Rose Weaver

2023-02-23

How to graph $y=\mathrm{tan}\left(2x\right)$?

Cheyenne Lynn

Beginner2023-02-24Added 10 answers

If the potential number is N, it can be written as

N = 9k + 6

and N = 21l + 12

$\therefore 9k+6=21l+12\phantom{\rule{0ex}{0ex}}\Rightarrow 9k-21l=6$

or 3(3k - 7l) = 6

or 3k = 7l + 2 or $k=\frac{7l+2}{3}$

Therefore, set l to its lowest possible value such that k's value is an integer, or the numerator, in other words. (i. e., 7l+ 2) will be divisible by 3.

Thus at I =1, we get k = 3 (an integer). So the least possible number $N=9\times 3+6=21\times 1+12=33$.

The higher numbers can now be achieved by multiplying the LCM of 9 and 21 by 33. i.e., The general form of the number is 63m + 33. So the other number in the given range including 33 are 96, 159, 222, 285, 348, ..., 1104. Hence there are total 18 numbers which satisfy the given condition.

N = 9k + 6

and N = 21l + 12

$\therefore 9k+6=21l+12\phantom{\rule{0ex}{0ex}}\Rightarrow 9k-21l=6$

or 3(3k - 7l) = 6

or 3k = 7l + 2 or $k=\frac{7l+2}{3}$

Therefore, set l to its lowest possible value such that k's value is an integer, or the numerator, in other words. (i. e., 7l+ 2) will be divisible by 3.

Thus at I =1, we get k = 3 (an integer). So the least possible number $N=9\times 3+6=21\times 1+12=33$.

The higher numbers can now be achieved by multiplying the LCM of 9 and 21 by 33. i.e., The general form of the number is 63m + 33. So the other number in the given range including 33 are 96, 159, 222, 285, 348, ..., 1104. Hence there are total 18 numbers which satisfy the given condition.