Exercise 2.2 - Chapter 2 Numbers and Sequences 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\operatorname{Ex} 2.2$
Question $1 .$
For what values of natural number $n, 4 n$ can end with the digit 6 ?
Solution:
$4^{\mathrm{n}}=(2 \times 2)^{\mathrm{n}}=2^{\mathrm{n}} \times 2^{\mathrm{n}}$
2 is a factor of $4^{n}$.
So, $4^{\mathrm{n}}$ is always even and end with 4 and 6 .
When $\mathrm{n}$ is an even number say $2,4,6,8$ then $4^{\mathrm{n}}$ can end with the digit 6 .
Example:
$\begin{aligned}
&4^{2}=16 \\
&4^{3}=64 \\
&4^{4}=256 \\
&4^{5}=1,024 \\
&4^{6}-4,096 \\
&4^{7}=16,384 \\
&4^{8}=65,536 \\
&4^{9}=262,144
\end{aligned}$

Question $2 .$
If $\mathrm{m}, \mathrm{n}$ are natural numbers, for what values of $\mathrm{m}$, does $2^{\mathrm{n}} \times 5^{\mathrm{m}}$ ends in 5 ?
Answer:
$2^{\mathrm{n}}$ is always even for any values of $\mathrm{n}$.
[Example. $2^{3}-4,2^{3}-8,2^{4}-16 \mathrm{elc}$ ]
$5^{\mathrm{m}}$ is always odd and it ends with 5 .
[Example. $5^{2}=25,5^{3}=125,5^{4}=625 \mathrm{etc}$ ]
But $2^{\mathrm{n}} \times 5^{\mathrm{m}}$ is always even and end in 0 .

[Example. $2^{3} \times 5^{3}=8 \times 125=1000$
$\left.2^{2} \times 5^{2}=4 \times 25=100\right]$
$\therefore 2^{\mathrm{n}} \times 5^{\mathrm{m}}$ cannot end with the digit 5 for any values of $\mathrm{m}$.
 

Question $3 .$
Find the H.C.F. of 252525 and $363636 .$
Solution:
To find the H.C.F. of 252525 and 363636
Using Euclid's Division algorithm
$363636=252525 \times 1+111111$
The remainder $111111 \neq 0$.
$\therefore$ Again by division algorithm
$252525=111111 \times 2+30303$
The remainder $30303 \neq 0$.
$\therefore$ Again by division algorithm.
$111111=30303 \times 3+20202$
The remainder $20202 \neq 0$.
$\therefore$ Again by division algorithm
$30303=20202 \times 1+10101$
The remainder $10101 \neq 0$.
$\therefore$ Again using division algorithm
$20202=10101 \times 2+0$
The remainder is 0 .
$\therefore 10101$ is the H.C.F. of 363636 and 252525 .
 

Question $4 .$
If $13824=2^{a} \times 3^{b}$ then find $a$ and $b$.
Solution:
If $13824=2^{a} \times 3^{b}$
Using the prime factorisation tree

$\begin{aligned}
&13824=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \\
&=2^{9} \times 3^{3}=2^{\mathrm{a}} \times 3^{\mathrm{b}} \\
&\therefore \mathrm{a}=9, \mathrm{~b}=3 .
\end{aligned}$
 

Question $5 .$
If $\mathrm{p}_{1}{ }_{1} \times \mathrm{p}_{2}{ }_{2}^{\mathrm{x}} \times \mathrm{p}_{3}{ }_{3}{ }^{\mathrm{x}} \times \mathrm{p}_{4}{ }_{4}^{\mathrm{x}}=113400$ where $\mathrm{p}_{1}, \mathrm{p}_{2}, \mathrm{p}_{3}, \mathrm{p}_{4}$ are primes in ascending order and $\mathrm{x}_{1}, \mathrm{x}_{2}$, $x_{3}, x_{4}$ are integers, find the value of $P_{1}, P_{2}, P_{3}, P_{4}$ and $x_{1}, x_{2}, x_{3}, x_{4}$.
Solution:
$\text { If } \mathrm{p}_{1}{ }_{1} \times \mathrm{p}_{2} \mathrm{x}_{2} \times \mathrm{p}_{3} \mathrm{x}_{3} \times \mathrm{p}_{4}{ }_{4}^{\mathrm{x}}=113400$
$\mathrm{p}_{1}, \mathrm{p}_{2}, \mathrm{p}_{3}, \mathrm{P}_{4}$ are primes in ascending order, $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \mathrm{x}_{4}$ are integers. using Prime factorisation tree.

$\begin{aligned}
&113400=2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 7 \\
&=23 \times 34 \times 52 \times 7 \\
&=\mathrm{p}_{1} \mathrm{x}_{1} \times \mathrm{p}_{2} \mathrm{x}_{2} \times \mathrm{p}_{3} \mathrm{x}_{3} \times \mathrm{p}_{4} \mathrm{x}_{4} \\
&\therefore \mathrm{p}_{1}=2, \mathrm{p}_{2}=3, \mathrm{p}_{3}=5, \mathrm{p}_{4}=7, \mathrm{x}_{1}=3, \mathrm{x}_{2}=4, \mathrm{x}_{3}=2, \mathrm{x}_{4}=1
\end{aligned}$

Question $6 .$
Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of arithmetic.
Solution:
408 and $170 .$

$\therefore \text { H.C.F. }=2^{1} \times 17^{1}=34$
To find L.C.M, we list all prime factors of 408 and 170 , and their greatest exponents as follows.

$\begin{aligned}
&\therefore \text { L.C.M. }=2^{3} \times 3^{1} \times 5^{1} \times 17^{1} \\
&=2040 .
\end{aligned}$

Question $7 .$
Find the greatest number consisting of 6 digits which is exactly divisible by $24,15,36 ?$
Solution:
To find L.C.M of $24,15,36$

$\therefore$ L.C.M $=2^{3} \times 3^{2} \times 5^{1}$ $=8 \times 9 \times 5$ $=360$ If a number has to be exactly divisible by 24,15 , and 36 , then it has to be divisible by 360 . Greatest 6 digit number is 999999 . Common multiplies of $24,15,36$ with 6 digits are $103680,116640,115520, \ldots 933120,999720$ with six digits. $\therefore$ The greatest number consisting 6 digits which is exactly divisible by $24,15,36$ is 999720 . Question 8 . What is the smallest number that when divided by three numbers such as 35,56 and 91 leaves remainder 7 in each case? Answer: Find the L.C.M of 35,56 , and 91 $35-5 \times 756$ $56=2 \times 2 \times 2 \times 7$ $91=7 \times 13$ L.C.M $=23 \times 5 \times 7 \times 13$ $=3640$ 6 digit number is 999999 .
Common multiplies of $24,15,36$ with 6 digits are $103680,116640,115520, \ldots 933120,999720$ with six digits.
$\therefore$ The greatest number consisting 6 digits which is exactly divisible by $24,15,36$ is 999720 .
 

Question 8.
What is the smallest number that when divided by three numbers such as 35,56 and 91 leaves remainder 7 in each case?
Answer:
Find the L.C.M of 35,56 , and 91
Since it leaves remainder 7
The required number $=3640+7$
$=3647$
The smallest number is $=3647$

Question $9 .$
Find the least number that is divisible by the first ten natural numbers.
Solution:
The least number that is divisible by the first ten natural numbers is 2520 .
Hint:
$1,2,3,4,5,6,7,8,9,10$
The least multiple of 2 \& 4 is 8
The least multiple of 3 is 9
The least multiple of 7 is 7
The least multiple of 5 is 5
$\therefore 5 \times 7 \times 9 \times 8=2520$.
L.C.M is $8 \times 9 \times 7 \times 5$
$=40 \times 63$
$-2520$