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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$5$
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$7325$
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$5$
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$1465$
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$293$
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$293$
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|
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$1$
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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:
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$3$
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$9999$
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$3$
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$3333$
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$11$
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$1111$
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$101$
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$101$
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$1$
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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:
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$3$
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$945$
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$3$
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$315$
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$3$
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$105$
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$5$
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$35$
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$7$
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$7$
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$1$
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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→