RS Aggarwal Solutions Class 6 Chapter 2 Factors And Multiples Exercise 2D

    

EXERCISE-2D

Question 1
First we find the H.C.F. of 161 and 207.Find the H.C.F. of the numbers in each of the following, using the prime factorization method :
84, 98
Ans.
We have
$\begin{array}{r|r}
2 & 84 \\
\hline 2 & 42 \\
\hline 3 & 21 \\
\hline 7 & 7 \\
\hline & 1
\end{array}$


$\begin{array}{r|r}
2 & 98 \\
\hline 7 & 49 \\
\hline 7 & 7 \\
\hline & 1
\end{array}$

So, $84=2 \times 2 \times 3 \times 7=22 \times 3 \times 7$
$98=2 \times 7 \times 7=2 \times 72$
$\therefore$ H.C.F. $=2 \times 7=14$.


Question 2

170, 238 
Ans.
We have
$\begin{array}{r|r}
2 & 170 \\
\hline 5 & 85 \\
\hline 17 & 17 \\
\hline & 1
\end{array}$

$\begin{array}{r|r}
2 & 238 \\
\hline 7 & 119 \\
\hline 17 & 17 \\
\hline & 1
\end{array}$

So, $170=2 \times 5 \times 17$
$238=2 \times 7 \times 17$
$\therefore$ H.C.F. of 170 and $238=2 \times 17=34$.

Question 3

504, 980
Ans. 
We have

$\begin{array}{l|r}
2 & 504 \\
\hline 2 & 252 \\
\hline 2 & 126 \\
\hline 3 & 63 \\
\hline 3 & 21 \\
\hline 7 & 7 \\
\hline & 1
\end{array}$

$\begin{array}{r|r}
2 & 980 \\
\hline 2 & 490 \\
\hline 5 & 245 \\
\hline 7 & 49 \\
\hline 7 & 7 \\
\hline & 1
\end{array}$

So, $504=2 \times 2 \times 2 \times 3 \times 3 \times 7=2^3 \times 3^2 \times 7$ $980=2 \times 2 \times 5 \times 7 \times 7=2^2 \times 5 \times 7^2$
$\therefore$ H.C.F. of 504 and $980=2 . \times 7$ $=4 \times 7=28$.

Question 4

 72,108,180
Ans. We have

$\begin{array}{l|r}
2 & 72 \\
\hline 2 & 36 \\
\hline 2 & 18 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$

$\begin{array}{l|r}
2 & 108 \\
\hline 2 & 54 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$

$\begin{array}{l|r}
2 & 180 \\
\hline 2 & 90 \\
\hline 3 & 45 \\
\hline 3 & 15 \\
\hline 5 & 5 \\
\hline & 1
\end{array}$

So, $72=2 \times 2 \times 2 \times 3 \times 3=2^3 \times 3^2$
$108=2 \times 2 \times 3 \times 3 \times 3=2^2 \times 3^3$
$180=2 \times 2 \times 3 \times 3 \times 5=2^2 \times 3^2 \times 5$
$\therefore$ H.C.F. of 72,108 ,
$180=22 \times 32$
$=4 \times 9=36$


Question 5

84, 120,138 
Ans. We have

$
\begin{aligned}
& \text { So, } 84=2 \times 2 \times 3 \times 7 \\
& =2^2 \times 3 \times 7 \\
& 120=2 \times 2 \times 2 \times 3 \times 5 \\
& =2^3 \times 3 \times 5 \\
& 138=2 \times 3 \times 23
\end{aligned}
$

$\therefore$ H.C.F. of 84,120 and

$138=2 \times 3=6$

Question 6

106, 159, 371
Ans. 
We have
$\begin{array}{l|r}
2 & 106 \\
\hline 53 & 53 \\
\hline & 1
\end{array}$

$\begin{array}{r|r}
3 & 159 \\
\hline 53 & 53 \\
\hline & 1
\end{array}$

$\begin{array}{r|r}
7 & 371 \\
\hline 53 & 53 \\
\hline & 1
\end{array}$


$\begin{aligned}
& \text { So, } 106=2 \times 53 \\
& 159=3 \times 53 \\
& 371=7 \times 53
\end{aligned}$

$\therefore$ H.C.F. of 106, 159, 371=53

Question 7

272, 425 
Ans. 
We have
$\begin{array}{r|r}
2 & 272 \\
\hline 2 & 136 \\
\hline 2 & 68 \\
\hline 2 & 34 \\
\hline 17 & 17 \\
\hline & 1
\end{array}$

$\begin{array}{r|r}
5 & 425 \\
\hline 5 & 85 \\
\hline 17 & 17 \\
\hline & 1
\end{array}$


So, $272=2 \times 2 \times 2 \times 2 \times 17=2^4 \times 17$
$425=5 \times 5 \times 17$
$=5^2 \times 17$
$\therefore$ H.C.F. of 272 and $425=17$.

Question 8

144, 252, 630
Ans. We have
$\begin{array}{r|r}
2 & 144 \\
\hline 2 & 72 \\
\hline 2 & 36 \\
\hline 2 & 18 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$

$\begin{array}{r|r}
2 & 252 \\
\hline 2 & 126 \\
\hline 3 & 63 \\
\hline 3 & 21 \\
\hline 7 & 7 \\
\hline & 1
\end{array}$

$\begin{array}{l|r}
2 & 630 \\
\hline 3 & 315 \\
\hline 3 & 105 \\
\hline 5 & 35 \\
\hline 7 & 7 \\
\hline & 1
\end{array}$


$
\begin{aligned}
& \text { So, } 144=2 \times 2 \times 2 \times 2 \times 3 \times 3=2^4 \times 3^2 \\
& 252=2 \times 2 \times 3 \times 3 \times 7=2^2 \times 3^2 \times 7 \\
& 630=2 \times 3 \times 3 \times 5 \times 7=2 \times 3^2 \times 5 \times 7 \\
& \therefore \text { H.C.F. of } 144,252 \text { and } 630=2 \times 3^2 \\
& =2 \times 9=18 .
\end{aligned}
$

Question 9

1197, 5320, 4389
Ans. We have


$\begin{array}{r|r}
3 & 1197 \\
\hline 3 & 399 \\
\hline 7 & 133 \\
\hline 19 & 19 \\
\hline & 1
\end{array}$


$\begin{array}{r|r}
2 & 5320 \\
\hline 2 & 2660 \\
\hline 2 & 1330 \\
\hline 5 & 665 \\
\hline 7 & 133 \\
\hline 19 & 19 \\
\hline & 1
\end{array}$

$\begin{array}{r|r}
7 & 4389 \\
\hline 3 & 627 \\
\hline 11 & 209 \\
\hline 19 & 19 \\
\hline & 1
\end{array}$

$
\begin{aligned}
& \text { So, } 1197=3 \times 3 \times 7 \times 19=3^2 \times 7 \times 19 \\
& 5320=2 \times 2 \times 2 \times 5 \times 7 \times 19 \\
& =2^3 \times 5 \times 7 \times 19 \\
& 4389=7 \times 3 \times 11 \times 19 \\
& \therefore \text { H.C.F. of } 1197,5320, \\
& 4389=7 \times 19=133
\end{aligned}
$

Find the H.C.F. of the numbers in each of the following using division method:

Question 10

58 , 70
Ans. By division method, we have :

to be added

$\therefore$ H.C.F. of 58 and 70=2

Question 11


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