Exercise 2.3 - Chapter 2 Numbers and Sequences 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $2.3$
Question $1 .$
Find the least positive value of $x$ such that
(i) $71 \equiv x(\bmod 8)$
(ii) $78+x \equiv 3(\bmod 5)$
(iii) $89 \equiv(x+3)(\bmod 4)$
(iv) $96=\frac{x}{7}(\bmod 5)$
(v) $5 x \equiv 4(\bmod 6)$
Solution:
To find the least value of $x$ such that
(i) $71 \equiv x(\bmod 8)$
$71 \equiv 7(\bmod 8)$
$\therefore \mathrm{x}=7$. $[\because 71-7=64$ which is divisible by 8$]$
(ii) $78+x \equiv 3(\bmod 5)$
$\Rightarrow 78+\mathrm{x}-3=5 \mathrm{n}$ for some integer $\mathrm{n}$.
$75+x=5 n$
$75+x$ is a multiple of 5 .
$75+5=80.80$ is a multiple of 5 .
Therefore, the least value of $x$ must be 5 .
(iii) $89 \equiv(x+3)(\bmod 4)$
$89-(x+3)=4 n$ for some integer $n .$
$86-x=4 n$
$86 x$ is a multiple of 4 .
$\therefore$ The least value of $x$ must be 2 then
$86-2=84$
84 is a multiple of 4 .
$\therefore \mathrm{x}$ value must be 2 .

(iv) $96 \equiv \frac{2}{7}(\bmod 5)$
$96-\frac{x}{7}=5 \mathrm{n}$ for some integer $\mathrm{n}$.
$\frac{672-x}{7}=5 n$ $672-x=35 n .$ $672-x$ is a multiple of $35 .$ $\therefore$ The least value of $x$ must
$\therefore$ The least value of $x$ must be 7 i.e. 665 is a multiple of 35 .
(v) $5 x \equiv 4(\bmod 6)$
$5 \mathrm{x}-4=6 \mathrm{M}$ for some integer $\mathrm{n}$.
$5 x=6 n+4$
$x=\frac{6 n+4}{5}$
When we put $1,6,11, \ldots$ as $n$ values in $x=\frac{6 n+4}{5}$ which is divisible by 5 .
When $\mathrm{n}=1, \mathrm{x}=\frac{10}{5}=2$
When $n=6, x=\frac{36+4}{5}=\frac{40}{5}=8$ and so on.
$\therefore$ The solutions are $2,8,14 \ldots \ldots$
$\therefore$ Least value is 2 .
 

Question 2.
If $x$ is congruent to 13 modulo 17 then $7 x-3$ is congruent to which number modulo $17 ?$
Answer:
Given $x \equiv 13(\bmod 17) \ldots \ldots(1)$
$7 \mathrm{x}-3 \equiv \mathrm{a}(\bmod 17) \ldots \ldots . .(2)$
From (1) we get
$x-13-17$ u (u may be any integes)
$x-13$ is a multiple of 17
$\therefore$ The least value of $x=30$
From (2) we get
$7(30)-3 \equiv \mathrm{a}(\bmod 17)$
$210-3 \equiv a(\bmod 17)$
$207 \equiv \mathrm{a}(\bmod 17)$
$207 \equiv 3(\bmod 17)$
$\therefore$ The value of $a=3$

Question $3 .$
Solve $5 x \equiv 4(\bmod 6)$
Solution:
$\begin{aligned}
&5 x \equiv 4(\bmod 6) \\
&5 x-4=6 M \text { for some integer } n \\
&5 x=6 n+4
\end{aligned}$
$\begin{aligned}
&x=\frac{6 n+4}{5} \text { where } n=1,6,11, \ldots \ldots \\
&\therefore x=2,8,14, \ldots
\end{aligned}$

Question $4 .$
Solve $3 x-2 \equiv 0(\bmod 11)$
Solution:
$3 x-2 \equiv 0(\bmod 11)$
$3 \mathrm{x}-2=11 \mathrm{n}$ for some integer $n$.
$3 x=1 \ln +2$
$x=\frac{11 n+2}{3}$ where $n=2,5,8, \ldots$
$x=\frac{11 \times 2+2}{3}=8$
$\therefore \quad x=\frac{11 \times 5+2}{3}=\frac{55+2}{3}$
$=\frac{57}{3}=19$

$\begin{aligned}
\vec{x} &=\frac{11 \times 8+2}{3}=\frac{88+2}{3} \\
&=\frac{90}{3}=30 . \\
\therefore \quad x &=8,19,30, \ldots .
\end{aligned}$
 

Question $5 .$
What is the time 100 hours after 7 a.m.?
Solution:
$100 \equiv \mathrm{x}(\bmod 12)$ ( $\because 7$ comes in every 12 hrs)
$100 \equiv 4(\bmod 12)(\because$ Least value of $x$ is 4$)$
 

Question 6.
What is time 15 hours before 11 p.m.?
Answer:
$15 \equiv x(\bmod 12)$
$15 \equiv 3(\bmod 12)$
The value of $\mathrm{x}$ must be 3 .

The time 15 hours before 11 o'clock is $(11-3) 8 \mathrm{pm}$

Question 7.
Today is Tuesday. My uncle will come after 45 days. In which day my uncle will be coming?
Solution:
No. of days in a week $=7$ days.
$45 \equiv \mathrm{x}(\bmod 7)$
$45-x=7 n$
$45-\mathrm{x}$ is a multiple of 7 .
$\therefore$ Value of $x$ must be 3 .
$\therefore$ Three days after Tuesday is Friday. Uncle will come on Friday.
 

Question $8 .$
Prove that $2^{\mathrm{n}}+6 \times 9 \mathrm{n}$ is always divisible by 7 for any positive integer $\mathrm{n}$.
Answer:
$9=2(\bmod 7)$
$9^{\mathrm{n}}=2^{\mathrm{n}}(\bmod 7)$ and $2^{\mathrm{n}}=2^{\mathrm{n}}(\bmod 7)$
$2^{\mathrm{n}}+6 \times 9^{\mathrm{n}}=2^{\mathrm{n}}(\bmod 7)+6\left[2^{\mathrm{n}}(\bmod 7)\right]$
$=2^{\mathrm{n}}(\bmod 7)+6 \times 2^{\mathrm{n}}(\bmod 7)$
$7 \times 2^{\mathrm{n}}(\bmod 7)$
It is always divisible for any positive integer $\mathrm{n}$
 

Question $9 .$
Find the remainder when $2^{81}$ is divided by 17 .
Solution:
$2^{81} \equiv x(\bmod 17)$
$2^{40} \times 2^{40} \times 2^{41} \equiv x(\bmod 17)$
$\left(2^{4}\right)^{10} \times\left(2^{4}\right)^{10} \times 2^{1} \equiv x(\bmod 17)$
$(16)^{10} \times(16)^{10} \times 2 \equiv x(\bmod 17)$
$\left(16^{5}\right)^{2} \times\left(16^{5}\right)^{2} \times 2$

$\begin{aligned}
&\left(16^{5}\right) \equiv 16(\bmod 17) \\
&\left(16^{5}\right)^{2} \equiv 16^{2}(\bmod 17) \\
&\left(16^{5}\right)^{2} \equiv 256(\bmod 17) \\
&\equiv 1(\bmod 17)[\because 255 \text { is divisible by } 17] \\
&\left(16^{5}\right)^{2} \times\left(16^{5}\right)^{2} \times 2 \equiv 1 \times 1 \times 2(\bmod 17) \\
&\therefore 2^{81} \equiv 2(\bmod 17) \\
&\therefore x=2
\end{aligned}$
 

Question $10 .$
The duration of flight travel from Chennai to London through British Airlines is approximately 11 hours. The aeroplane begins its journey on Sunday at $23: 30$ hours. If the time at Chennai is four and a half hours ahead to that of London's time, then find the time in London, when will the flight lands at London Airport.
Solution:
The duration of the flight from Chennai to London is 11 hours.
Starting time at Chennai is $23.30 \mathrm{hrs} .=11.30 \mathrm{p} . \mathrm{m}$.
Travelling time $=11.00$ hrs. $=22.30 \mathrm{hrs}=10.30 \mathrm{a} . \mathrm{m}$.
Chennai is $4 \frac{1}{2} \mathrm{hrs}$ ahead to London.
$=10.30-4.30=6.00$
$\therefore$ At 6 a.m. on Monday the flight will reach at London Airport.