Question 12 Marks
Determine the $L.C.M$ of the numbers given below: $18, 17$
Answer$18, 17$ Prime factorization of $18 = 2 \times 3 \times 3$
Prime factorization of $17 = 17$
Therefore, Required $LCM = 2 \times 3 \times 3 \times 17 = 306$
View full question & answer→Question 22 Marks
Determine the $L.C.M$ of the numbers given below: $180, 384, 144$
Answer$180, 384, 144$ Prime factorization of $180 = 2 \times 2 \times 3 \times 3 \times 5$
Prime factorization of $384 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3$
Therefore, Required $LCM = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 5,760$
View full question & answer→Question 32 Marks
Find the $H.C.F$ of the following numbers using prime factorization method: $150, 140, 210$
Answer$504$ and $980$ Prime factorization of $504 = 2 \times 2 \times 2 \times 3 \times 3 \times 7$
Prime factorization of $980 = 2 \times 2 \times 5 \times 7 \times 7$
Therefore, $HCF = 2 \times 2 \times 7 = 28$
View full question & answer→Question 42 Marks
Which of the following numbers are divisible by $21?\ 21063$
Answer$21063$ Sum of the digits of the given number $= 2 + 1 + 0 + 6 + 3 = 12$
which is divisible by $3.$
Hence, $21,063$ is divisible by $3.$
Again, a number is divisible by $7$ if the difference between twice the one’s digit and the number formed by the other digits is either $0$ or a multiple of $7.$
$2,106 - (2 \times 3) = 2,100$ which is a multiple of $7.$
Thus, $21,063$ is divisible by $21.$
View full question & answer→Question 52 Marks
Determine prime factorization of the following numbers: $468$
Answer$468$ We have:
|
$2$
|
$468$
|
|
$2$
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$234$
|
|
$3$
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$117$
|
|
$3$
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$39$
|
|
$13$
|
$13$
|
|
|
$1$
|
Therefore, Prime factorization of $468 = 2 \times 2 \times 3 \times 3 \times 13$ View full question & answer→Question 62 Marks
What are prime numbers? List all primes between $1$ and $30.$
AnswerThose numbers with only two factors, i.e., $1$ and the number itself,
are known as prime numbers. Examples: $2, 3, 5, 7. 11$ and $13 $
The prime numbers between $1$ and $30$ are $2, 3, 5, 7, 11, 13, 17, 19, 23$ and $29.$
View full question & answer→Question 72 Marks
Determine prime factorization of the following numbers: $13915$
Answer$13915 $ We have:
|
$5$
|
$13915$
|
|
$11$
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$2783$
|
|
$11$
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$253$
|
|
$23$
|
$23$
|
|
|
$1$
|
Therefore, Prime factorization of $13915 = 5 \times 11 \times 11 \times 23$ View full question & answer→Question 82 Marks
What are composite numbers? Can a composite number be odd? If yes, write the smallest odd composite number.
AnswerA number which has more than two factors is called a composite number.
For example, the numbers $4, 6, 8, 9 10$ and $15$ are composite numbers.
Yes, a composite number can be an odd number. The smallest odd number is $9.$
View full question & answer→Question 92 Marks
Find the common factors of: $35$ and $50$
Answer$35$ and $50 $
$35 = 1 \times 35$
$35 = 5 \times 7$ i.e.,
the factors of $35$ are $1, 5, 7 $ and $35.$
Again, $50 = 1 \times 50$
$50 = 2 \times 25$
$50 = 5 \times 10$ i.e.,
the factors of $50$ are $1, 2, 5, 10, 25$ and $50.$
Therefore, the common factors of the two numbers are $1$ and $5.$
View full question & answer→Question 102 Marks
Determine the $L.C.M$ of the numbers given below:
$28, 36, 45, 60$
Answer$28, 36, 45, 60$
Prime factorization of $28 = 2 \times 2 \times 7$
Prime factorization of $36 = 2 \times 2 \times 3 \times 3$
Prime factorization of $45 = 3 \times 3 \times 5$
Prime factorization of $60 = 2 \times 2 \times 3 \times 5$
Therefore, Required $LCM = 2 \times 2 \times 3 \times 3 \times 5 \times 7 = 1,260$
View full question & answer→Question 112 Marks
Find the $H.C.F$ of the following numbers using prime factorization method: $225,450$
Answer$225$ and $450$ Prime factorization of $225 = 3 \times 3 \times 5 \times 5$
Prime factorization of $198 = 2 \times 3 \times 3 \times 5 \times 5$
Therefore, $HCF = 3 \times 3 \times 5 \times 5 = 225$
View full question & answer→Question 122 Marks
Find the common factors of: $5, 15$ and $25$
Answer$5, 15$ and $25$
Factors of $5$ are $1$ and $5$
Factors of $15$ are $1, 3, 5$ and $15$
Factors of $25$ are $1, 5$ and $25$
Therefore, the common factors of $5, 15,$ and $25$ are $1$ and $5.$
View full question & answer→Question 132 Marks
Without actual division show that $11$ is a factor of the following numbers: $110011$
Answer$110011$ The sum of the digits at the odd places $= 1 + 0 + 1 = 2$
The sum of the digits at the even places $= 1 + 0 + 1 = 2$
The difference of the two sums $= 2 - 2 = 0$
Therefore, $1, 10,011$ is divisible by $11$ because the difference of the sums is zero.
View full question & answer→Question 142 Marks
Determine the $H.C.F$ of the following numbers by using Euclid's algorithm $(i-x):$
$399,437$
Answer$399$ and $437$
We have dividend $= 399$ and divisor $= 437$
Clearly, the last divisor is $19.$
Hence, $HCF$ of the given number is $19$ View full question & answer→Question 152 Marks
Determine prime factorization of the following numbers: $240$
Answer$420$ We have:
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$2$
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$420$
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|
$2$
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$210$
|
|
$3$
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$105$
|
|
$5$
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$35$
|
|
$7$
|
$7$
|
|
|
$1$
|
Therefore, Prime factorization of $420 = 2 \times 2 \times 3 \times 5 \times 7$ View full question & answer→Question 162 Marks
Find the common factors of: $15$ and $25$
Answer$15$ and $25$
$15 = 1 \times 15$
$15 = 3 \times 5$
i.e., the factors of $15$ are $1, 3, 5$ and $15.$
Again, $25 = 1 \times 25$
$25 = 5 \times 5$
i.e., the factors of $25$ are $1, 5$ and $25. $
Therefore, the common factors of the two numbers are $1$ and $5.$
View full question & answer→Question 172 Marks
Test the divisibility of the following numbers by $6:$
$7020$
AnswerRule: A number is divisible by $6$ if it is divisible by $2$ as well as $3.$
$7020$
Here, the units digit $= 0$
Thus, the given number is divisible by $2.$
Also, the sum of the digits $= 7 + 0 + 2 + 0 = 9$
which is divisible by $3.$
So, the given number is
divisible by $3.$
Hence, $7,020$ is divisible by $6.$
View full question & answer→Question 182 Marks
In the following numbers, replace $*$ by the smallest number to make it divisible by $9: 66784 *$
Answer$66784 *$ Sum of the given digits $= 6 + 6 + 7 + 8 + 4 = 31$
The multiple of $9$ which is greater than $31$ is $36.$
Therefore, the smallest required number $= 36 - 31 = 5$
View full question & answer→Question 192 Marks
What is the smallest prime number? Is it an even number?
AnswerThe number $2$ is the smallest prime number. It is an even prime number. Except $2,$ all other even numbers are composite numbers.
View full question & answer→Question 202 Marks
What is the smallest odd prime? Is every odd number a prime number? If not, give an example of an odd number which is not prime.
AnswerThe smallest odd prime number is $3.$ No, every odd number is not a prime number. For example, $9$ is an odd number but it is not a prime number because its three factors are $1, 3$ and $9.$
View full question & answer→Question 212 Marks
Test the divisibility of the following numbers by $6: 56423$
AnswerRule: A number is divisible by $6$ if it is divisible by $2$ as well as $3.$
$56423$ Here, the units digit $= 3$
Thus, the given number is not divisible by $2.$
Also, the sum of the digits $= 5 + 6 + 4 + 2 + 3 = 20$
which is not divisible by $3.$
So, the given number is not divisible by $3.$
Since $3,56,423$ is neither divisible by $2$ nor by $3,$
it is not divisible by $6.$
View full question & answer→Question 222 Marks
Find the common factors of:
$20$ and $28$
Answer$20$ and $28$
$20 = 1 \times 20$
$20 = 2 \times 10$
$20 = 4 \times 5$
i.e., the factors of $20$ are $1, 2, 4, 5, 10$ and $20.$
Again, $28 = 1 \times 28$
$28 = 2 \times 14$
$28 = 7 \times 4$
i.e., the factors of $28$ are $1, 2, 4, 7, 14$ and $28.$
Therefore, the common factors of the two numbers are $1, 2$ and $4.$
View full question & answer→Question 232 Marks
Find the $H.C.F$ of the following numbers using prime factorization method: $106, 159, 265$
Answer$106, 159$ and $265$
Prime factorization of $106 = 2 \times 53$
Prime factorization of $159 = 2 \times 53$
Prime factorization of $265 = 5 \times 53$
Therefore, $HCF = 53$
View full question & answer→Question 242 Marks
What are the twin-primes? Write all pairs of twin-primes between $50$ and $100.$
AnswerTwin primes: Two prime numbers are said to be twin primes if there is only one composite number between them.
For example, $(3, 5)$ and $(5, 7)$ are twin primes.
Twin primes between $50$ and $100$ are $(59, 61)$ and $(71, 73).$
View full question & answer→Question 252 Marks
Find the $H.C.F $ of the following numbers using prime factorization method:
$84, 120, 138$
Answer$84, 120$ and $138$
Prime factorization of $84 = 2 \times 2 \times 3 \times 7$
Prime factorization of $120 = 2 \times 2 \times 2 \times 3 \times 5$
Prime factorization of $138 = 2 \times 3 \times 23$
Therefore, $HCF = 2 \times 3 = 6$
View full question & answer→Question 262 Marks
Find the $H.C.F$ of the following numbers using prime factorization method: $144,198$
Answer$144$ and $198$
Prime factorization of $144 = 2 \times 2 \times 2 \times 3 \times 3$
Prime factorization of $198 = 2 \times 3 \times 3 \times 11$
Therefore, $HCF = 2 \times 2 \times 3 = 18$
View full question & answer→Question 272 Marks
Determine the $L.C.M$ of the numbers given below: $48, 60$
Answer$48, 60$
Prime factorization of $48 = 2 \times 2 \times 2 \times 2 \times 3$
Prime factorization of $60 = 2 \times 2 \times 3 \times 5$
Therefore, Required $LCM = 2 \times 2 \times 2 \times 2 \times 3 \times 5 = 240$
View full question & answer→Question 282 Marks
A number is divisible by $24.$ By what other numbers will that number be divisible$?$
AnswerSince the number is divisible by $24,$
it will be divisible by all the factors of $24.$
The factors of $24$ are $1, 2, 3, 4, 6, 8, 12$ and $24.$
Hence, the number is also divisible by $1, 2, 3, 4, 6, 8$ and $12.$
View full question & answer→Question 292 Marks
A list consists of the following pairs of numbers: $51, 53; 55, 57; 59, 61; 63, 65; 67, 69; 71, 73$ Categorize them as pairs of: Primes
AnswerPrimes: Natural numbers which have exactly two distinct factors, i.e., $1$ and the number itself are called prime numbers. Hence, $(59, 61)$ and $(71, 73)$ are pairs of prime numbers.
View full question & answer→Question 302 Marks
In the following numbers, replace $*$ by the smallest number to make it divisible by $9: 67 * 19$
Answer$67 *19$
Sum of the given digits$ = 6 + 7 + 1 + 9 = 23$
The multiple of $9$ which is greater than $23$ is $27.$
Therefore, the smallest required number$ = 27 - 23 = 4$
View full question & answer→Question 312 Marks
Test the divisibility of the following numbers by $11:$
$10000001$
AnswerThe given number is $1,00,00,001.$
The sum of the digit at the odd places $= 1 + 0 + 0 + 0 = 1$
The sum of the digits at the even places $= 0 + 0 + 0 + 1 = 1$
Their difference $= 1 - 1 = 0$
Therefore, $1,00,00,001$ is divisible by $11.$
View full question & answer→Question 322 Marks
Write first five multiples of the following numbers: $25$
Answer$25$ The first five multiples of $25$ are as follows:
$25 \times 1 = 25$
$25 \times 2 = 50$
$25 \times 3 = 75$
$25 \times 4 = 100$
$25 \times 5 = 125$
View full question & answer→Question 332 Marks
Which factors are not included in the prime factorization of a composite number$?$
Answer$1$ and the number itself are not included in the prime factorization of a composite number.
Example: $4$ is a composite number.
Prime factorization of $4 = 2 \times 2.$
View full question & answer→Question 342 Marks
Find the $H.C.F$ of the following numbers using prime factorization method:
$504, 980$
Answer$504$ and $980$
Prime factorization of $504 = 2 \times 2 \times 2 \times 3 \times 3 \times 7$
Prime factorization of $980 = 2 \times 2 \times 5 \times 7 \times 7$
Therefore, $HCF = 2 \times 2 \times 7 = 28$
View full question & answer→Question 352 Marks
Write the smallest $4-$digit number and express it as a product of primes.
AnswerThe smallest $4-$digit number is $1000.$
$1000 = 2 \times 500$
$=2 \times 2 \times 250$
$=2 \times 2 \times 2 \times 125$
$=2 \times 2 \times 2 \times 5 \times 25$
$=2 \times 2 \times 2 \times 5 \times 5 \times 5$
Therefore, $1000=2 \times 2 \times 2 \times 5 \times 5 \times 5$
View full question & answer→Question 362 Marks
Find the $H.C.F$ and $L.C.F$ of the following pairs of numbers: $145,232$
Answer$145$ and $232$
Prime factorization of $145 = 5 \times 29$
Prime factorization of $232 = 2 \times 2 \times 2 \times 29$
Therefore, Required $HCF$ of $145$ and $232 = 289$
Therefore, Required $LCM$ of $145$ and $232 = 2 \times 2 \times 2 \times 5 \times 29 = 1160$
View full question & answer→Question 372 Marks
Which of the following pairs are always co-primes? One prime and one composite number
AnswerOne prime and one composite number One prime and one composite number are not always co-prime Example: $3$ and $21$ are not co-primes to each other.
View full question & answer→Question 382 Marks
Test the divisibility of the following numbers by $11: 5335$
AnswerThe given number is $5,335.$
The sum of the digit at the odd places $= 5 + 3 = 8$
The sum of the digits at the even places $= 3 + 5 = 8$
Their difference $= 8 - 8 = 0$
Therefore, $5,335$ is divisible by $11.$
View full question & answer→Question 392 Marks
Write all factors of the following numbers: $125$
Answer$125$
$125 = 1 \times 125$
$125 = 5 \times 25$
Therefore, the factors of $125$ are $1, 5, 25$ and $125.$
View full question & answer→Question 402 Marks
$H.C.F$ of co-prime numbers $4$ and $15$ was found as follow: $4 = 2 \times 2$ and $15 = 3 \times 5$ Since there is no common prime factor. So, $H.C.F$ of $4$ and $15$ is $0.$ Is the answer correct$?$ If not, what is the correct $H.C.F?$
AnswerNo, it is not correct. We know that $HCF$ of two co-prime number is $1.$
$4$ and $15$ are co-prime numbers because the only factor common to them is $1.$
Thus, $HCF$ of $4$ and $15$ is $1.$
View full question & answer→Question 412 Marks
Test the divisibility of the following numbers by $11: 70169803$
AnswerThe given number is $7,01,69,803. $
The sum of the digit at the odd places $= 7 + 1 + 9 + 0 = 17$
The sum of the digits at the even places $= 0 + 6 + 8 + 3 = 17$
Their difference $= 17 - 17 = 0$
Therefore, $7,01,69,803$ is divisible by $11.$
View full question & answer→Question 422 Marks
Find the $H.C.F$ of the following numbers using prime factorization method: $81,117$
Answer$81$ and $117$
Prime factorization of $81 = 3 \times 3 \times 3 \times 3$
Prime factorization of $117 = 3 \times 3 \times 13$
Therefore, $HCF = 3 \times 3 = 9$
View full question & answer→Question 432 Marks
Find the $H.C.F$ and $L.C.F$ of the following pairs of numbers: $234,572$
Answer$234$ and $572.$
Prime factorization of $234 = 2 \times 3 \times 3 \times 13$
Prime factorization of $572 = 2 \times 2 \times 11 \times 13$
Therefore, Required $HCF$ of $234$ and $572 = 226$
Therefore, Required $LCM$ of $117$ and $221 = 2 \times 2 \times 3 \times 3 \times 11 \times 13 = 5148$
View full question & answer→Question 442 Marks
Determine the $H.C.F$ of the following numbers by using Euclid's algorithm $(i-x): 300,450$
Answer$300$ and $450$
Dividend $= 450$ and divisor $= 300$
Clearly, the last divisor is $150.$
Hence, $HCF$ of the given number is $150.$ View full question & answer→Question 452 Marks
In the following numbers, replace $*$ by the smallest number to make it divisible by $9: 538 * 8$
Answer$538 * 8$ Sum of the given digits $= 5 + 3 + 8 + 8 = 24$
The multiple of $9$ which is greater than $24$ is $27.$
Therefore, the smallest required number $= 27 - 24 = 3$
View full question & answer→Question 462 Marks
Write all factors of the following numbers: $60$
Answer$60$
$60 = 1 \times 60$
$60 = 2 \times 30$
$60 = 3 \times 20$
$60 = 4 \times 15$
$60 = 5 \times 12$
$60 = 6 \times 10$
The factors of $60$ are $1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30$ and $60.$
View full question & answer→Question 472 Marks
Here are two different factor trees for $60.$ Write the missing numbers:
AnswerSince $60 = 30 \times 2.$
$30 = 10 \times 3$ and $10 = 5 \times 2$ we have:
View full question & answer→Question 482 Marks
Find the $H.C.F$ of the following numbers using prime factorization method: $170, 238$
Answer$170$ and $238$
Prime factorization of $170 = 2 \times 5 \times 17$
Prime factorization of $238 = 2 \times 7 \times 17$
Therefore, $HCF = 2 \times 17 = 34$
View full question & answer→Question 492 Marks
Without actual division show that $11$ is a factor of the following numbers:$ 1100011$
Answer$1100011$
the sum of the digits at the odd places $= 1 + 0 + 0 + 1 = 2$
The sum of the digits at the even places $= 1 + 0 + 1 = 2$
The difference of the two sums $= 2 - 2 = 0$
Therefore, $11, 00,011$ is divisible by $11$ because the difference of the sums is zero.
View full question & answer→Question 502 Marks
Determine prime factorization of the following numbers: $216$
Answer$216$ We have:
|
$2$
|
$216$
|
|
$2$
|
$108$
|
|
$2$
|
$54$
|
|
$3$
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$27$
|
|
$3$
|
$9$
|
|
$3$
|
$3$
|
|
|
$1$
|
Therefore, Prime factorization of $216 = 2 \times 2 \times 2 \times 3 \times 3$ View full question & answer→Question 512 Marks
The $HCF$ of two numbers is $23$ and their product is $55545.$ Find their $LCM.$
AnswerProduct of two numbers $= HCF$ of two numbers $\times \ LCM$ of two numbers
$\Rightarrow 55545 = 23 \times LCM$ of two numbers
$\Rightarrow LCM$ of two numbers $= 5554523 = 2415$
View full question & answer→Question 522 Marks
Write first five multiples of the following numbers: $45$
Answer$45$ The first five multiples of $45$ are as follows:
$45 \times 1 = 45$
$45 \times 2 = 90$
$45 \times 3 = 135$
$45 \times 4 = 180$
$45 \times 5 = 225$
View full question & answer→Question 532 Marks
Sort out even and odd numbers:
$i.\ 42$
$ii.\ 89$
$iii.\ 144$
$iv.\ 321$
AnswerA number which is exactly divisible by $2$ is called an even number. Therefore, $42$ and $144$ are even numbers. A number which is not exactly divisible by $2$ is called an odd number. Therefore, $89$ and $321$ are odd numbers.
View full question & answer→Question 542 Marks
Write all factors of the following numbers: $729$
Answer$729$
$729 = 1 \times 729$
$729 = 3 \times 243$
$729 = 9 \times 81$
$729 = 27 \times 27$
Therefore, the factors of $729$ are $1, 3, 9, 27, 81, 243$ and $729.$
View full question & answer→Question 552 Marks
Can two numbers have $16$ as their $H.C.F$ and $380$ as their $L.C.M.?$ Give reason.
AnswerNo. We know that $HCF$ of the given two numbers must exactly divide their $LCM.$ But $16$ does not divide $380$ exactly. Hence, there can be no two numbers with $16$ as their $HCF$ and $380$ as their $LCM.$
View full question & answer→Question 562 Marks
Determine prime factorization of the following numbers: $7325$
Answer$7325$ We have:
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$5$
|
$7325$
|
|
$5$
|
$1465$
|
|
$293$
|
$293$
|
|
|
$1$
|
Therefore, Prime factorization of $732$
$5= 5 \times 5 \times 293$ View full question & answer→Question 572 Marks
Write first five multiples of the following numbers:
$35$
Answer$35$
The first five multiples of $35$ are as follows:
$35 \times 1 = 35$
$35 \times 2 = 70$
$35 \times 3 = 105$
$35 \times 4 = 140$
$35 \times 5 = 175$
View full question & answer→Question 582 Marks
Which of the following pairs are always co-primes? two prime numbers.
AnswerTwo prime numbers. Two prime numbers are always co-primes to each other. Example: $7$ and $11$ are co-primes to each other.
View full question & answer→Question 592 Marks
Determine the $L.C.M$ of the numbers given below:
$42, 63$
Answer$42, 63$
Prime factorization of $42 = 2 \times 3 \times 7$
Prime factorization of $63 = 3 \times 3 \times 7$
Therefore, Required $LCM = 2 \times 3 \times 3 \times 7 = 126$
View full question & answer→Question 602 Marks
Write the largest $4-$digit number and give its prime factorization.
AnswerThe largest $4-$digit number is $9999.$ We have:
|
$3$
|
$9999$
|
|
$3$
|
$3333$
|
|
$11$
|
$1111$
|
|
$101$
|
$101$
|
|
|
$1$
|
Hence, the largest $4-$digit number $9999$ can be expressed in the form of its prime factors as: $3 \times 3 \times 11 \times 101.$ View full question & answer→Question 612 Marks
Find the $H.C.F$ of the following numbers using prime factorization method: $84,98$
Answer$84$ and $98$
Prime factorization of $84 = 2 \times 2 \times 3 \times 7$
Prime factorization of $98 = 2 \times 7 \times 7$
Therefore, $HCF = 2 \times 7 = 14$
View full question & answer→Question 622 Marks
What are co-primes? Give examples of five of co-primes. Are co-primes always prime? If no, illustrate your answer by an examples.
AnswerTwo numbers are said to be co-primes if they do not have any common factors other than $1.$
For example, $(2, 3), (3, 4), (4, 5), (5, 7)$ and $(13, 17)$ are co-primes.
Two co-primes numbers need not be both prime numbers. e.g., $(3, 4), (6, 7)$ and $(4, 13).$
View full question & answer→Question 632 Marks
Find numbers between $1$ and $100$ having exactly three factors.
AnswerThe numbers between $1$ and $100$ having exactly three factors are $4, 9, 25,$ and $49.$
The factors of $4$ are $1, 2$ and $4.$
The factors of $9$ are $1, 3$ and $9.$
The factors of $25$ are $1, 5$ and $25.$
The factors of $49$ are $1, 7$ and $49.$
View full question & answer→Question 642 Marks
Determine the $L.C.M$ of the numbers given below: $108, 135, 162$
Answer$108, 135, 162$
Prime factorization of $108 = 2 \times 2 \times 3 \times 3 \times 3$
Prime factorization of $135 = 3 \times 3 \times 3 \times 5$
Prime factorization of $162 = 2 \times 3 \times 3 \times 3 \times 3 $T
herefore, Required $LCM = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 = 1,620$
View full question & answer→Question 652 Marks
For a number, greater than $10,$ to be prime what may be the possible digit in the unit's place $?$
AnswerFor a number $($greater than $10)$ to be a prime number,
the possible digit in the unit’s place may be $1, 3, 7$ or $9.$
Example: $11, 13, 17$ and $19$ are prime numbers greater than $10.$
View full question & answer→Question 662 Marks
Write first five multiples of the following numbers: $40$
Answer$40$ The first five multiples of $40$ are as follows:
$40 \times 1 = 40$
$40 \times 2 = 80$
$40 \times 3 = 120$
$40 \times 4 = 160$
$40 \times 5 = 200$
View full question & answer→Question 672 Marks
A list consists of the following pairs of numbers:
$51, 53; 55, 57; 59, 61; 63, 65; 67, 69; 71, 73$
Categorize them as pairs of:
Co-primes
AnswerCo-primes: Two natural numbers are said to be co-primes numbers if they have $1$ as their only common factor.
Hence, all the given pairs of numbers are co-primes.
View full question & answer→Question 682 Marks
Here are two different factor trees for $60.$ Write the missing numbers:
AnswerWe have: Since $6 = 2 \times 3$ and $10 = 5 \times 2.$
View full question & answer→Question 692 Marks
Determine prime factorization of the following numbers: $945$
Answer$945$ We have:
|
$3$
|
$945$
|
|
$3$
|
$315$
|
|
$3$
|
$105$
|
|
$5$
|
$35$
|
|
$7$
|
$7$
|
|
|
$1$
|
Therefore, Prime factorization of $945 = 3 \times 3 \times 3 \times 5 \times 7$ View full question & answer→Question 702 Marks
A list consists of the following pairs of numbers: $51, 53; 55, 57; 59, 61; 63, 65; 67, 69; 71, 73$ Categorize them as pairs of: Composites
AnswerComposite numbers: Natural numbers which have more than two factors are called composite numbers. Hence, $(55, 57)$ and $(63, 65)$ are pairs of composite numbers.
View full question & answer→Question 712 Marks
Which of the following numbers are divisible by $21?\ 20163$
Answer$20163$ Sum of the digits of the given number $= 2 + 0 + 1 + 6 + 3 = 12$
which is divisible by $3.$
Hence, $20,163$ is divisible by $3.$
Again, a number is divisible by $7$
if the difference between twice the one’s digit and the number formed by the other digits is either $0$ or multiple of $7.$
$2016 - (2 \times 3) = 2010$ which is not a multiple of $7.$
Thus, $20,163$ is not divisible by $21.$
View full question & answer→Question 722 Marks
Determine the $L.C.M$ of the numbers given below:
$15, 30, 90$
Answer$15, 30, 90$
Prime factorization of $15 = 3 \times 5$
Prime factorization of $30 = 2 \times 3 \times 5$
Prime factorization of $90 = 2 \times 3 \times 3 \times 5$
Therefore, Required $LCM = 2 \times 3 \times 3 \times 5 = 90$
View full question & answer→Question 732 Marks
Which of the following pairs are always co-primes? Two composite numbers.
AnswerTwo composite numbers Two composite numbers are not always co-primes to each other. Example: $4$ and $6$ are not co-primes to each other.
View full question & answer→Question 742 Marks
Find the common factors of: $2, 6$ and $8$
Answer$2, 6$ and $8$
Factors of $2$ are $1$ and $2$
Factors of $6$ are $1, 2, 3$ and $6$
Factors of $8$ are $1, 2, 4$ and $8 $
Therefore, the common factors of $2, 6$ and $8$ are $1$ and $2.$
View full question & answer→Question 752 Marks
Write all factors of the following numbers: $76$
Answer$76$
$76 = 1 \times 76$
$76 = 2 \times 38$
$76 = 4 \times 19$
Therefore, The factors of $76$ are $1, 2, 4, 19, 38$ and $76.$
View full question & answer→Question 762 Marks
Find first two common multiples of $12$ and $18.$
AnswerMultiples of $12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132…$
Multiples of $18: 18, 36, 54, 72, 90, 108, 126, 144, 162, 180, 198…$
Therefore, the first two common multiples of $12$ and $18$ are $36$ and $72.$
View full question & answer→Question 772 Marks
Find first three common multiples of $6$ and $8$
AnswerMultiples of $6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84…$
Multiples of $8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96…$
Therefore, the first three common multiples of $6$ and $8$ are $24, 48$ and $72.$
View full question & answer→Question 782 Marks
Without actual division show that $11$ is a factor of the following numbers:
$1111$
Answer$1,111$
The sum of the digits at the odd places $= 1 + 1 = 2$
The sum of the digits at the even places $= 1 + 1 = 2$
The difference of the two sums $= 2 - 2 = 0$
Therefore, $1,111$ is divisible by $11$ because the difference of the sums is zero.
View full question & answer→Question 792 Marks
Find the $HCF$ of all natural numbers from $200$ to $478.$
AnswerThe $HCF$ of all natural numbers from $200$ to $478$ is $1$ because there are some prime numbers like $211, 233$ and so on which can't have common factor other than $1.$
View full question & answer→Question 802 Marks
The $LCM$ of two numbers is $1024$ and one of them is a prime number. Find their $HCF.$
Answer$LCM$ of two numbers is $1024 = 2^{10}$
Since, the other is prime number.
Hence, the other must be $2.$
$HCF$ of $2$ and $1024$ is $2.$
View full question & answer→Question 812 Marks
Find the $H.C.F$ and $L.C.F$ of the following pairs of numbers:
$861,1353$
Answer$861$ and $1353$
Prime factorization of $861 = 3 \times 7 \times 41$
Prime factorization of $1353 = 3 \times 11 \times 41$
Therefore, Required $HCF$ of $861$ and $1353 = 123$
Therefore, Required $LCM$ of $861$ and $1353 = 3 \times 7 \times 11 \times 41 = 9471$
View full question & answer→Question 822 Marks
Determine the $H.C.F$ of the following numbers by using Euclid's algorithm $(i-x): 1045,1520$
Answer$1045$ and $1520$ We have dividend $= 1045$ and divisor $= 1520$
Clearly, the last divisor is $95.$ Hence, $HCF$ of given numbers is $95.$ View full question & answer→Question 832 Marks
Determine the $L.C.M$ of the numbers given below: $56, 65, 85$
Answer$56, 65, 85$
Prime factorization of $56 = 2 \times 2 \times 2 \times 7$
Prime factorization of $65 = 5 \times 13$
Prime factorization of $85 = 5 \times 17$
Therefore, Required $LCM = 2 \times 2 \times 2 \times 5 \times 7 \times 13 \times 17 = 61,880$
View full question & answer→Question 842 Marks
Find the $H.C.F$ and $L.C.F$ of the following pairs of numbers: $117,221$
Answer$174$ and $221$
Prime factorization of $117 = 3 \times 3 \times 13$
Prime factorization of $221 = 13 \times 17$
Therefore, Required $HCF$ of $117$ and $221 = 13$
Therefore, Required $LCM$ of $117$ and $221 = 3 \times 3 \times 13 \times 17 = 1989$
View full question & answer→Question 852 Marks
Without actual division show that $11$ is a factor of the following numbers: $11011$
Answer$11011$ The sum of the digits at the odd places $= 1 + 0 + 1 = 2$
The sum of the digits at the even places $= 1 + 1 = 2$
The difference of the two sums $= 2 – 2 = 0$
Therefore, $11,011$ is divisible by $11$ because the difference of the sums is zero.
View full question & answer→