Maths 3 Mark Question

Given numbers are 87 and 145.
Prime factorization of 87 = 3 × 29
Prime factorization of 145 = 5 × 29
HCF of 87 and 145 = 29
LCM of 87 and 145 = 3 × 5 × 29 = 435
Product of the given number = 87 × 145 = 12615
Product of their HCF and LCM = 29 × 435 = 12615
Therefore, Product of the number = Product of their HCF and LCM (Verified)

We know that two numbers are co-primes if their HCF is 1.
875 and 1859
Here, dividend = 1859 and divisor = 875

Clearly, the last divisor is 1.
Hence, the given numbers are co-prime.

The maximum capacity of the required tin is the HCF of the three quantities of oil.
Prime factorization of 120 = 2 × 2 × 2 × 3 × 5
Prime factorization of 180 = 2 × 2 × 3 × 3 × 5
Prime factorization of 240 = 2 × 2 × 2 × 2 × 3 × 5
Therefore, HCF of 120, 180, and 240 = 2 × 2 × 3 × 5 = 60
Hence, the required greatest capacity of the tin must be 60 liters.

Rule: A number is divisible by 11 if the difference of the sums of the alternate digits is either 0 or a multiple of 11.
9 × 8071
Sum of the digits at the odd places = 9 + 8 + 7 = 24
Sum of the digits at the even places = missing number + 0 + 1 = missing number + 1
Difference = 24 - [missing number + 1] = 23 - missing number
According to rule, 23 - missing number = 22 [Because 22 is a multiple of 11 and the missing number is a single digit]
Thus, missing number = 1
Hence, the smallest required number is 1.

To find the required number of pencils and crayons, we need to find the LCM of 24 and 32.
Prime factorization of 24 = 2 × 2 × 2 × 3
Prime factorization of 32 = 2 × 2 × 2 × 2 × 2
Required LCM of 24 and 32 =2 × 2 × 2 × 2 × 3 = 96
Thus, number of pencils and crayons needed to be bought is 96 each, i.e. 96 ÷ 24 = 4 packs of color pencils and 96 ÷ 32 = 3 packs of crayons.

Given: Length of the first tape = 7m = 700cm
Length of the second tape = 3m 85cm = 385cm
Length of the third tape = 12m 95cm = 1,295cm
The length of the longest tape will be the HCF of 700, 385, and 1,295.
Prime factorization of 700 = 2 × 2 × 5 × 5 × 7
Prime factorization of 385 = 5 × 7 × 11
Prime factorization of 1,295 = 5 × 7 × 37 I-ICF of 700, 385, and 1,295 = 5 × 7 = 35
Required length of the longest tape = 35cm

A perfect number is a positive number that equals the sum of its divisors, excluding itself.
Divisors of 6 = 1, 2, 3
Divisors of 28 = 1, 2, 4, 7, 14, 28
In 6, the sum of divisors except itself, 1 + 2 + 3 is 6.
In 28, the sum of divisors except itself, 1 + 2 + 4 + 7 + 14 is 28.
Two perfect numbers are 6 and 28.

HCF of two numbers is a factor of the LCM of those numbers
Thu, we cannot have two numbers whose HCF is 12 and LCM is 512.
Because, when we divide 512 by 12, we get a remainder of 42.68
Thus, 12 is not a factor of 512.
Hence, we cannot have two numbers of whose HCF is 12 and LCM is 512.

Given numbers are 117 and 221.
Prime factorization of 117 = 3 × 3 × 13
Prime factorization of 221 = 13 × 17
HCF of 117 and 221 = 13
LCM of 117 and 221 = 3 × 3 × 13 × 17 = 1, 989
Product of the given number = 117 × 221 = 12, 857
Product of their HCF and LCM = 13 × 1,989 = 12,857
Therefore, Product of the number = Product of their HCF and LCM (verified)

Given: HCF of two numbers = 6
LCM of two numbers = 180
One of the given number = 30
Product of the two numbers = Product of their HCF and LCM
Therefore, 30 x other number = 6 × 180
Other number = 6 × 18030 = 36
Thus, the required number is 36.

The HCF of two consecutive numbers is 1.
Example:
D = 4 and d= 5 are two consecutive numbers.
Here, we have dividend = 5 and divisor = 4

Clearly, the last divisor is 1.
Hence, HCF of 4 and 5 is 1.

We have to find the greatest number which divides (285 - 9) and (1,249 - 7) exactly.
The required number will be given by the HCF of 276 and 1242.
Resolving 276 and 1242 into prime factors, we have:
276 =2 × 2 × 3 × 23
1242 =2 × 3 × 3 × 3 × 23
HCF of 276 and 1242 is 2 × 3 × 23 = 138.

The given number is 1729.
We have:

7	1729
13	247
19	19

Thus, the number 1729 can be expressed in the form of its prime factors ass 7 × 13 × 19.
Relation between its two consecutive prime factors:
The consecutive prime factors of the given number are 7, 13 and 19.
Clearly, 13 - 7 = 6 and 19 - 13 =6
Here, in two consecutive prime factors, the latter is 6 more than the previous one.

First bus stop at which both the buses will stop together = LCM of 6th block and 8th block
Prime factorization of 6 = 2 × 3
Prime factorization of 8 = 2 × 2 × 2
Therefore, Required LCM = 2 × 2 × 2 × 3 = 24
Hence, the first bus stop at which both the buses will stop together will be at the 24th block.

We have to find prime factorization of 33 and 39.
Prime factorization of 33 = 3 × 11
Prime factorization of 39 = 3 × 13
Therefore, Required LCM = 3 × 11 × 13 = 429
Thus, 429 is the smallest number exactly divisible by 33 and 39.
To get the remainder as 5: Smallest number = 429 + 5 = 434
Thus, the required number is 434.

We have to find the LCM of 80cm, 85cm, and 90cm.
Prime factorization of 80 = 2 × 2 × 2 × 2 × 5
Prime factorization of 85 = 5 × 17
Prime factorization of 90 = 2 × 3 × 3 × 5
Therefore, Required LCM = 2 × 2 × 2 × 2 × 3 ×3 × 5 × 17 = 12,240
Therefore, Required minimum distance = LCM of 80cm, 85cm, and 90cm
= 12,240cm
= 122m 40cm (since 1m =100cm)

HCF of two numbers is 4
So, the numbers can be 4a and 4b.
Now, 4a × 4b = 4 × 400
⇒ ab = 100
So, we can get the pairs
a = 25 and b = 4
a = 4 and b = 25
Thus, the numbers are 4 × 25 = 100 and 4 × 4 = 16.
Also, we can get the other pair 4 × 1 = 4 and 400.
Hence, there are two pairs.

To find the required least number, we have to find the LCM of the numbers from 1 to 10.
We know that 2, 3, 5, and 7 are prime number.
Prime factorization of 4 = 2 × 2
Prime factorization of 6 = 2 × 3
Prime factorization of 8= 2 × 2 × 2
Prime factorization of 9 = 3 × 3
Prime factorization of 10 = 2 × 5
Therefore, Required least number = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2,520

Given numbers are 25 and 65
Prime factorization of 25 = 5 × 5
Prime factorization of 65 = 5 × 13
HCF of 25 and 65 = 5
LCM of 25 and 65= 5 × 5 × 13 = 325
Product of the given numbers = 25 × 65 = 1,625
Product of their HCF and LCM = 5 × 325 = 1,625
Therefore, Product of the number = Product of their HCF and LCM (Verified)

We know that two numbers are co-primes if their HCF is 1.
59 and 97
Here, dividend = 97 and divisor = 59

Clearly, the last divisor is 1.
Hence, the given numbers are co-primes.

We have to find the largest number which divides (615 - 6) and (963 - 6) exactly.
Therefore, the required number = HCF of 609 and 957
Resolving 609 and 957 into prime factors, we have:
609 = 3 × 7 × 29
957 = 3 × 11 × 29
Therefore, HCF of 609 and 957 = 29 × 3 = 87
Hence, the required largest number is 87.

Given: HCF of two numbers = 16
Product of these two numbers = 3,072
Product of the two numbers = Product of their HCF and LCM
Therefore, 3,072 = 16 × LCM
LCM = 307216 = 192
Thus, the required LCM is 192.

Rule: A number is divisible by 11 if the difference of the sums of the alternate digits is either 0 or a multiple of 11.
467 × 91
Sum of the digits at the odd places = 4 + 7 + 9 = 20
Sum of the digits at the even places = 6 + missing number + 1 = missing number + 7 Difference = 20 - [missing number + 7] = 13 - missing number
According to rule, 13 - missing number = 11 [Because the missing number is a single digit]
Thus, missing number = 2
Hence, the smallest required number is 2.

Given numbers are 490 and 1155.
Prime factorization of 490 = 2 × 5 × 7 × 7
Prime factorization of 1155 = 3 × 5 × 7 × 11
HCF of 490 and 1155 = 35
LCM of 490 and 1155 = 2 × 3 × 3 × 5 × 7 × 7 × 11 = 16710
Product of the given number = 490 × 1155 = 5,65,950
Product of their HCF and LCM = 35 × 16,170 = 5,65,950
Therefore, Product of the number = Product of their HCF and LCM (Verified)

Given numbers are 35 and 40.
Prime factorization of 35 = 5 × 7
Prime factorization of 40 = 2 × 2 × 2 × 5
HCF of 35 and 40 = 5
LCM of 35 and 40 = 2 × 2 × 2 × 5 × 7 = 280
Product of the given number = 35 × 40 = 1400
Product of their HCF and LCM = 5 × 280 = 1400
Therefore, Product of the number = Product of their HCF and LCM (Verified)

HCF of two numbers = 145
LCM of two numbers = 2,175
One of the given numbers = 725
Product of the given two numbers = Product of their LCM and HCF
Therefore, 725 x other number = 145 × 2,175
Other number = 145 × 2175725 = 435
Thus, the required number is 435.

First, we have to find the LCM of 28 and 32.
Prime factorization of 28 = 2 × 2 × 7
Prime factorization of 32 = 2 × 2 × 2 × 2 × 2
Therefore, Required LCM = 2 × 2 × 2 × 2 × 2 × 7 = 224
It is given that when we divide the number by 28, the remainder is 8 and when we divide the number by 32, the remainder is 12.
We observe:
28 - 8 = 20
32 - 12 = 20
Therefore, Required number = 224 - 20 = 204

We have to find the LCM of 220 m and 300 m.
Prime factorization of 220 = 2 × 2 × 5 × 11
Prime factorization of 300 = 2 × 2 × 3 × 5 × 5
Therefore, Required LCM = 2 × 2 × 3 × 5 × 5 × 11 = 3,300
Hence, 3,300 m far is the next heap that lies at the foot of a pole.

Multiple:When a number ‘a’ is multiplied by another number ‘b’, the product is the multiple of both the numbers ‘a’ and ‘b’.
Examples of multiples:
6 is a multiple of 2 because 2 × 3 = 6
8 is a multiple of 4 because 4 × 2 = 8
12 is a multiple of 6 because 6 × 2 = 12
21 is a multiple of 7 because 7 × 3 = 21

We have to find the largest number which divides (626 - 1), (3,127 - 2), and (15,628 - 3) exactly.
The required number will be given by the HCF of 625, 3,125 and 15,625.
Resolving 625, 3125, and 15625 into prime factors, we have:
625 = 5 × 5 × 5 × 5 3,
125 = 5 × 5 × 5 × 5 × 5 15,
625 = 5 × 5 × 5 × 5 × 5 × 5
Therefore, HCF of 625, 3125 and 15625 = 5 × 5 × 5 × 5 = 625 Hence, the required largest number is 625.

We have to find the largest possible number of animals. Thus, we will have to find the HCF of 105, 140, and 175.
Prime factorization of 105 = 3 × 5 × 7
Prime factorization of 140 = 2 × 2 × 5 × 7
Prime factorization of 175 = 5 × 5 × 7
Required HCF = 5 × 7 = 35 Hence, 35 animals went in each trip.

We have to find prime factorization of 24, 36, and 54.
Prime factorization of 24 = 2 × 2 × 2 × 3
Prime factorization of 36 = 2 × 2 × 3 × 3
Prime factorization of 54 = 2 × 3 × 3 × 3
Therefore, Required LCM=2 × 2 × 2 × 3 × 3 × 3 = 216
Thus, 216 is the smallest number exactly divisible by 24, 36, and 54.
To get the remainder as 5:
Smallest number = 216 + 5 = 221
Thus, the required number is 221.

Rule: A number is divisible by 11 if the difference of the sums of the alternate digits is either 0 or a multiple of 11.
86 × 72
Sum of the digits at the odd places = 8 + missing number + 2 = missing number + 10
Sum of the digits at the even places = 6 + 7 = 13
Difference = [missing number + 10 ] - 13 = Missing number – 3
According to the rule, missing number - 3 = 0 [Because the missing number is a single digit]
Thus, missing number = 3
Hence, the smallest required number is 3.

Since the number is divisible by 7 and 16, they are the factors of that number.
So, the number will be divisible by the common factor of 7 and 16.
The factors of 7 are 1 and 7.
The factors of 16 are 1, 2, 4, 8, and 16.
Therefore, the common factor of 7 and 16 is 1 and the number is divisible by 1.

We know that two numbers are co-primes if their HCF is 1.
288 and 1375
Here, dividend = 288 and divisor = 1375

Clearly, the last divisor is 1.
Hence, the given numbers are co-prime.

We have to find the prime factorization of 35, 56, and 91.
Prime factorization of 35 = 5 × 7
Prime factorization of 56 = 2 × 2 × 2 × 7
Prime factorization of 91 = 7 × 13
Therefore, Required LCM = 2 × 2 × 2 × 5 × 7 × 13 = 3,640
Thus, 3,640 is the smallest number exactly divisible by 35, 56, and 91.
To get the remainder as 7:
Smallest number = 3,640 + 7 = 3,647
Thus, the required number is 3,647.

We have to find the LCM of 32 and 36.
Prime factorization of 32 = 2 × 2 × 2 × 2 × 2
Prime factorization of 36 = 2 × 2 × 3 × 3
Required LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288
Therefore, Minimum number of books required = LCM of 32 and 36 = 288 books