Playing With Numbers — Exercise 6(C)

Find the H.C.F. of the following numbers using prime factorisation method :

32, 56

Answer

By prime factorisation method, we get:

$\begin{array}{r|r} 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 56 \\ \hline 2 & 28 \\ \hline 2 & 14 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

So,

32 = 2 × 2 × 2 × 2 × 2;

56 = 2 × 2 × 2 × 7

The common factors are 2, 2 and 2.

∴ H.C.F. = 2 × 2 × 2 = 8

Hence, the H.C.F. of 32 and 56 is 8.

Find the H.C.F. of the following numbers using prime factorisation method :

42, 63

Answer

By prime factorisation method, we get:

$\begin{array}{r|r} 2 & 42 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

So,

42 = 2 × 3 × 7;

63 = 3 × 3 × 7

The common factors are 3 and 7.

∴ H.C.F. = 3 × 7 = 21

Hence, the H.C.F. of 42 and 63 is 21.

Find the H.C.F. of the following numbers using prime factorisation method :

72, 96

Answer

By prime factorisation method, we get:

$\begin{array}{r|r} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So,

72 = 2 × 2 × 2 × 3 × 3;

96 = 2 × 2 × 2 × 2 × 2 × 3

The common factors are 2, 2, 2 and 3.

∴ H.C.F. = 2 × 2 × 2 × 3= 24

Hence, the H.C.F. of 72 and 96 is 24.

Find the H.C.F. of the following numbers using prime factorisation method :

81, 108

Answer

By prime factorisation method, we get:

$\begin{array}{r|r} 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \qquad \text{and} \quad \begin{array}{r|r} 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So,

81 = 3 × 3 × 3 × 3;

108 = 2 × 2 × 3 × 3 × 3

The common factors are 3, 3 and 3.

∴ H.C.F. = 3 × 3 × 3 = 27

Hence, the H.C.F. of 81 and 108 is 27.

Find the H.C.F. of the following numbers using prime factorisation method :

135, 180

Answer

By prime factorisation method, we get:

$\begin{array}{r|r} 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 180 \\ \hline 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So,

135 = 3 × 3 × 3 × 5;

180 = 2 × 2 × 3 × 3 × 5

The common factors are 3, 3 and 5.

∴ H.C.F. = 3 × 3 × 5 = 45

Hence, the H.C.F. of 135 and 180 is 45.

Find the H.C.F. of the following numbers using prime factorisation method :

168, 216

Answer

By prime factorisation method, we get:

$\begin{array}{r|r} 2 & 168 \\ \hline 2 & 84 \\ \hline 2 & 42 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So,

168 = 2 × 2 × 2 × 3 × 7;

216 = 2 × 2 × 2 × 3 × 3 × 3

The common factors are 2, 2, 2, and 3.

∴ H.C.F. = 2 × 2 × 2 × 3 = 24

Hence, the H.C.F. of 168 and 216 is 24.

Find the H.C.F. of the following numbers using prime factorisation method :

144, 180, 198

Answer

By prime factorisation method, we get:

$\begin{array}{r|r} 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 180 \\ \hline 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 198 \\ \hline 3 & 99 \\ \hline 3 & 33 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

So,

144 = 2 × 2 × 2 × 2 × 3 × 3;

180 = 2 × 2 × 3 × 3 × 5;

198 = 2 × 3 × 3 × 11

The common factors are 2, 3 and 3.

∴ H.C.F. = 2 × 3 × 3 = 18

Hence, the H.C.F. of 144, 180 and 198 is 18.

Find the H.C.F. of the following numbers using prime factorisation method :

66, 102, 138

Answer

By prime factorisation method, we get:

$\begin{array}{r|r} 2 & 66 \\ \hline 3 & 33 \\ \hline 11 & 11 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 102 \\ \hline 3 & 51 \\ \hline 17 & 17 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 138 \\ \hline 3 & 69 \\ \hline 23 & 23 \\ \hline & 1 \end{array}$

So,

66 = 2 × 3 × 11;

102 = 2 × 3 × 17;

138 = 2 × 3 × 23

The common factors are 2 and 3.

∴ H.C.F. = 2 × 3 = 6

Hence, the H.C.F. of 66, 102 and 138 is 6.

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

144, 312

Answer

Rule : To find the H.C.F., we perform long division by dividing the larger number by the smaller one and then repeatedly dividing each divisor by its remainder until the remainder is zero. The last divisor is the H.C.F.

We have:

$\begin{array}{l} 144\overline{\smash{\big)}312 \smash{\big(}}2 \\ \phantom{144}\phantom{)}{288 } \\ \phantom{14444}\overline{24\smash{\big)}144 \smash{\big(}6} \\ \phantom{space 143 }\underline{144\phantom{1}} \\ \phantom{space 1431}0\ \end{array}$

Hence, H.C.F. of 144 and 312 is 24.

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

252, 576

Answer

We have:

$\begin{array}{l} 252\overline{\smash{\big)}576 \smash{\big(}}2 \\ \phantom{144}\phantom{)}{504 } \\ \phantom{14444}\overline{72\smash{\big)}252 \smash{\big(}3} \\ \phantom{space 143 }{216\phantom{1}} \\ \phantom{space 1444}\overline{36\smash{\big)}72 \smash{\big(}2} \\ \phantom{space 1444 545 }\underline{72\phantom{1}} \\ \phantom{space 1444 545 )}0\ \end{array}$

Hence, H.C.F. of 252 and 576 is 36.

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

245, 315

Answer

We have:

$\begin{array}{l} 245\overline{\smash{\big)}315 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{245 } \\ \phantom{14444}\overline{70\smash{\big)}245 \smash{\big(}3} \\ \phantom{space 143 }{210\phantom{1}} \\ \phantom{space 1444}\overline{35\smash{\big)}70 \smash{\big(}2} \\ \phantom{space 1444 545 }\underline{70\phantom{1}} \\ \phantom{space 1444 545 )}0\ \end{array}$

Hence, H.C.F. of 245 and 315 is 35.

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

575, 920

Answer

We have:

$\begin{array}{l} 575\overline{\smash{\big)}920 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{575} \\ \phantom{1444}\overline{345\smash{\big)}575 \smash{\big(}1} \\ \phantom{space 14) }{345\phantom{1}} \\ \phantom{space 144}\overline{230\smash{\big)}345 \smash{\big(}1} \\ \phantom{space 1444 545 }{230\phantom{1}} \\ \phantom{space 144 5456 }\overline{115\smash{\big)}230 \smash{\big(}2} \\ \phantom{space 1444 5456778 }\underline{230\phantom{1}} \\ \phantom{space 1444 5456778 )}0\ \end{array}$

Hence, H.C.F. of 575 and 920 is 115.

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

605, 935

Answer

We have:

$\begin{array}{l} 605\overline {\smash{\big)}935 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{605} \\ \phantom{1444}\overline{330\smash{\big)}605 \smash{\big(}1} \\ \phantom{space 14) }{330\phantom{1}} \\ \phantom{space 144}\overline{275\smash{\big)}330 \smash{\big(}1} \\ \phantom{space 1444 545 }{275\phantom{1}} \\ \phantom{space 144 54567 }\overline{55\smash{\big)}275 \smash{\big(}5} \\ \phantom{space 1444 5456778 }\underline{275\phantom{1}} \\ \phantom{space 1444 5456778 )}0\ \end{array}$

Hence, H.C.F. of 605 and 935 is 55.

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

60, 96, 150

Answer

First, we will find the H.C.F. of 60 and 96.

$\begin{array}{l} 60\overline{\smash{\big)}96 \smash{\big(}}1 \\ \phantom{14}\phantom{)}{60 } \\ \phantom{144}\overline{36\smash{\big)}60 \smash{\big(}1} \\ \phantom{space 1 }{36\phantom{1}} \\ \phantom{space 4}\overline{24\smash{\big)}36 \smash{\big(}1} \\ \phantom{space 1441 }{24\phantom{1}} \\ \phantom{space 4444}\overline{12\smash{\big)}24 \smash{\big(}2} \\ \phantom{space 1444 545 }\underline{24\phantom{1}} \\ \phantom{space 1444 545 )}0\ \end{array}$

∴ H.C.F. of 60 and 96 is 12.

Next, we find the H.C.F. of 12 and 150.

$\begin{array}{l} 12\overline{\smash{\big)}150 \smash{\big(}}12 \\ \phantom{14}\phantom{)}{144} \\ \phantom{1444)}\overline{6\smash{\big)}12 \smash{\big(}2} \\ \phantom{space 14 }\underline{12\phantom{1}} \\ \phantom{space 14)}0\ \end{array}$

The H.C.F. of 12 and 150 is 6.

Hence, H.C.F. of 60, 96 and 150 is 6.

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

112, 140, 168

Answer

First, we will find the H.C.F. of 112 and 140.

$\begin{array}{l} 112\overline{\smash{\big)}140 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{112} \\ \phantom{1444)}\overline{28\smash{\big)}112 \smash{\big(}4} \\ \phantom{space 144 }\underline{112\phantom{1}} \\ \phantom{space 1444)}0\ \end{array}$

∴ H.C.F. of 112 and 140 is 28.

Next, we find the H.C.F. of 28 and 168.

$\begin{array}{l} 28\overline{\smash{\big)}168 \smash{\big(}}6 \\ \phantom{14}\phantom{)}\underline{168} \\ \phantom{ 114)}0\ \end{array}$

The H.C.F. of 28 and 168 is 28.

Hence, H.C.F. of 112, 140 and 168 is 28.

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

147, 210, 294

Answer

First, we will find the H.C.F. of 147 and 210.

$\begin{array}{l} 147\overline{\smash{\big)}210 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{147} \\ \phantom{1444)}\overline{63\smash{\big)}147 \smash{\big(}2} \\ \phantom{144 4444}\phantom{)}{126} \\ \phantom{1444 5678)}\overline{21\smash{\big)}63 \smash{\big(}3} \\ \phantom{space 144 5678 }\underline{63\phantom{1}} \\ \phantom{space 1444 567)}0\ \end{array}$

∴ H.C.F. of 147 and 210 is 21.

Next, we find the H.C.F. of 21 and 294.

$\begin{array}{l} 21\overline{\smash{\big)}294 \smash{\big(}}14 \\ \phantom{14}\phantom{)}\underline{21} \\ \phantom{14}\phantom{1)}{84} \\ \phantom{14}\phantom{2)}\underline{84} \\ \phantom{ 1141)}0\ \end{array}$

The H.C.F. of 21 and 294 is 21.

Hence, H.C.F. of 147, 210 and 294 is 21.

Find the greatest number that exactly divides 385 and 735.

Answer

The required number is the H.C.F. of 385 and 735.

By long division, we get:

$\begin{array}{l} 385\overline{\smash{\big)}735 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{385} \\ \phantom{1444}\overline{350\smash{\big)}385 \smash{\big(}1} \\ \phantom{144 4444}\phantom{)}{350} \\ \phantom{1444 5678)}\overline{35\smash{\big)}350 \smash{\big(}10} \\ \phantom{space 144 5678 }\underline{350\phantom{1}} \\ \phantom{space 1444 567)}0\ \end{array}$

So the H.C.F. of 385 and 735 is 35.

Hence, the required number is 35.

Find the greatest number that exactly divides 306, 450 and 540.

Answer

The required number is the H.C.F. of 306, 450 and 540.

First, we will find the H.C.F. of 306 and 450.

$\begin{array}{l} 306\overline{\smash{\big)}450 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{306} \\ \phantom{1444}\overline{144\smash{\big)}306 \smash{\big(}2} \\ \phantom{144 4444}\phantom{)}{288} \\ \phantom{1444 5678)}\overline{18\smash{\big)}144 \smash{\big(}8} \\ \phantom{space 144 5678 }\underline{144\phantom{1}} \\ \phantom{space 1444 567)}0\ \end{array}$

Therefore, the H.C.F. of 306 and 450 is 18.

Next, we need to find the H.C.F. of 18 and 540.

$\begin{array}{l} 18\overline{\smash{\big)}540 \smash{\big(}}30 \\ \phantom{14}\phantom{)}\underline{54} \\ \phantom{14}\phantom{1)}{00} \\ \phantom{14}\phantom{2)}\underline{00} \\ \phantom{ 1141)}0\ \end{array}$

So the H.C.F. of 306, 450 and 540 is 18.

Hence, the required number is 18.

Find the greatest number that will divide 37 and 53 leaving 5 as remainder in each case.

Answer

The required number is the H.C.F. of {(37 - 5) , (53 - 5)}. This implies that we need to find the H.C.F. of 32 and 48.

$\begin{array}{l} 32\overline{\smash{\big)}48 \smash{\big(}}1 \\ \phantom{144}{32} \\ \phantom{144}\overline{16\smash{\big)}32 \smash{\big(}2} \\ \phantom{144 12) }\underline{32\phantom{1}} \\ \phantom{space 14}0\ \end{array}$

The H.C.F. of 32 and 48 is 16.

Hence, the required number is 16.

Find the greatest number that will divide 138, 183 and 423 leaving remainder 3 in each case.

Answer

The required number is the H.C.F. of {(138 - 3), (183 - 3), (423 - 3)}. This implies that we need to find the H.C.F. of 135, 180 and 420.

First find H.C.F. of 135 and 180:

$\begin{array}{l} 135\overline{\smash{\big)}180 \smash{\big(}}1 \\ \phantom{1441}{135} \\ \phantom{144 12}\overline{45\smash{\big)}135 \smash{\big(}3} \\ \phantom{144 1223) }\underline{135\phantom{1}} \\ \phantom{space 1444}0\ \end{array}$

The H.C.F. of 135 and 180 is 45. Next, find the H.C.F. of 45 and 420.

$\begin{array}{l} 45\overline{\smash{\big)}420 \smash{\big(}}9 \\ \phantom{144}{405} \\ \phantom{144 1}\overline{15\smash{\big)}45 \smash{\big(}3} \\ \phantom{144 122) }\underline{45\phantom{1}} \\ \phantom{space 144}0\ \end{array}$

The H.C.F. of 45 and 420 is 15. Therefore, the H.C.F. of 135, 180 and 420 is 15.

Hence, the required number is 15.

Find the greatest number that will divide 76, 114 and 152 leaving the remainders 2, 3 and 4 respectively.

Answer

The required number = H.C.F. of {(76 - 2), (114 - 3), (152 - 4)} = H.C.F. of { 74, 111, 148}.

First find H.C.F. of 74 and 111:

$\begin{array}{l} 74\overline{\smash{\big)}111 \smash{\big(}}1 \\ \phantom{1441}{74} \\ \phantom{144 1}\overline{37\smash{\big)}74 \smash{\big(}2} \\ \phantom{144 1221 }\underline{74\phantom{1}} \\ \phantom{space 144}0\ \end{array}$

The H.C.F. of 74 and 111 is 37.

Now find H.C.F. of 37 and 148:

$\begin{array}{l} 37\overline{\smash{\big)}148 \smash{\big(}}4 \\ \phantom{144 }\underline{148\phantom{1}} \\ \phantom{space }0\ \end{array}$

The H.C.F. of 37 and 148 is 37. Therefore the H.C.F. of 74, 111 and 148 is 37.

Hence, the required number is 37.

Two vessels contain 104 litres and 91 litres of milk respectively. Find the measure of a bucket of maximum capacity which can measure the milk of either vessel an exact number of times.

Answer

The required measure of the bucket = H.C.F. of 104 and 91

By long division method, we get:

$\begin{array}{l} 91\overline{\smash{\big)}104 \smash{\big(}}1 \\ \phantom{1441}{91} \\ \phantom{144 1}\overline{13\smash{\big)}91 \smash{\big(}7} \\ \phantom{144 1221 }\underline{91\phantom{1}} \\ \phantom{space 144}0\ \end{array}$

The H.C.F. of 104 and 91 is 13.

Hence, the required measure of the bucket is 13 litres.

Use H.C.F. to show that 357 and 1625 are co-primes.

Answer

Two numbers are co-prime if their Highest Common Factor is 1.

By prime factorisation method, we get:

$\begin{array}{r|r} 3 & 357 \\ \hline 7 & 119 \\ \hline 17 & 17 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 5 & 1625 \\ \hline 5 & 325 \\ \hline 5 & 65 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$

Here,

357 = 3 × 7 × 17;

1625 = 5 × 5 × 5 × 13

By comparing the prime factors, we can see there are no common factors other than 1. So H.C.F. of 357 and 1625 is 1.

Since the H.C.F. of 357 and 1625 is 1, the numbers are co-primes.

Hence, proved that 357 and 1625 are co-primes.