Question 1: Use prime factorization method to find the H.C.F. of the following:

i) $204, 1190$      ii) $1445, 1785$      iii) $1512, 4212$

iv) $280, 315, 385$      v) $576, 792, 1512$      vi) $1197, 5320, 4389$

Note: Prime Factorization Method:

Step 1: Express each one of the given numbers as a product of prime factors

Step 2: The product of terms containing least power of common prime factors gives the HCF of the given numbers.

i) $204, 1190$

$\begin{array}{r | r} 2 & 204 \\ \hline 2 & 102 \\ \hline 3 & 51 \\ \hline 17 & 17 \\ \hline {\ } & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r} 2 & 1190 \\ \hline 5 & 595 \\ \hline 7 & 119 \\ \hline 17 & 17 \\ \hline {\ } & 1 \end{array}$

$204 = 22 \times 3 \times 17$

$1190 = 2 \times 5 \times 7 \times 17$

Therefore HCF $= 2 \times 17 = 34$

ii) $1445, 1785$

$\begin{array}{r | r} 5 & 1445 \\ \hline 17 & 289 \\ \hline 17 & 17 \\ \hline {\ } & 1 \\ \hline {\ } & {\ } \end{array} \hspace{1.5cm} \begin{array}{r | r} 5 & 1785 \\ \hline 3 & 357 \\ \hline 7 & 119 \\ \hline 17 & 17 \\ \hline {\ } & 1 \end{array}$

$1445 = 5 \times 172$

$1785 = 5 \times 3 \times 7 \times 17$

Therefore HCF $= 5 \times 17 = 85$

iii) $1512, 4212$

$\begin{array}{r | r} 2 & 1512 \\ \hline 2 & 756 \\ \hline 2 & 378 \\ \hline 3 & 189 \\ \hline 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline {\ } & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r} 2 & 4212 \\ \hline 2 & 2106 \\ \hline 3 & 1053 \\ \hline 3 & 351 \\ \hline 3 & 117 \\ \hline 3 & 39 \\ \hline 13 & 13 \\ \hline {\ } & 1 \end{array}$

$1512 = 23 \times 33 \times 7$

$4212 = 22 \times 34 \times 13$

Therefore HCF $= 22 \times 33 = 108$

iv)  iv) $280, 315, 385$

$\begin{array}{r | r} 2 & 280 \\ \hline 2 & 140 \\ \hline 2 & 70 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline {\ } & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r} 5 & 315 \\ \hline 5 & 63 \\ \hline 5 & 21 \\ \hline 7 & 7 \\ \hline {\ } & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r} 5 & 385 \\ \hline 7 & 77 \\ \hline 11 & 77 \\ \hline {\ } & 1 \end{array}$

$280 = 2^3 \times 5 \times 7$

$315 = 5 \times 3^2 \times 7$

$385 = 5 \times 7 \times 11$

Therefore HCF $= 5 \times 7 = 35$

v) $576, 792, 1512$

$\begin{array}{r | r} 2 & 576 \\ \hline 2 & 288 \\ \hline 2 & 144 \\ \hline 2 & 72 \\ \hline 3 & 36 \\ \hline 3 & 12 \\ \hline 2 & 4 \\ \hline 2 & 1 \\ \hline {\ } & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r} 2 & 792 \\ \hline 2 & 396 \\ \hline 2 & 198 \\ \hline 3 & 99 \\ \hline 3 & 33 \\ \hline 11 & 11 \\ \hline {\ } & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r} 2 & 1512 \\ \hline 2 & 756 \\ \hline 2 & 378 \\ \hline 3 & 189 \\ \hline 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline {\ } & 1 \end{array}$

$576 = 2^7 \times 3^2$

$792 = 2^3 \times 3^2 \times 11$

$1512 = 2^3 \times 3^3 \times 7$

Therefore HCF $= 2^3 \times 3^2 = 72$

vi) $1197, 5320, 4389$

$\begin{array}{r | r} 3 & 1197 \\ \hline 3 & 399 \\ \hline 7 & 133 \\ \hline 19 & 19 \\ \hline {\ } & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r} 2 & 5320 \\ \hline 2 & 2660 \\ \hline 7 & 1330 \\ \hline 2 & 190 \\ \hline 5 & 95 \\ \hline 19 & 19 \\ \hline {\ } & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r} 7 & 4389 \\ \hline 3 & 627 \\ \hline 19 & 209 \\ \hline 11 & 11 \\ \hline {\ } & 1 \end{array}$

$1197 = 3^2 \times 7 \times 19$

$5320 = 2^3 \times 7 \times 5 \times 19$

$4389 = 3 \times 7 \times 19 \times 11$

Therefore HCF $= 7 \times 19 = 133$

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Question 2: Find the HCF of the following numbers by using the long division method:

i) $837, 992$           ii) $2923, 3239$           ii) $2923, 3239$

iv) $204, 1190, 1445$           v) $370, 592, 1036$         vi) $1701, 2106, 2754$

i) $837, 992$

$\begin{array}{r r r r r r} 837) & \overline{992} & (1 & & & \\ & 837 & & & & \\ & \overline{155)} & \overline{837} & (5 & & \\ & & 775 & & & \\ & & \overline{62)} & \overline{155} & (2 & \\ & & & 124 & & \\ & & & \overline{31)} & \overline{62} & (2 \\ & & & & 62 & \\ & & & & \overline{0} & \end{array}$

Therefore HCF $= 31$

ii) $2923, 3239$

$\begin{array}{r r r r r } 2923) & \overline{3239} & (1 & & \\ & 2923 & & & \\ & \overline{316)} & \overline{2923} & (9 & \\ & & 2844 & & \\ & & \overline{79)} & \overline{316} & (4 \\ & & & 316 & \\ & & & \overline{0} & \end{array}$

Therefore HCF $= 79$

iii) $5508, 9282$

$\begin{array}{r r r r r r r r } 5508) & \overline{9282} & (1 & & & & & \\ & 5508 & & & & & & \\ & \overline{3774)} & \overline{5508} & (1 & & & & \\ & & 3774 & & & & & \\ & & \overline{1734)} & \overline{3774} & (2 & & & \\ & & & 3468 & & & & \\ & & & \overline{306)} & \overline{1734} & (5 & & \\ & & & & 1530 & & & \\ & & & & \overline{204)} & \overline{306} & (1 & \\ & & & & & 204 & & \\ & & & & & \overline{102)} & \overline{204} & (2 \\ & & & & & & 204 & \\ & & & & & & \overline{0} & \end{array}$

Therefore HCF $= 102$

iv) $204, 1190, 1445$

$\begin{array}{r r r r r r} 1190) & \overline{1445} & (1 & & & \\ & 1190 & & & & \\ & \overline{255)} & \overline{1190} & (4 & & \\ & & 1020 & & & \\ & & \overline{170)} & \overline{255} & (1 & \\ & & & 170 & & \\ & & & \overline{85)} & \overline{170} & (2 \\ & & & & 170 & \\ & & & & \overline{0} & \end{array}$

Therefore HCF of 1445 and 1190 = 85

$\begin{array}{r r r r r } 85) & \overline{204} & (2 & & \\ & 170 & & & \\ & \overline{34)} & \overline{85} & (2 & \\ & & 68 & & \\ & & \overline{17)} & \overline{34} & (2 \\ & & & 34 & \\ & & & \overline{0} & \end{array}$

Therefore HCF $= 17$

v) $370, 592, 1036$

$\begin{array}{r r r r r } 592) & \overline{1036} & (1 & & \\ & 592 & & & \\ & \overline{444)} & \overline{592} & (1 & \\ & & 44 & & \\ & & \overline{148)} & \overline{444} & (3 \\ & & & 444 & \\ & & & \overline{0} & \end{array}$

Therefore HCF of $1036$ and $592= 148$

$\begin{array}{r r r r } 148) & \overline{370} & (1 & \\ & 296 & & \\ & \overline{74)} & \overline{148} & (2 \\ & & 148 & \\ & & \overline{0} & \end{array}$

Therefore HCF $= 74$

vi) $1701, 2106, 2754$

$\begin{array}{r r r r r } 2106) & \overline{2754} & (1 & & \\ & 2106 & & & \\ & \overline{648)} & \overline{2106} & (3 & \\ & & 1944 & & \\ & & \overline{162)} & \overline{648} & (4 \\ & & & 648 & \\ & & & \overline{0} & \end{array}$

HCF of $2754$ and $2106 = 162$

$\begin{array}{r r r r } 162) & \overline{1701} & (10 & \\ & 1620 & & \\ & \overline{81)} & \overline{162} & (2 \\ & & 162 & \\ & & \overline{0} & \end{array}$

Therefore HCF $= 81$

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Question 3: Reduce each of the following fractions to its lowest terms:

i) $\frac{682}{868}$    ii) $\frac{777}{1147}$    iii) $\frac{1095}{1168}$

i) $\frac{682}{868}$

$\begin{array}{r r r r r r} 682) & \overline{868} & (1 & & & \\ & 682 & & & & \\ & \overline{186)} & \overline{682} & (3 & & \\ & & 558 & & & \\ & & \overline{124)} & \overline{186} & (1 & \\ & & & 124 & & \\ & & & \overline{62)} & \overline{124} & (2 \\ & & & & 124 & \\ & & & & \overline{0} & \end{array}$

Therefore $\frac{682}{868}$ $=$ $\frac{682 \div 62}{868 \div 62}$ $=$ $\frac{11}{14}$

ii) $\frac{777}{1147}$

$\begin{array}{r r r r r } 777) & \overline{1147} & (1 & & \\ & 777 & & & \\ & \overline{370)} & \overline{777} & (2 & \\ & & 740 & & \\ & & \overline{37)} & \overline{370} & (10 \\ & & & 370 & \\ & & & \overline{0} & \end{array}$

Therefore $\frac{777}{1147}$ $=$ $\frac{777 \div 37}{1147 \div 37}$ $=$ $\frac{21}{31}$

iii) $\frac{1095}{1168}$

$\begin{array}{r r r r } 1095) & \overline{1168} & (1 & \\ & 1095 & & \\ & \overline{73)} & \overline{1095} & (15 \\ & & 1095 & \\ & & \overline{0} & \end{array}$

Therefore $\frac{1095}{1168}$ $=$ $\frac{1095 \div 73}{1068 \div 73}$ $=$ $\frac{15}{16}$

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Question 4. Which of the following numbers are co-primes?

i) $18, 25$ ii) $62, 81$ iii) $69, 92$

Note: The natural numbers are said to be co-primes if the HCF of the numbers is $1$.
You can calculate the HCF by using any of the two methods…prime factorization or long division method.

i) $18, 25$

HCF of $18, 25 = 1$

Hence, $18$ and $25$ are co-primes.

ii) $62, 81$

HCF of $62, 81 = 1$

Hence, $62$ and $81$ are co-primes.

iii) $69, 92$

HCF of $69, 92 = 23$

Hence, $69$ and $92$ are not co-primes.

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Question 5: Find the greatest number that exactly divides $105, 1001$ and $2436$.

We need to find the HCF of $105, 1001$ and $2436$.

$\begin{array}{r r r r r r r r } 1001) & \overline{2436} & (2 & & & & & \\ & 2002 & & & & & & \\ & \overline{434)} & \overline{1001} & (2 & & & & \\ & & 868 & & & & & \\ & & \overline{133)} & \overline{434} & (3 & & & \\ & & & 399 & & & & \\ & & & \overline{35)} & \overline{133} & (3 & & \\ & & & & 105 & & & \\ & & & & \overline{28)} & \overline{35} & (1 & \\ & & & & & 28 & & \\ & & & & & \overline{7)} & \overline{28} & (4 \\ & & & & & & 28 & \\ & & & & & & \overline{0} & \end{array}$

$\begin{array}{r r r } 7) & \overline{105} & (15 \\ & 105 & \\ & \overline{0} & \end{array}$

Hence the HCF of the three numbers is $7$. That means, $7$ is the largest number that would divide all the three numbers perfectly.

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Question 6: Find the largest number which can divide $290, 460$ and $552$ leaving the remainders $4, 5$ and $6$ respectively.

The largest number should divide $(290-4), (460-5)$ and $(552-6)$ or $286, 455$ and $546$ perfectly. Now find the HCF of $286, 455$ and $546$.

$\begin{array}{r r r r } 455) & \overline{546} & (1 & \\ & 455 & & \\ & \overline{91)} & \overline{455} & (5 \\ & & 455 & \\ & & \overline{0} & \end{array}$

$\begin{array}{r r r r } 91) & \overline{286} & (3 & \\ & 273 & & \\ & \overline{13)} & \overline{91} & (7 \\ & & 91 & \\ & & \overline{0} & \end{array}$

Hence $13$ is the number that can divide $290, 460$ and $552$ leaving the remainders $4, 5$ and $6$ respectively.

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Question 7: Find the largest number which can divide $1354, 1866$ and $2762$ leaving the same remainder $10$ in each case.

The largest number should divide $(1354 - 10), (1866 - 10)$ and $(2762 - 10)$ or $1344, 1856$ and $2752$ perfectly. Now find the HCF of $1344, 1856$ and $2752$.

$\begin{array}{r r r r r } 1856) & \overline{2752} & (1 & & \\ & 1856 & & & \\ & \overline{896)} & \overline{1856} & (2 & \\ & & 1792 & & \\ & & \overline{64)} & \overline{896} & (14 \\ & & & 896 & \\ & & & \overline{0} & \end{array}$

$\begin{array}{r r r } 64) & \overline{1344} & (21 \\ & 1344 & \\ & & \overline{0} \end{array}$

HCF is $64$. Hence $64$ is the largest number which can divide $1354, 1866$ and $2762$ leaving the same remainder $10$ in each case.

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Question 8: Find the greatest number that will divide $1305, 4665$ and $6905$ so as to leave the same remainder in each case.

Required number

$=$ HCF of $(4665 - 1305), (6905 - 4665)$ and $(6905 - 1305)$

$=$ HCF of $3360, 2240, 5600$.

$\begin{array}{r r r r r } 3360) & \overline{5600} & (1 & & \\ & 3360 & & & \\ & \overline{2240)} & \overline{3360} & (1 & \\ & & 2240 & & \\ & & \overline{1120)} & \overline{2240} & (2 \\ & & & 2240 & \\ & & & \overline{0} & \end{array}$

$\begin{array}{r r r } 1120) & \overline{3360} & (3 \\ & 3360 & \\ & \overline{0} & \end{array}$

HCF is $1120$.

The required number is $112$ that will divide $1305, 4665$ and $6905$ so as to leave the same remainder in each case.

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Question 9: Three pieces of timber $13$ m $44$ cm, $18$ m $56$ cm and $27$ m $52$ cm have to be divided into planks of same lengths. What is the greatest possible length of each plank?

The lengths are $1344$ cm, $1856$ cm and $2752$ cm. Find HCF.

$\begin{array}{r r r r r } 1856) & \overline{2752} & (1 & & \\ & 1856 & & & \\ & \overline{896)} & \overline{1856} & (2 & \\ & & 1792 & & \\ & & \overline{64)} & \overline{896} & (14 \\ & & & 896 & \\ & & & \overline{0} & \end{array}$

$\begin{array}{r r r } 64) & \overline{1344} & (21 \\ & 1344 & \\ & \overline{0} & \end{array}$

HCF is $64$. Hence greatest length is $64$ cm.

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Question 10: An NGO wishes to distribute $1651$ pencils and $2032$ erasers among poor children in such a way that each child gets the same number of pencils and the same number of erasers. Find the maximum number of children among whom these pencils and erasers are distributed.

Take HCF of $1651$ and $2032$.

$\begin{array}{r r r r r } 1651) & \overline{2032} & (1 & & \\ & 1651 & & & \\ & \overline{381)} & \overline{1651} & (4 & \\ & & 1524 & & \\ & & \overline{127)} & \overline{381} & (3 \\ & & & 381 & \\ & & & \overline{0} & \end{array}$

HCF $= 127$.

The maximum number of children among whom these pencils and erasers are distributed $= 127$

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Question 11: Find the greatest possible length of a rope which can be used to measure exactly the lengths $5$ m $13$ cm, $7$ m $83$ cm and $10$ m $80$ cm.

Lengths are $513$ cm, $783$ cm and $1080$ cm.

HCF of $513, 783$ and $1080$

$\begin{array}{r r r r r r r r } 783) & \overline{1083} & (1 & & & & & \\ & 783 & & & & & & \\ & \overline{297)} & \overline{783} & (2 & & & & \\ & & 594 & & & & & \\ & & \overline{189)} & \overline{297} & (1 & & & \\ & & & 189 & & & & \\ & & & \overline{108)} & \overline{189} & (1 & & \\ & & & & 108 & & & \\ & & & & \overline{81)} & \overline{108} & (1 & \\ & & & & & 81 & & \\ & & & & & \overline{27)} & \overline{81} & (3 \\ & & & & & & 81 & \\ & & & & & & \overline{0} & \end{array}$

$\begin{array}{r r r } 27) & \overline{513} & (19 \\ & 513 & \\ & \overline{0} & \end{array}$

Hence HCF $= 27$. Therefore the length of the rope is $27$ cm.

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Question 12: Three different containers contain different quantity of mixture of milk and water whose measurements are $403$ kg, $465$ kg, and $527$ kg. What biggest measure must be there to measure all different quantities an exact number of times.

Required Number is the HCF of $403, 465$ and $527$.

$\begin{array}{r r r r r } 465) & \overline{527} & (1 & & \\ & 465 & & & \\ & \overline{62)} & \overline{465} & (7 & \\ & & 434 & & \\ & & \overline{31)} & \overline{62} & (2 \\ & & & 62 & \\ & & & \overline{0} & \end{array}$

$\begin{array}{r r r } 31) & \overline{403} & (13 \\ & 403 & \\ & \overline{0} & \end{array}$

HCF $= 31$. Hence $31$ liter measure would be required to measure the three given quantities.

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Question 13: Find the least number of square tiles required to pave the ceiling of a room $15$ m $17$ cm long and $9$ m $2$ cm broad.

The dimensions of the ceiling are $1517$ cm and $902$ cm.

The required number of times $=$ HCF of $1517$ and $902$.

$\begin{array}{r r r r r r} 902) & \overline{1517} & (1 & & & \\ & 902 & & & & \\ & \overline{615)} & \overline{902} & (1 & & \\ & & 615 & & & \\ & & \overline{287)} & \overline{615} & (2 & \\ & & & 574 & & \\ & & & \overline{41)} & \overline{287} & (7 \\ & & & & 287 & \\ & & & & \overline{0} & \end{array}$

HCF $= 41$

Dimensions of tiles $= 41$ cm.

Therefore the number of tiles $=$ $\frac{area \ of \ ceiling}{area \ of \ tile}$ $=$ $\frac{1517 \times 902}{41 \times 41}$ $= 814$

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Question 14: A school admitted $1190$ boys and $204$ girls in a year. The authorities decided to divide these students into classes in such a way that each class gets the same number of boys and the same number of girls. Find the maximum number of classes that can be formed.

Required number is the HCF of $1190$ and $204$.

$\begin{array}{r r r r r } 204) & \overline{1190} & (5 & & \\ & 1020 & & & \\ & \overline{170)} & \overline{204} & (1 & \\ & & 170 & & \\ & & \overline{34)} & \overline{170} & (5 \\ & & & 170 & \\ & & & \overline{0} & \end{array}$

HCF $= 34$.

Number of classes $= 34$

Size of the class $= 41$ students

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Question 15: In a training camp, there were $195$ boys and $143$ girls. The coordinator instructed the pupils to form separate teams for boys and girls such that each team includes equal number of pupils. Find the maximum number of pupils that each team can have and also the number of teams so formed.

Required number is HCF of $195$ and $143$.
$\begin{array}{r r r r r r} 143) & \overline{195} & (1 & & & \\ & 143 & & & & \\ & \overline{52)} & \overline{143} & (2 & & \\ & & 104 & & & \\ & & \overline{39)} & \overline{52} & (1 & \\ & & & 39 & & \\ & & & \overline{13)} & \overline{39} & (3 \\ & & & & 39 & \\ & & & & \overline{0} & \end{array}$
HCF $= 13$.
Number of teams $= 13$.
Number of pupil in each team $=$ $\frac{(195 + 143)}{13}$ $= 26$.