Required Number = H.C.F of (91- 43), (183- 91) and (183-43)
= H.C.F of 48, 92, and 140 = 4
Required number of students = H.C.F of 1001 and 910 = 91
The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.
Number of cans of Maaza = 368/16 = 23
Number of cans of Pepsi = 80/16 = 5
Number of cans of Sprite = 144/16 = 9
The total number of cans required = 23 + 5 + 9 = 37 cans.
LCM of (80, 85, 90) can be found by prime factorizing them.
80 ? 2 × 2 × 2 × 2 × 5
85 ? 17 × 5
90 ? 2 × 3 × 3 × 5
L.C.M of (80,85,90) = 2 × 2 x 2 × 2 × 3 × 3 × 5 × 17
= 16 x 9 x 85
= 144 x 85
= 12240
L.C.M of (80,85,90) = 12240.
To get the least number of coconuts :
LCM = 30 => 30 + 1 = 31
L.C.M of 5, 6, 7, 8 = 840
Therefore, Required Number is of the form 840k+3.
Least value of k for which (840k+3) is divisible by 9 is k = 2
Therefore, Required Number = (840 x 2+3)=1683
LCD is nothing but Lowest or Least Common Denominator
Here LCD of 12 and 18 means LCD of two fractions with denominators 12 and 18 respectively.
Therefore, LCM of 12 & 18 = 6 x 3 x 2 = 36
The lowest common denominator or least common denominator (LCD) is the least common multiple (LCM) of the denominators of a set of fractions.
99 = 1 x 3 x 3 x 11
101 = 1 x 101
176 = 1 x 2 x 2 x 2 x 2 x 11
182 = 1 x 2 x 7 x 13
So, divisors of 99 are 1, 3, 9, 11, 33, .99
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.
Hence, 176 has the most number of divisors.
HCF of 210 and 55 is 5
Now, 210x5 + 55P = 5
=> 1050 + 55P = 5
=> 55P = -1045
=> P = -1045/55
=> P = -19.
Since their HCFs are 7, numbers are divisible by 7 and are of the form 7x and 7y
Difference = 14
=> 7x - 7y = 14
=> x - y = 2
product of numbers = product of their hcf and lcm
=> 7x * 7y = 441 * 7
=> x * y = 63
Now, we have
x * y = 63 , x - y = 2
=> x = 9 , y = 7
The numbers are 7x and 7y
=> 63 and 49
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