Required number = H.C.F. of $\left(91 - 43\right)$, $\left(183 - 91\right)$ and $\left(183 - 43\right)$
= H.C.F. of 48, 92 and 140 = 4.
Since the numbers are co-prime, their HCF = 1
Product of first two numbers = 119
Product of last two numbers = 391
The middle number is common in both of these products. Hence, if we take HCF of 119
and 391, we get the common middle number.
HCF of 119 and 391 = 17
=> Middle Number = 17
First Number $=\dfrac{119}{17}=7$
Last Number $=\dfrac{391}{17}=23$
Sum of the three numbers = 7 + 17 + 23 = 47
Let the numbers be 37$ a $ and 37$ b $.
Then, 37$ a $ x 37$ b $ = 4107
$\Rightarrow$ ab = 3.
Now, co-primes with product 3 are $\left(1, 3\right)$.
So, the required numbers are $\left(37 \times 1, 37 \times 3\right)$ i.e., $\left(37, 111\right)$.
$\therefore$ Greater number = 111.
Let the numbers be 3$ x $, 4$ x $ and 5$ x $.
Then, their L.C.M. = 60$ x $.
So, 60$ x $ = 2400 or x = 40.
$\therefore$ The numbers are $\left(3 \times 40\right)$, $\left(4 \times 40\right)$ and $\left(5 \times 40\right)$.
Hence, required H.C.F. = 40.
Other number =$ \left(\dfrac{11 \times 7700}{275} \right) $= 308.
LCM of 4, 8, 10, 12, 15 and 20 = 120
120 seconds = 2 minutes
Hence all the six bells will ring together in every 2 minutes
Hence, number of times they will ring together in 60 minutes $=1+\dfrac{60}{2}=31$
If the remainder is same in each case and remainder is not given, HCF of the
differences of the numbers is the required greatest number.
34 - 24 = 10
34 - 28 = 6
28 - 24 = 4
Hence, the greatest number which divides 24, 28 and 34 and gives the same remainder
= HCF of 10, 6, 4
= 2
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
$\therefore$ Required number $\left(9999 - 399\right)$ = 9600.
LCM of fractions $=\dfrac{\text{LCM of Numerators}}{\text{HCF of Denominators}}$
LCM of $\dfrac{2}{3}$, $\dfrac{5}{6}$ and $\dfrac{4}{9}$
$=\dfrac{\text{LCM (2, 5, 4)}}{\text{HCF (3, 6, 9)}}$
=$\dfrac{20}{3}$
L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e.,46 min. 12 sec.
LCM of 12, 14, 18 and 22 = 2772
Hence the least number which will be exactly divisible by 12, 14, 18 and 22 = 2772
2772 ÷ 2 = 1386
1386 is the number which when doubled, we get 2772
Hence, 1386 is the least number which when doubled will be exactly divisible by 12, 14,
18 and 22.