\longrightarrow(-1)^r+1=\left\{\begin{array}{ll}0,&r=2k-1\\2,&r=2k\end{array}\right. }$

So we can avoid all terms having odd value of r (since they are zero each) and only consider the terms having even value of r, then the sum becomes,

$\small\displaystyle\longrightarrow a=\sum_{k=0}^{101}\,^{203}C_{2k}\,7^{203-2k}\cdot2$

[The limit changes such that r = 0 for k = 0 and r = 202 for k = 0. Since r = 203 is odd, the term having this value of r is zero.]

$\small\displaystyle\longrightarrow a=2\sum_{k=0}^{101}\,^{203}C_{2k}\,7^2\cdot7^{201-2k}$

$\small\displaystyle\longrightarrow a=2\cdot49\sum_{k=0}^{101}\,^{203}C_{2k}\,7^{201-2k}$

Since the exponent of 7, 201 - 2k ≥ 0 ⇒ k ≤ 100, we take the term having k = 101 out of the sum.

$\small\displaystyle\longrightarrow a=2\cdot49\left[\sum_{k=0}^{100}\,^{203}C_{2k}\,7^{201-2k}+\,^{203}C_{202}\,7^{201-202}\right]$

$\small\displaystyle\longrightarrow a=2\cdot49\left[\sum_{k=0}^{100}\,^{203}C_{2k}\,7^{201-2k}+203\cdot\dfrac{1}{7}\right]$

$\small\displaystyle\longrightarrow a=2\cdot49\left[\sum_{k=0}^{100}\,^{203}C_{2k}\,7^{201-2k}+29\right]$

Taking $\small\displaystyle2\left[\sum_{k=0}^{100}\,^{203}C_{2k}\,7^{201-2k}+29\right]=m,$

$\small\displaystyle\longrightarrow a=49m$

Hence the remainder is 0.