$12= 2 \times 2 \times 3$
$16 = 2 \times 2 \times 2 \times 2$
$24 = 2 \times 2 \times 2 \times 3$
$36= 2 \times 2 \times 3 \times 3$
LCM = $2 \times 2 \times 2 \times 2 \times 3 \times 3 =144$
$70 = 2 \times 5 \times 7$
$105 = 3 \times 7 \times 5$
$175 = 5 \times 5 \times 7$
HCF =$5 \times 7 =35$
Two prime numbers whose difference is 2 are called twin primes.
For example:
3 and 5 are twin primes
5 and 7 are twin primes
A prime triplet is a set of three prime numbers of the form:
(p,p+2,p+6) or (p,p+4,p+6)
Some Examples are (5, 7, 11), (7, 11, 13), (11, 13, 17)
$1350 = 2 \times 3 \times 3 \times 3 \times 5 \times 5$
Prime factors in descending order
5 > 3 >2
a. 44
b. 125
c. 48
d. 64
53 = 13 + 17 + 23
or 53 = 7 + 17 + 29
We have to find the LCM of 80 cm, 85 cm and 90 cm
LCM =12240 cm
LCM of 12,16,24 and 36 =144
Least Number = 144 + 7 = 151
Multiples of 11 = 11,22,33,44,55,66,77,88
So multiples greater than 56 but less than 79 are 66 and 77
1.
a.
11= 11
44 =11 X 2 X 2
HCF =11
b.
35 = 5 x 7
42 = 7 x 3 x 2
HCF =7
c.
128= 2 x 2 x 2 x 2 x 2 x 2 x 2
60 = 2 x 2 x 3 x 5
HCF =4
d. 81 = 3 x 3 x3 x3
63 = 7 x 3 x 3
HCF =9
e.
36 =2 x 2 x3 x3
180 = 2 x 2 x 5 x 3 x 3
HCF = 12
f.
1152=2 x 2 x 2 x 2 x 2 x 2 x 2 x3 x3
1664=2 x 2 x 2 x 2 x 2 x 2 x 2 x 13
HCF= 128
g.
70 =2 x 5 x 3 x 3
115= 23 x 5
175 = 5 x 5 x7
HCF = 5
h.7
i. 9
j. 1
a.
$4=2 \times 2$
$7= 7 \times 1$
$12 =2 \times 2 \times 3$
$84= 2 \times 2 \times 3 \times 7$
LCM =$2 \times 2 \times 3 \times 7 =84$
b.
$25 = 5 \times 5$
$15 = 5 \times 3$
$36 = 2 \times 2 \times 3 \times 3$
LCM =$2 \times 2 \times 3 \times 3 \times 5 \times 5 =900$
c.
$24=2 \times 2 \times 2 \times 3$
$36= 2 \times 2 \times 3 \times 3$
$40= 2 \times 2 \times 2 \times 5$
LCM =$2 \times 2 \times 2 \times 3 \times 3 \times 5= 360$
d.
$27 = 3 \times 3 \times 3$
$36=2 \times 2 \times 3 \times 3$
LCM =$2 \times 2 \times 3 \times 3 \times 3 = 108$
e. 72
f. 420
24 small bottles and 36 large bottles
HCF of 24 and 36 = 12
Largest number of equal-sized groups = 12
30 cm and 45 cm ribbons
HCF of 30 and 45 = 15
Greatest possible length of each piece = 15 cm
84 science books and 108 mathematics books
HCF of 84 and 108 = 12
Maximum number of books on each shelf = 12
To solve this problem, we need to find the Highest Common Factor (HCF) of the three given quantities: 391 litres, 483 litres, and 667 litres. The HCF will give us the maximum capacity of a container that can measure the diesel in each of these tankers an exact number of times.
391 = 17 × 23
483 = 3 × 7 × 23
667 = 23 × 29
So the HCF is 23.
Prime factorization of 1080: $2^3 \times 3^3 \times 5$
Prime factorization of 792: $2^3 \times 3^2 \times 11$
Therefore, the HCF is =8×9=72.
The minimum number of each subject that should be placed on each shelf is 72 books.
To find when they all blink together again, find the LCM of 4, 6, and 8.
Prime factorization of 4: 2 x 2
Prime factorization of 6: 2 x 3
Prime factorization of 8: 2 x 2 x 2
LCM = 24 seconds.
All lights will blink together again after 24 seconds.
Six bells commence tolling together and toll at intervals of 2,4,6,8,10,12 minutes respectively.
Find LCM of using prime factorization :
2=2
4=2 x 2
6 =2 x 3
8 =2 x 2 x2
10= 2 x 5
12= 2 x 2 x 3
Therefore LCM (2,4,6,8,10,12) = 120 min
After every 120 minutes = 2 hours , bells will toil together.
So required Number of times = 13
To find the minimum number of square slabs to cover the floor, we have to find the greatest size of each such slab. For this purpose, we have to find the HCF of 450 and 300.
(Since 4.5m = 450cm and 3m = 300cm)
Now HCF of 450 and 300 = 150
So the required size of the slab must be 150cm × 150cm.
Hence, the number of slabs required =Area of the floor/ Area of 1 Slab= 6
a.
$9=3 \times 3$
$12 =2 \times 2 \times 3$
HCF =3
b.
$10=2 \times 5$
$11 =1 \times 11$
LCM =110
c.
$4=2 \times 2$
$13 =1 \times 13$
LCM =52
d.
$6=2 \times 3$
$16 =2 \times 2 \times 2 \times 2$
HCF =2
e.
$17=1 \times 17$
$15 =3 \times 5$
LCM =255
f.
$6=2 \times 3$
$22 =2 \times 11$
HCF =2
g.
$5=5 \times 1$
$20 =2 \times 2 \times 5$
LCM =20
h.
$22=2 \times 11$
$66 =3 \times 2 \times 11$
HCF =22
i.
$8=2 \times 2 \times 2$
$16 =2 \times 2 \times 2 \times 2$
HCF =8
j.
$2=2$
$132 =2 \times 2 \times 3 \times 11$
HCF =2
k.
$30=2 \times 3 \times 5$
$6=2 \times 3$
LCM =30
k.
$25=5 \times 5$
$32 =2 \times 2 \times 2 \times 2 \times 2$
LCM =800