First, find all the prime factors of 210.
Prime factors of 210 are : 2, 3, 5, 7
Given, The GCF of M and 210 is 14. And GCF is obtained by multiplying common prime factors,
So, M must be a number with 2 and 7 as a prime factor because (2 * 7) =14 **
** So, any number with 2 and 7 as its prime factors are possible values of M. Some of them are 70, 98, 154, .............
Hello,
we know that to find the GCF we descompose the numbers and then we multiply the common terms with the lowest exponent. So we first descompose 210:
210 | 2 105 | 3 35 | 5 7 | 7 1
210= 2x3x5x7
Now one possible solution is 14 because:
14 | 2 7 | 7 1
As you can see 2 and 7 are common terms, so:
GCF= 2x7 GCF=14 --> It's correct
Then, if we analyze it, we'll realize that we have an infinitive number of solutions. The only requirement is that "M" must have 2 and 7 when we descompose it.
For example:
M= 2²x7x3x11 M=924
924 is another possible solution, it doesn't matter if 2 or 7 repeat, since I said the GCF is the multiplication of the common terms with the LOWEST exponent. That's all.
The possible values of M include multiples of 14, such as 14, 42, 70, and more, which maintain the GCF of 14 with 210. To find M , note that it must have 2 and 7 as factors alongside possible additional primes from 210. This means M can be expressed as M = 14 k , where k is an integer that incorporates factors of 3 and 5 as necessary.
;
[tex]\frac{3}{4} + \frac{2}{3} = \frac{(3 \times 3) + (2 \times 4)}{12} = \frac{9 + 8}{12} = \frac{17}{12} = 1 \frac{5}{12} [/tex]