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Question 12 Marks
In the following numbers, replace * by the smallest number to make it divisible by 9:
835*86
Answer
835686
Here, 8 + 3 + 5 + * + 8 + 6 = 30 + * should be a multiple of 9.
To be divisible of 9, the least value of * should be 6, i.e., 30 + 6 = 36, which is a multiple of 9.
$\therefore$ * = 6
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Question 22 Marks
Test the divisibility of the following numbers by 6:
251780
Answer
A number is divisible by 6 if it is divisible by both 2 and 3.
Since 251780 is not divisible by 3, it is not divisible by 6.
Checking the divisibility by 3: The sum of the digits of the number, 2 + 5 + 1 + 7 + 8 + 0, is 23, which is not divisible by 3.
So, the number is not divisible by 3.
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Question 32 Marks
Test the divisibility of:
12030624 by 8
Answer
12030624 by 8
12030624 is divisible by 8.
It is because the number formed by its hundreds, tens and ones digits is 624, which is divisible by 8.
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Question 42 Marks
Find the HCF of the numbers in the following using the prime factorization method:504, 980
Answer
The given numbers are 504 and 980.We have:
$\begin{array}{c|c}2&504\\\hline2&252\\\hline2&126\\\hline3&63\\\hline3&21\\\hline7&7\\\hline&1\end{array}$
$\begin{array}{c|c}2&980\\\hline2&490\\\hline5&245\\\hline7&49\\\hline7&7\\\hline&1\end{array}$
$504=2\times2\times2\times3\times3\times7=2^3\times3^2\times7$
$980=2\times2\times5\times7\times7=2^2\times5\times7^2$
$\therefore$ HCF of the given numbers = $2^2$ × $7$ = $28$
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Question 52 Marks
Give the prime factorization of the following number:
1323
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}3&1323\\\hline3&441\\\hline3&147\\\hline7&49\\\hline7&7\\\hline&1\end{array}$
$\therefore1323=3\times3\times3\times7\times7\times1$
$=3^3\times7^2$
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Question 62 Marks
Give the prime factorization of the following number:
18
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}2&18\\\hline2&9\\\hline3&3\\\hline&1\end{array}$
$\therefore18=2\times3\times3$
$=2\times3^2$
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Question 72 Marks
Which of the following are prime numbers?
137
Answer
A number between 100 and 200 is a prime number if it is not divisible by any prime number less than 15.
Similarly, a number between 200 and 300 is a prime number if it is not divisible by any prime number less than 20.
137 is a prime number, because it is not divisible by 2, 3, 5, 7 and 11.
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Question 82 Marks
The product of two numbers is 2560 and their LCM is 320. Find their HCF.
Answer
Product of the two numbers = 2560.HCF = 320
We know that,
LCM × HCF = Product of two numbers
$\therefore$ HCF $=\frac{2560}{320}=8$
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Question 92 Marks
Test the divisibility of the following numbers by 7:
2345
Answer
To determine if a number is divisible by 7, double the last digit of the number and subtract it from the number formed by the remaining digits.
2345 is divisible by 7.
We have 234 - 2 × 5 = 224, which is a multiple of 7.
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Question 102 Marks
Which of the following are prime numbers?
331
Answer
A number between 100 and 200 is a prime number if it is not divisible by any prime number less than 15.
Similarly, a number between 200 and 300 is a prime number if it is not divisible by any prime number less than 20.
331 is a prime number, because it is not divisible by 2, 3, 5, 7, 11, 13, 17 and 19.
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Question 112 Marks
In the following numbers, replace * by the smallest number to make it divisible by 3:
8*711
Answer
81711
Here, 8 + * + 7 + 1 + 1 = 17 + * should be a multiple of 3.
To be divisible by 3, the least value of * should be 1, i.e., 17 + 1 = 18, which is a multiple of 3.
$\therefore$ * = 1
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Question 122 Marks
Find the LCM of the numbers given below:60, 75
Answer
The given numbers are 60 and 75.We have:
$\begin{array}{c|c}3&60,75\\\hline5&20,25\\\hline5&4,5\\\hline2&4,5\\\hline2&2,1\\\hline&1,1\end{array}$
$\therefore$ LCM = 3 × 5 × 5 × 2 × 2
= 300
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Question 132 Marks
Test the divisibility of the following numbers by 7:
6021
Answer
To determine if a number is divisible by 7, double the last digit of the number and subtract it from the number formed by the remaining digits.
6021 is divisible by 7.
We have 602 - 2 × 1 = 600, which is not a multiple of 7.
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Question 142 Marks
Write all prime numbers between 50 and 100.
Answer
53, 59, 61, 67, 71, 73, 79, 83, 89, 97 are the prime numbers between 50 and 100.
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Question 152 Marks
Which of the following are prime numbers?
161
Answer
A number between 100 and 200 is a prime number if it is not divisible by any prime number less than 15.
Similarly, a number between 200 and 300 is a prime number if it is not divisible by any prime number less than 20.
161 is a not prime number, because it is divisible by 7.
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Question 162 Marks
Show that the following pairs are co-primes:59, 97
Answer
The given numbers are 59 and 97.
59 = 59 × 1
97 = 97 × 1
$\therefore$ HCF = 1
Since 59 and 97 does not have any common factor other than 1, the two numbers are co-primes.
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Question 172 Marks
Which of the following are prime numbers?
397
Answer
A number between 100 and 200 is a prime number if it is not divisible by any prime number less than 15.
Similarly, a number between 200 and 300 is a prime number if it is not divisible by any prime number less than 20.
397 is a prime number, because it is not divisible by 2, 3, 5, 7, 11, 13, 17 and 19.
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Question 182 Marks
Test the divisibility of the following numbers by 7:
14126
Answer
To determine if a number is divisible by 7, double the last digit of the number and subtract it from the number formed by the remaining digits.
14126 is divisible by 7.
We have 1412 - 2 × 6 = 1400, which is a multiple of 7.
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Question 192 Marks
In the following numbers, replace * by the smallest number to make it divisible by 3:
6*1054
Answer
621054
Here, 6 + * + 1 + 0 + 5 + 4 = 16 + * should be a multiple of 3.
To be divisible by 3, the least value of * should be 2, i.e., 16 + 2 = 18, which is a multiple of 3.
$\therefore$ * = 2
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Question 202 Marks
Test the divisibility of:
10001001 by 3
Answer
10001001 by 3
10001001 is divisible by 3.
It is because the sum of its digits, 1 + 0 + 0 + 0 + 1 + 0 + 0 + 1, is 3, which is divisible by 3.
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Question 212 Marks
In the following numbers, replace * by the smallest number to make it divisible by 9:
6678*1
Answer
667881
Here, 6 + 6 + 7 + 8 + * + 1 = 28 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 8, i.e., 28 + 8 = 36, which is a multiple of 9.
$\therefore$ * = 8
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Question 222 Marks
What are composite numbers? Can a composite number be odd? If yes, write the smallest odd composite number.
Answer
COMPOSITE NUMBERS: Numbers having more than two factors are known as composite numbers.
Yes a composite number can odd. The smallest odd composite number is 9.
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Question 232 Marks
Find the LCM of the numbers given below:42, 63
Answer
The given numbers are 42 and 63.We have:
$\begin{array}{c|c}7&42,63\\\hline3&6,9\\\hline3&2,3\\\hline2&2,1\\\hline&1,1\end{array}$
$\therefore$ LCM = 7 × 3 × 3 × 2 × 1
= 126
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Question 242 Marks
Test the divisibility of the following numbers by 11:
66311
Answer
A number is divisible by 11 if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.
66311 is not divisible by 11.
Sum of the digits at odd places = (1 + 3 + 6) = 10
Sum of the digits at even places = (1 + 6) = 7
Difference of the two sums = (10 - 7) = 3, which is not divisible by 11.
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Question 252 Marks
Write seven consecutive composite numbers less than 100 having no prime number between them.
Answer
Seven consecutive composite numbers less than 100 having no prime number between them are 90, 91, 92, 93, 94, 95 and 96.
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Question 262 Marks
Give the prime factorization of the following number:
1035
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}3&1035\\\hline3&345\\\hline5&115\\\hline23&23\\\hline&1\end{array}$
$\therefore1035=3\times3\times5\times23$
$=3^2\times5\times23$
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Question 272 Marks
In the following numbers, replace * by the smallest number to make it divisible by 9:
2*135
Answer
27135
Here, 2 + * + 1 + 3 + 5 = 11 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 7, i.e., 11 + 7 = 18, which is a multiple of 9.
$\therefore$ * = 7
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Question 282 Marks
Test the divisibility of the following numbers by 11:
137269
Answer
A number is divisible by 11 if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.
137269 is divisible by 11.
Sum of the digits at odd places = (9 + 2 + 3) = 14
Sum of the digits at even places = (6 + 7 + 1) = 14
Difference of the two sums = (14 - 14) = 0, which is a divisible by 11.
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Question 292 Marks
Test the divisibility of the following numbers by 7:
826
Answer
To determine if a number is divisible by 7, double the last digit of the number and subtract it from the number formed by the remaining digits.
If their difference is a multiple of 7, the number is divisible by 7.
826 is divisible by 7.
We have 82 - 2 × 6 = 70, which is a multiple of 7.
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Question 302 Marks
In the following numbers, replace * by the smallest number to make it divisible by 3:
27*4
Answer
2724
Here, 2 + 7 + * + 4 = 13 + * should be a multiple of 3.
To be divisible by 3, the least value of * should be 2, i.e., 13 + 2 = 15, which is a multiple of 3.
$\therefore$ * = 2
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Question 312 Marks
Test the divisibility of:
1000001 by 11
Answer
10000001 by 11
10000001 is divisible by 11.
Sum of digits at odd places = (1 + 0 + 0 + 0) = 1
Sum of digits at even places = (0 + 0 + 0 + 1) = 1
Difference of the two sums = (1 - 1) = 0, which is divisible by 11.
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Question 322 Marks
Test the divisibility of the following numbers by 8:
901674
Answer
A number is divisible by 8 if the number formed by the last three digits (digits in the hundreds, tens and units places) is divisible by 8.
901674 is not divisible by 8.
It is because the number formed by its hundreds, tens and ones digits, i.e., 674, is not divisible by 8.
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Question 332 Marks
The HCF of two numbers is 145 and their LCM is 2175. If one of the numbers is 725, find the other.
Answer
HCF = 145LCM = 2175
One of the numbers = 725
We know that,
HCF × LCF = Product of two numbers
$\therefore$ Other number $=\frac{145\times2175}{725}=435$
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Question 342 Marks
Find the LCM of the numbers given below:36, 60, 72
Answer
The given numbers are 36, 60 and 72.We have:
$\begin{array}{c|c}2&36,60,72\\\hline2&18,30,36\\\hline3&9,15,18\\\hline3&3,5,6\\\hline5&1,5,2\\\hline2&1,1,2\\\hline&1,1,1\end{array}$
$\therefore$ LCM = 2 × 2 × 2 × 3 × 3 × 5
= 360
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Question 352 Marks
Give the prime factorization of the following number:
9317
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}7&9317\\\hline11&1331\\\hline11&121\\\hline11&11\\\hline&1\end{array}$
$\therefore9317=7\times11\times11\times11$
$=7\times11^3$
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Question 362 Marks
Test the divisibility of the following numbers by 8:
36792
Answer
A number is divisible by 8 if the number formed by the last three digits (digits in the hundreds, tens and units places) is divisible by 8.
36792 is divisible by 8.
It is because the number formed by its hundreds, tens and ones digits, i.e., 792, is divisible by 8
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Question 372 Marks
Test the divisibility of the following numbers by 8:
1790184
Answer
A number is divisible by 8 if the number formed by the last three digits (digits in the hundreds, tens and units places) is divisible by 8.
1790184 is divisible by 8.
It is because the number formed by its hundreds, tens and ones digits, i.e., 184, is divisible by 8.
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Question 382 Marks
Find the HCF of:2 and an even number.
Answer
2 and 4 are two prime numbers.
Now, HCF of 2 and 4 is as follows:
2 = 2 × 1
4 = 2 × 2 × 1
$\therefore$ HCF =2 × 1 = 2
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Question 392 Marks
Find the HCF of the numbers in the following using the division method:1965, 2096
Answer
The given numbers are 1965 and 2096.We have:

$\therefore$ The HFC is 131.
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Question 402 Marks
Which of the following are prime numbers?
217
Answer
A number between 100 and 200 is a prime number if it is not divisible by any prime number less than 15.Similarly, a number between 200 and 300 is a prime number if it is not divisible by any prime number less than 20.
217 is a not prime number, because it is divisible by 7.
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Question 412 Marks
Find the HCF of the numbers in the following using the division method:2241, 2324
Answer
The given numbers are 2241 and 2341.
We have:

$\therefore$ The HFC = 83.
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Question 422 Marks
Test the divisibility of the following numbers by 8:
2138
Answer
A number is divisible by 8 if the number formed by the last three digits (digits in the hundreds, tens and units places) is divisible by 8.
2138 is not divisible by 8.
It is because the number formed by its hundreds, tens and ones digits, i.e., 138, is not divisible by 8.
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Question 432 Marks
Test the divisibility of the following numbers by 11:
4334
Answer
A number is divisible by 11 if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.
4334 is divisible by 11.
Sum of the digits at odd places = (4 + 3) = 7
Sum of the digits at even places = (3 + 4) = 7
Difference of the two sums = (7 - 7) = 0, which is divisible by 11.
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Question 442 Marks
Test the divisibility of the following numbers by 6:
872536
Answer
A number is divisible by 6 if it is divisible by both 2 and 3.Since 872536 is not divisible by 3, it is not divisible by 6.
Checking the divisibility by 3: The sum of the digits of the number, 8 + 7 + 2 + 5 + 3 + 6, is 31, which is not divisible by 3.
So, the number is not divisible by 3.
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Question 452 Marks
Give the prime factorization of the following number:
2907
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}3&2907\\\hline3&969\\\hline17&323\\\hline19&19\\\hline&1\end{array}$
$\therefore4641=3\times3\times17\times19$
$=3^2\times17\times19$
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Question 462 Marks
Test the divisibility of:
2134563 by 9
Answer
2134563 by 9
2134563 is not divisible by 9.
It is because the sum of its digits, 2 + 1 + 3 + 4 + 5 + 6 + 3, is 24, which is not divisible by 9.
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Question 472 Marks
Which of the following are prime numbers?
103
Answer
A number between 100 and 200 is a prime number if it is not divisible by any prime number less than 15.
Similarly, a number between 200 and 300 is a prime number if it is not divisible by any prime number less than 20.
103 is a prime number, because it is not divisible by 2, 3, 5, 7, 11 and 13.
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Question 482 Marks
In the following numbers, replace * by the smallest number to make it divisible by 3:
234*17
Answer
234117
Here, 2+ 3 +4 + * + 1 + 7 = 17 + * should be a multiple of 3.
To be divisible by 3, the least value of * should be 1, i.e., 17 + 1 = 18, which is a multiple of 3.
$\therefore$ * = 1
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Question 492 Marks
Give the prime factorization of the following number:
252
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}2&252\\\hline2&126\\\hline3&63\\\hline3&21\\\hline7&7\\\hline&1\end{array}$
$\therefore252=2\times2\times3\times3\times7\times1$
$=2^2\times3^2\times7\times1$
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Question 502 Marks
Give the prime factorization of the following number:
945
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}3&945\\\hline3&315\\\hline3&105\\\hline5&35\\\hline7&7\\\hline&1\end{array}$
$\therefore945=3\times3\times3\times5\times7\times1$
$=3^3\times5\times7$
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Question 512 Marks
Test the divisibility of:
10203574 by 4
Answer
10203574 by 4
10203574 is not divisible by 4.
It is because the number formed by its tens and the ones digits is 74, which is not divisible by 4.
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Question 522 Marks
Test the divisibility of the following numbers by 7:
117
Answer
To determine if a number is divisible by 7, double the last digit of the number and subtract it from the number formed by the remaining digits.
117 is not divisible by 7.
We have 11 - 2 × 7 = -3, which is not a multiple of 7.
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Question 532 Marks
In the following numbers, replace * by the smallest number to make it divisible by 9:
65*5
Answer
6525
Here, 6 + 5 + * + 5 = 16 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 2, i.e., 16 + 2 = 18, which is a multiple of 9.
$\therefore$ * = 2
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Question 542 Marks
Test the divisibility of the following numbers by 11:
83721
Answer
A number is divisible by 11 if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.
83721 is divisible by 11.
Sum of the digits at odd places = (1 + 7 + 8) = 16
Sum of the digits at even places = (2 + 3) = 5
Difference of the two sums = (16 - 5) = 11, which is divisible by 11.
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Question 552 Marks
Test the divisibility of the following numbers by 7:
25368
Answer
To determine if a number is divisible by 7, double the last digit of the number and subtract it from the number formed by the remaining digits.
25368 is divisible by 7.
We have 2536 - 2 × 8 = 2520, which is a multiple of 7.
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Question 562 Marks
The HCF of two number is 15 and their product is 1650. Find their LCM.
Answer
HCF × LCM = Products of the two numbers.
Product of the two numbers = 1650
HCF = 15
Required LCM $=\frac{1650}{15}$
= 110
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Question 572 Marks
Find the LCM of the numbers given below:12, 18, 20
Answer
The given numbers are 12, 18 and 20.We have:
$\begin{array}{c|c}2&12,18,20\\\hline2&6,9,10\\\hline3&3,9,5\\\hline3&1,3,5\\\hline5&1,1,5\\\hline&1,1,1\end{array}$
$\therefore$ LCM = 2 × 2 × 3 × 3 × 5
= 180
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Question 582 Marks
Give the prime factorization of the following number:
1224
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}2&1224\\\hline2&612\\\hline2&306\\\hline3&153\\\hline3&51\\\hline17&17\\\hline&1\end{array}$
$\therefore1224=2\times2\times2\times3\times3\times17$
$=2^3\times3^2\times17$
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Question 592 Marks
Test the divisibility of the following numbers by 8:
9364
Answer
A number is divisible by 8 if the number formed by the last three digits (digits in the hundreds, tens and units places) is divisible by 8.
9364 is not divisible by 8.
It is because the number formed by its hundreds, tens and ones digits, i.e., 364, is not divisible by 8.
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Question 602 Marks
In the following numbers, replace * by the smallest number to make it divisible by 9:
6702*
Answer
67023
Here, 6 + * + 7 + 0 + 2 = 15 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 3, i.e., 15 + 3 = 18, which is a multiple of 9.
$\therefore$ * = 3
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Question 612 Marks
Test the divisibility of the following numbers by 11:
901351
Answer
A number is divisible by 11 if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.
901351 is divisible by 11.
Sum of the digits at odd places = (0 + 3 + 1) = 4
Sum of the digits at even places = (9 + 1 + 5) = 15
Difference of the two sums = (4 - 15) = -11, which is divisible by 11.
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Question 622 Marks
Give the prime factorization of the following number:
1197
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}3&1197\\\hline3&399\\\hline7&113\\\hline19&19\\\hline&1\end{array}$
$\therefore1197=3\times3\times7\times19$
$=3^2\times7\times19$
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Question 632 Marks
Give the prime factorization of the following number:
136
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}2&136\\\hline2&68\\\hline2&34\\\hline17&17\\\hline&1\end{array}$
$\therefore136=2\times2\times2\times17$
$=2^3\times17$
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Question 642 Marks
Give the prime factorization of the following number:
48
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}2&48\\\hline2&24\\\hline2&12\\\hline2&6\\\hline3&3\\\hline&1\end{array}$
$\therefore48=2\times2\times2\times2\times3$
$=2^4\times3$
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Question 652 Marks
Find the HCF of the numbers in the following using the prime factorization method:170, 238
Answer
The given numbers are 170 and 238.We have:
$\begin{array}{c|c}2&170\\\hline5&85\\\hline17&17\\\hline&1\end{array}$
$\begin{array}{c|c}2&238\\\hline7&119\\\hline17&17\\\hline&1\end{array}$
$170=2\times5\times17$
$238=2\times7\times17$
$\therefore$ HCF of the given numbers = 2 × 17 = 34
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Question 662 Marks
Test the divisibility of the following numbers by 6:
71232
Answer
A number is divisible by 6 if it is divisible by both 2 and 3.
Since 71232 is divisible by both 2 and 3, it is divisible by 6.
Checking the divisibility by 2: Since the number has 2 in its units place, it is divisible by 2.
Checking the divisibility by 3: The sum of the digits of the number, 7 + 1 + 2 + 3 + 2, is 15, which is divisible by 3.
So, the number is divisible by 3.
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Question 672 Marks
Find the HCF of the numbers in the following using the prime factorization method:84, 98
Answer
The given numbers are 84 and 98.We have:
$\begin{array}{c|c}2&84\\\hline2&42\\\hline3&21\\\hline7&7\\\hline&1\end{array}$
$\begin{array}{c|c}2&98\\\hline7&49\\\hline7&7\\\hline&1\end{array}$
$84=2\times2\times3\times7=2^2\times3\times7$
$98=2\times7\times7=2\times7^2$
$\therefore$ HCF of the given numbers = 2 × 7 = 14
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Question 682 Marks
Test the divisibility of the following numbers by 6:
934706
Answer
A number is divisible by 6 if it is divisible by both 2 and 3.
Since 934706 is not divisible by 3, it is not divisible by 6.
Checking the divisibility by 3: Since the sum of the digits of the number, 9 + 3 + 4 + 7 + 0 + 6, is 29, which is not divisible by 3.
So, the number is not divisible by 3.
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Question 692 Marks
Test the divisibility of 67529124 by 8.
Answer
67529124A number is divisible by 8 if the number formed by the hundreds, tens and ones digits is divisible by 8.
Since the digits at the hundred’s, ten’s and unit places are 124, which is not divisible by 8, 67529124 is not divisible by 8.
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Question 702 Marks
Give the prime factorization of the following number:
56
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}2&56\\\hline2&28\\\hline2&14\\\hline&7\end{array}$
$\therefore56=2\times2\times2\times7$
$=2^3\times7$
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Question 712 Marks
Test the divisibility of the following numbers by 8:
136976
Answer
A number is divisible by 8 if the number formed by the last three digits (digits in the hundreds, tens and units places) is divisible by 8.
136976 is divisible by 8.
It is because the number formed by its hundreds, tens and ones digits, i.e., 976, is divisible by 8.
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Question 722 Marks
In the following numbers, replace * by the smallest number to make it divisible by 3:
62*35
Answer
62235
Here, 6 + 2 + * + 3 + 5 = 16 + * should be a multiple of 3.
To be divisible by 3, the least value of * should be 2, i.e., 16 + 2 = 18, which is a multiple of 3.
$\therefore$ * = 2
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Question 732 Marks
Test the divisibility of the following numbers by 6:
2070
Answer
A number is divisible by 6 if it is divisible by both 2 and 3.
Since 2070 is divisible by 2 and 3, it is divisible by 6.
Checking the divisibility by 2: Since the number 2070 has 0 in its units place, it is divisible by 2.
Checking the divisibility by 3: The sum of the digits of 2070, 2 + 0 + 7 + 0, is 9, which is divisible by 3.
So, it is divisible by 3.
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Question 742 Marks
Test the divisibility of the following numbers by 6:
46523
Answer
A number is divisible by 6 if it is divisible by both 2 and 3.
Since 46523 is not divisible by 2, it is not divisible by 6.
Checking the divisibility by 2: Since the number 46523 has 3 in its units place, it is not divisible by 2.
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Question 752 Marks
In the following numbers, replace * by the smallest number to make it divisible by 3:
53*46
Answer
53046
Here, 5 + 3 + * + 4 + 6 = 18 + * should be a multiple of 3.
As 18 is divisible by 3, the least value of * should be 0, i.e., 18 + 0 = 18.
$\therefore$ * = 0
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Question 762 Marks
Can two numbers have 12 as their HCF and 512 as their LCM? Justify your answer.
Answer
No, they cannot have 512 as their LCM.
We know that the HCF is one of the factors of the LCM.
Here, 3, which is a factor of 12, is not a factor of 512.
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Question 772 Marks
Test the divisibility of the following numbers by 11:
8790322
Answer
A number is divisible by 11 if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.
8790322 is not divisible by 11.
Sum of the digits at odd places = (2 + 3 + 9 + 8) = 22.
Sum of the digits at even places = (2 + 0 + 7) = 9
Difference of the two sums = (22 - 9) = 13, which is not divisible by 11.
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Question 782 Marks
Test the divisibility of:
19083625 by 11
Answer
19083625 by 11
19083625 is divisible by 11.
Sum of digits at odd places = (5 + 6 + 8 + 9) = 28
Sum of digits at even places = (2 + 3 + 0 + 1) = 6
Difference of the two sums = (28 - 6) = 22, which is divisible by 11.
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Question 792 Marks
Give the prime factorization of the following number:
637
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}7&637\\\hline7&91\\\hline13&13\\\hline&1\end{array}$
$\therefore637=7\times7\times13$
$=7^2\times13$
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Question 802 Marks
Find the HCF of:Two consecutive numbers.
Answer
4 and 5 are two consecutive numbers.
Now, HCF of 4 and 5 is as follows:
$4=2 \times 2 \times 1=2^2 \times 1$
$5=5 \times 1$
$\therefore H C F=1$
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Question 812 Marks
Give the prime factorization of the following number:
90
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}2&90\\\hline3&45\\\hline3&15\\\hline5&5\\\hline&1\end{array}$
$\therefore90=2\times3\times3\times5$
$=2\times3^2\times5$
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Question 822 Marks
Give the prime factorization of the following number:
4641
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}3&4641\\\hline7&1547\\\hline13&221\\\hline17&17\\\hline&1\end{array}$
$\therefore4641=3\times7\times13\times17$
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Question 832 Marks
The product of two numbers is 2160 and their HCF is 12. Find their LCM.
Answer
Product of the two numbers = 2160.HCF = 12
We know that,
LCM × HCF = Product of two numbers
$\therefore$ LCM $=\frac{2160}{12}$
$=180$
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Question 842 Marks
What are co-primes? Give examples of five pairs of five pairs of co-primes. Are co-primes always primes? If no, illustrate your answer by an example.
Answer
CO-PRIMES: Two numbers are said to be co-primes if they do not have a common factor other than 1.
No, co-primes are not be always primes.
For example 6, 7 are co-primes, while 6 is not a prime number.
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Question 852 Marks
Give the prime factorization of the following number:
420
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}2&420\\\hline2&210\\\hline3&105\\\hline7&35\\\hline5&5\\\hline&1\end{array}$
$\therefore420=2\times2\times3\times7\times5\times1$
$=2^2\times3\times7\times5$
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Question 862 Marks
Which of the following are prime numbers?
277
Answer
A number between 100 and 200 is a prime number if it is not divisible by any prime number less than 15.
Similarly, a number between 200 and 300 is a prime number if it is not divisible by any prime number less than 20.
277 is a prime number, because it is not divisible by 2, 3, 5, 7, 11, 13, 17 and 19.
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Question 872 Marks
Give the prime factorization of the following number:
12
Answer
We use the didvision method as shown below:
$\begin{array}{c|c}2&12\\\hline2&6\\\hline3&3\\\hline&1\end{array}$
$\therefore12=2\times2\times3$
$=2^2\times3$
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Question 882 Marks
What are twin primes? Write all the pairs of twin primes between 50 and 100.
Answer
TWIN PRIMES: Two consecutive odd prime numbers are known as twin primes.
(59, 61), (71, 73) are the pairs of twin primes between 50 and 100.
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Question 892 Marks
Give the prime factorization of the following number:
4335
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}3&4335\\\hline5&1445\\\hline17&289\\\hline17&17\\\hline&1\end{array}$
$\therefore4641=3\times5\times17\times17$
$=3\times5\times17^2$
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Question 902 Marks
Which of the following are prime numbers?
179
Answer
A number between 100 and 200 is a prime number if it is not divisible by any prime number less than 15.
Similarly, a number between 200 and 300 is a prime number if it is not divisible by any prime number less than 20.
179 is a prime number, because it is not divisible by 2, 3, 5, 7, 11 and 13.
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Question 912 Marks
What are Prime Numbers? Give ten examples.
Answer
Each of the numbers which have exactly two factors, namely, 1 and itself, is called a Prime Number.
Example- The numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, etc. are all prime number.
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Question 922 Marks
In the following numbers, replace * by the smallest number to make it divisible by 9:
91*67
Answer
91467
Here, 9 + 1 * + 6 + 7 = 23 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 4, i.e., 23 + 4 = 27, which is a multiple of 9.
$\therefore$ * = 4
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Question 932 Marks
Find the HCF of:Two prime numbers.
Answer
2 and 3 are two prime numbers.
Now, HCF of 2 and 3 is as follows:
2 = 2 × 1
3 = 3 × 1
$\therefore$ HCF = 1
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Question 942 Marks
Find the HCF of the numbers in the following using the prime factorization method:272, 425
Answer
The given numbers are 272 and 425.We have:
$\begin{array}{c|c}2&272\\\hline2&136\\\hline2&68\\\hline2&34\\\hline17&17\\\hline&1\end{array}$
$\begin{array}{c|c}5&425\\\hline5&85\\\hline17&17\\\hline&1\end{array}$
Now, $272=2\times2\times2\times2\times17$
$425=5\times5\times17$
$\therefore$ The required HCF is 17.
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Question 952 Marks
Find the HCF of:Two co-primes.
Answer
2 and 3 are two prime numbers.
Now, HCF of 2 and 3 is as follows:
2 = 2 × 1
3 = 3 × 1
$\therefore$ HCF = 1
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Question 962 Marks
The HCF and LCM of two numbers are 131 and 8253 respectively. If one of the numbers is 917, find the other.
Answer
HCF = 131LCM = 8253
One of the numbers = 917
We know that,
HCF × LCF = Product of two numbers
Other number $=\frac{8253\times131}{917}$
$\therefore$ The other number is 1179.
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