Let the numbers be 3$ x $ and 4$ x $. Then, their H.C.F. = $ x $. So, $ x $ = 4.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.
Let the numbers be $2x$, $3x$ and $4x$
LCM of $2x$, $3x$ and $4x$ = $12x$
12x=240
$\Rightarrow x=\dfrac{240}{12}=20$
H.C.F of $2x$, $3x$ and $4x$ $=x=20$
L.C.M. of 8, 16, 40 and 80 = 80.
$ \dfrac{7}{8} $=$ \dfrac{70}{80} $; $ \dfrac{13}{16} $=$ \dfrac{65}{80} $; $ \dfrac{31}{40} $=$ \dfrac{62}{80} $
Since,$ \dfrac{70}{80} $>$ \dfrac{65}{80} $>$ \dfrac{63}{80} $>$ \dfrac{62}{80} $, so$ \dfrac{7}{8} $>$ \dfrac{13}{16} $>$ \dfrac{63}{80} $>$ \dfrac{31}{40} $
So,$ \dfrac{7}{8} $is the largest.
N = H.C.F. of $\left(4665 - 1305\right)$, $\left(6905 - 4665\right)$ and $\left(6905 - 1305\right)$
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = $\left( 1 + 1 + 2 + 0 \right)$ = 4
Required H.C.F. =$ \dfrac{H.C.F. of 9, 12, 18, 21}{L.C.M. of 10, 25, 35, 40} $=$ \dfrac{3}{1400} $
Let the numbers be $ a $ and $ b $.
Then, $ a $ + $ b $ = 55 and ab = 5 x 120 = 600.
$\therefore$ The required sum =$ \dfrac{1}{a} $+$ \dfrac{1}{b} $=$ \dfrac{a + b}{ab} $=$ \dfrac{55}{600} $=$ \dfrac{11}{120} $
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.
L.C.M. of 5, 6, 7, 8 = 840.
$\therefore$ Required number is of the form 840 k + 3
Least value of $ k $ for which [840$ k $ + 3] is divisible by 9 is $ k $ = 2.
$\therefore$ Required number = $\left(840 \times 2 + 3\right)$ = 1683.
L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
$\therefore$ Number to be added = $\left(60 - 37\right)$ = 23.
LCM of 8, 12, 15 and 20 = 120
Required Number = 120 + 5 = 125
Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;
First number =$ \left(\dfrac{551}{29} \right) $= 19; Third number =$ \left(\dfrac{1073}{29} \right) $= 37.
$\therefore$ Required sum = $\left(19 + 29 + 37\right)$ = 85.
Solution 1
LCM of 5, 6, 7 and 8 = 840
Hence the number can be written in the form (840k + 3) which is divisible by 9.
If k = 1, number = (840 × 1) + 3 = 843 which is not divisible by 9.
If k = 2, number = (840 × 2) + 3 = 1683 which is divisible by 9.
Hence 1683 is the least number which when divided by 5, 6, 7 and 8 leaves a remainder
3, but when divided by 9 leaves no remainder.
Solution 2 - Hit and Trial Method
Just see which of the given choices satisfy the given condtions.
Take 3363. This is not even divisible by 9. Hence this is not the answer.
Take 1108. This is not even divisible by 9. Hence this is not the answer.
Take 2007. This is divisible by 9.
2007 ÷ 5 = 401, remainder = 2 . Hence this is not the answer
Take 1683. This is divisible by 9.
1683 ÷ 5 = 336, remainder = 3
1683 ÷ 6 = 280, remainder = 3
1683 ÷ 7 = 240, remainder = 3
1683 ÷ 8 = 210, remainder = 3
Hence 1683 is the answer
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90$ k $ + 4, which is multiple of 7.
Least value of $ k $ for which [90$ k $ + 4] is divisible by 7 is $ k $ = 4.
$\therefore$ Required number = $\left(90 \times 4\right)$ + 4 = 364.
[Refer " How to calculate LCM and HCF of Decimals " here]
Step 1: Make the same number of decimal places in all the given numbers by suffixing
zero(s) in required numbers as needed.
=> 2.04, 0.24 and 0.80
Step 2: Now find the HCF of these numbers without decimal.
=> HCF of 204, 24 and 80 = 4
Step 3: Put the decimal point in the result obtained in step 2 leaving as many digits on its
right as there are in each of the numbers.
i.e., here we need to put decimal point in the result obtained in step 2 leaving two digits
on its right.
=> HCF of 2.04, 0.24 and 0.8 = 0.04
Given numbers are 1.08, 0.36 and 0.90. H.C.F. of 108, 36 and 90 is 18,
$\therefore$ H.C.F. of given numbers = 0.18.
Let the numbers 13$ a $ and 13$ b $.
Then, 13$ a $ x 13$ b $ = 2028
$\Rightarrow$ ab = 12.
Now, the co-primes with product 12 are $\left(1, 12\right)$ and $\left(3, 4\right)$.
[Note: Two integers $ a $ and $ b $ are said to be co-prime or relatively prime if they have no
common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are $\left(13 \times 1, 13 \times 12\right)$ and $\left(13 \times 3, 13 \times 4\right)$.
Clearly, there are 2 such pairs.