Ans .
3
Using Rule 1,
Required number
= LCM X HCF
First number
= 864 x 144 = 432
288
Ans .
3
Using Rule 1, LCM × HCF = 1st Number x 2nd Number
225 x 5 = 25 x X
X = 225 x 5 = 45
25
Ans .
3
Given that L.C.M. of two numbers = 1820
H.C.F. of those numbers = 26
One of the number is 130
Another number = 1820 x 26 = 364
130
Ans .
2
Using Rule 1, We have,
First number × second number
= LCM × HCF
Second number = 1920 x 16 = 240
128
Ans .
1
Using Rule 1,
Product of two numbers = HCF × LCM
= 12906 × 14818
= LCM x 478
LCM = 12906 x 14818
478
= 400086
Ans .
4
Using Rule 1, H.C.F. of the two 2-digit numbers = 16
Hence, the numbers can be expressed as 16x and 16y, where x and y are prime to each other.
Now, First number × second number
= H.C.F. × L.C.M.
16x × 16y = 16 × 480
xy = 16 480 = 30
16 x 16
The possible pairs of x andy, satisfying the conditionxy = 30 are : (3, 10), (5, 6), (1, 30), (2, 15)
Since the numbers are of 2-digits each. Hence, admissible pair is (5, 6)
Numbers are : 16 × 5 = 80
Ans .
2
Using Rule 1,
We know that,
First number × Second number = LCM × HCF
Second number
16 x 160 = 80
32
Ans .
2
LCM = Product of two numbers
HCF
4107 = 111
37
Obviously, numbers are 111 and 37 which satisfy the given condition. Hence, the greater number = 111
Ans .
2
Using Rule 1,
First number × Second number = HCF × LCM
Second number
15 x 300 = 75
60
Ans .
3
Let the numbers be 12x and 12y where x and y are prime to each other.
LCM = 12xy
12xy = 924
xy = 77
Possible pairs = (1,77) and (7,11)
Ans .
3
Using Rule 1,
First number × second number = LCM × HCF Let the second number be x.
10x = 30 × 5
x = 30 x 5 = 15
10
Ans .
1
Using Rule 1,
HCF × LCM = Product of two numbers
8 × LCM = 1280
LCM = 1280 = 160
160
Ans .
4
Using Rule 1,
First number × second number = HCF × LCM
24 × second number = 8 × 48
Second number = 8 x 48 = 16
24
Ans .
2
Using Rule 1,
First number × second number = HCF × LCM
84 × second number = 12 × 336
Second number
= 12 x 336 = 48
84
Ans .
4
Let the numbers be 6x and 6y where x and y are prime to each other.
6x × 6y = 216
xy = 216 = 6
6 x 6
LCM = 6xy = 6 × 6 = 36
Ans .
1
Using Rule 1,
Second number = HCF x LCM
First number
18 x 378 = 126
54
Ans .
3
Let the number be 15x and 15y, where x and y are co –prime.
15x × 15y = 6300
xy = 6300 = 28
15 x 15
So, two pairs are (7, 4) and (14, 2)
Ans .
4
Using Rule 1,
First number × Second number = HCF × LCM
75 × Second number = 15 × 225
Second number = 15 x 225 = 45
75
Ans .
1
Using Rule 1,
First number × second number = HCF × LCM
52 × second number = 4 x 520 = 40
52
Ans .
4
Using Rule 1,
First number × Second number = HCF × LCM
864 × Second number = 96 × 1296
Second number = 96 x 1296 = 144
864
Ans .
2
Using Rule 1,
Let LCM be L and HCF be H, then L = 4H
H + 4H = 125
5H = 125
H = 125 = 25
5
L = 4 × 25 = 100
Second number = L x H
First number
= 100 x 25 = 25
100
Ans .
1
HCF of two-prime numbers = 1
Product of numbers = their LCM = 117
117 = 13 × 9 where 13 & 9 are co-prime. L.C.M (13,9) = 117.
Ans .
2
HCF = 12 Numbers = 12x and 12y where x and y are prime to each other.
12x × 12y = 2160
xy = 2160
12 x 12
= 15 = 3 × 5, 1 × 15 Possible pairs = (36, 60) and (12, 180)
Ans .
1
Using Rule 1, Second number
= H.C.F. x L.C.M.
First Number
= 27 x 2079 = 297
189
Ans .
2
Here, HCF = 13 Let the numbers be 13x and 13y where x and y are Prime to each other.
Now, 13x × 13y = 2028
xy = 2028 = 12
13 x 13
The possible pairs are : (1, 12), (3, 4), (2, 6) But the 2 and 6 are not co-prime.
The required no. of pairs = 2
Ans .
2
HCF = 13 Let the numbers be 13x and 13y.
Where x and y are co-prime.
LCM = 13 xy
13 xy = 45
xy = 455 = 35 = 5 x 7
13
Numbers are 13 × 5 = 65 and 13 × 7 = 91
Ans .
4
HCF of two numbers is 8. This means 8 is a factor common to both the numbers. LCM is common multiple for the two numbers, it is divisible by the two numbers. So, the required answer = 60
Ans .
4
Let the numbers be 23x and 23y where x and y are co-prime.
LCM = 23 xy As given, 23xy = 23 × 13 × 14
x = 13, y = 14
The larger number = 23y = 23 × 14 = 322
Ans .
4
LCM = 2 × 2 × 2 × 3 × 5 Hence, HCF = 4, 8, 12 or 24 According to question 35 cannot be H.C.F. of 120.
Ans .
3
Using Rule 1,
First number = 2 × 44 = 88
First number × Second number = H.C.F. × L.C.M.
88 × Second numebr = 44 × 264
Second number = 44 x 264 = 13
88
Ans .
3
Using Rule 4, L.C.M. of 4, 6, 8, 12 and 16 = 48
Required number = 48 + 2 = 50
Ans .
4
Using Rule 4, LCM of 15, 12, 20, 54 = 540
Then number = 540 + 4 = 544 [4 being remainder]
Ans .
4
Using Rule 4, The greatest number of five digits is 99999. LCM of 3, 5, 8 and 12
2 3, 5, 8, 12
2 3, 5, 4, 6
3 3, 5, 2, 3
1, 5, 2, 1
LCM = 2 × 2 × 3 × 5 × 2 = 120 After dividing 99999 by 120, we get 39 as remainder 99999 – 39 = 99960 = (833 × 120) 99960 is the greatest five digit number divisible by the given divisors. In order to get 2 as remainder in each case we will simply add 2 to 99960.
Greatest number = 99960 + 2 = 99962
Ans .
3
Using Rule 4, LCM of 4, 5, 6, 7 and 8
= 2 4, 5, 6, 7, 8
2 2, 5, 3, 7, 4
1, 5, 3, 7, 2
= 2 × 2 × 2 × 3 × 5 × 7 = 840.
let required number be 840 K + 2 which is multiple of 13. Least value of K for which (840 K + 2) is divisible by 13 is K = 3
Required number = 840 × 3 + 2 = 2520 + 2 = 2522
Ans .
4
Required time = LCM of 252, 308 and 198 seconds
2 252, 308, 198
2 126, 154, 99
7 63, 77, 99
9 9, 11, 99
11 1, 11, 11
1, 1, 1
LCM = 2 × 2 × 7 × 9 × 11 = 2772 seconds = 46 minutes 12 seconds
Ans .
2
15 = 3 × 5
18 = 32 × 2
21 = 3 × 7
24 = 23 × 3
LCM = 8 × 9 × 5 × 7 = 2520 The largest number of four digits = 9999
2520 ) 9999 ( 3
7560
2439
Required number = 9999 – 2439 – 4 = 7556 (Because 15 – 11 = 4
18 – 14 = 4
21 – 17 = 4
24 – 20 = 4)
Ans .
3
LCM of 21, 36 and 66
LCM = 3 × 2 × 7 × 6×11 = 3 × 3 × 2 × 2 × 7 × 11
Required number = 32 × 22 × 72 × 112 = 213444
Ans .
1
Using Rule 5, Here 4 – 1 = 3, 5 – 2 = 3, 6 – 3 = 3
The required number =
LCM of (4, 5, 6) – 3
= 60 – 3 = 57
Ans .
2
LCM of 4, 6, 10, 15 = 60 Least number of 6 digits = 100000
The least number of 6 digits which is exactly divisible by 60 = 100000 + (60 – 40) = 100020
Required number (N) = 100020 + 2 = 100022 Hence,
the sum of digits = 1 + 0 + 0 + 0 + 2 + 2 = 5
Ans .
3
The LCM of 12, 18, 21, 30
2 12, 18, 21, 30
3 6, 9, 21, 15
2, 3, 7, 5
LCM = 2 × 3 × 2 × 3 × 7 × 5 = 1260
The required number
1260 = 630
2
Ans .
4
2 10, 16, 24
2 5, 8, 12
2 5, 4, 6
2 5, 2, 3
3 5, 1, 3
5 5, 1, 1
1, 1, 1
LCM = 22 × 22 × 3 × 5
Required number = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 3600
Ans .
2
Required number of students =
LCM of 6, 8, 10 = 120
Ans .
2
LCM of 4, 6, 8, 9
2 4, 6, 8, 9
3 2, 3, 4, 9
3 1, 3, 2, 9
1, 1, 2, 3
LCM = 2 × 2 × 3 × 2 × 3 = 72
Required number = 72, because it is exactly divisible by 4, 6, 8 and 9 and it leaves remainder 7 when divided by 13
Ans .
2
Using Rule 5, Here, 12 – 5 = 7, 16 – 9 = 7
Required number = (L.C.M. of 12 and 16) – 7 = 48 – 7 = 41
Ans .
3
Using Rule Number 5,
Here, Divisor – remainder = 1 e.g., 10 – 9 = 1, 9 – 8 = 1, 8 – 7 = 1
Required number = (L.C.M. of 10, 9, 8) –1 = 360 – 1 = 359
Ans .
4
We find LCM of 5, 6 and 8
5=5
6=3×2
8=23
= 23×3 × 5 = 8 × 15 = 120 Required number = 120K + 3
when K = 2, 120 × 2 + 3 = 243 required no. It is completely divisible by 9
Ans .
4
LCM of 16, 18, 20 and 25 = 3600
Required number = 3600K + 4 which is exactly divisible by 7 for certain value of K. When K = 5, number = 3600 × 5 + 4 = 18004 which is exactly divisible by 7.
Ans .
2
LCM of 3, 5, 6, 8, 10 and 12 = 120
Required number = 120x + 2, which is exactly divisible by 13. 120x + 2 = 13 × 9x + 3x + 2 Clearly 3x + 2 should be divisible by 13. For x=8,3x + 2 is divisible by 13.
Required number = 120x + 2 = 120 × 8 + 2 = 960 + 2 = 962
Ans .
4
LCM of 6, 9, 15 and 18
2 6, 9, 15, 18
3 3, 9, 15, 9
3 1, 3, 5, 3
1, 1, 5, 1
LCM = 2 × 3 × 3 × 5 = 90
Required number = 90k + 4, which must be a multiple of 7 for some value of k. For k = 4, Number = 90 × 4 + 4 = 364, which is exactly divisible by 7.
Ans .
2
Using Rule 9, We will find the LCM of 16, 24, 30 and 36.
2 16, 24, 30, 36
2 8, 12, 15, 18
2 4, 6, 15, 9
3 2, 3, 15, 9
2, 1, 5, 3
LCM = 2 × 2 × 2 × 3 × 2 × 5 × 3 = 720 The largest number of five digits = 99999 On dividing 99999 by 720, the remainder = 639
The largest five-digit number divisible by 720 = 99999 – 639 = 99360
Required number = 99360 + 10 = 99370
Ans .
2
LCM of 5, 10, 12, 15
2 5, 10, 12, 15
3 5, 5, 6, 15
5 5, 5, 2, 5
1, 1, 2, 1
LCM = 2 × 3 × 5 × 2 = 60
Number = 60k + 2 Now, the required number should be divisible by 7. Now, 60k + 2 = 7 × 8k + 4k + 2 If we put k = 3, (4k + 2) is equal to 14 which is exactly divisible by 7.
Required number = 60 × 3 + 2 = 182
Ans .
3
LCM of 9, 10 and 15 = 90 Þ The multiple of 90 are also divisible by 9, 10 or 15.
21 × 90 = 1890 will be divisible by them.
Now, 1897 will be the number that will give remainder 7. 1936 – 1897 Required number = 1936 – 1897 = 39
Ans .
1
The difference between the divisor and the corresponding remainder is same in each case ie. 18 – 5 = 13, 27 – 14 = 13, 36 – 23 = 13
Required number = (LCM of 18, 27, and 36 ) – 13 = 108 – 13 = 95