Find the least possible value of n such that 300 is the LCM of N 15 and 10
Answers
Answer:
The prime factors of 45 are 3, 3, 5. That means the two numbers for which the LCM is 45 have a combined set of prime factors - discarding duplicates - of 3, 3, 5.
The prime factors of 15 are 3, 5. That explains where the 5 and one of the 3’s came from in the LCM of 45.
What remains is what requires not one, but two 3’s. Fortunately, it’s not difficult to figure out. 3 x 3 = 9.
The other number that you’re looking for is 9.
3x3x5 = 45 (LCM)
----------------
3 x 5 = 15
3x3 = 9 (n)
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n=9.
3x15=45
15 is a multiple of 3, so it’s not a valid solution.
Prime factors of 45 are 1, 3 and 5, so we’re looking for a multiple of one of these. This is because prime numbers are numbers which have no factors except 1 and themselves, so they give us “atoms” to build with, while looking for a number of the given description.
10 is even, 15 is a multiple of 5, so those are out. 6 is even, but 9 is odd and, when multiplied by 5, is equal to 45. Thus, 9 is our solution.
We know that LCM(x,n) = x*n / GCD(x,n)
So, n = LCM(x,n) * GCD(x,n) / x
Since x and LCM(x,n) are given, we can minimize n by minimizing GCD(x,n)
We can do this by ensuring n has only the necessary factors to produce 45 that 15 does not.
45/15 = 3. So, the only prime factor n should have in this case is 3.
Since 45 is divisible by 9 but not 27, the smallest value for n in this case would be 9.
There are 2 possible answers:
45 = 3 * 3 * 5
and
15 = 3 * 5
so n must be a multiple of 9 (to produce the extra factor of 3) and can, if we so choose, also be a multiple of 5.
So the possible answers are 9 and 45.
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Smallest value of n such that the LCM of n and 15 is 45.
Step 1: Prime factorization of 45:
45 →9 * 5 →3 * 3 * 5
Step 2: LCM of 15 and 3 = 15
Step 3: LCM of 15 and 5 = 15
Step 4: LCM of 15 and 9 = 45 (check)
QED
LCM of any 2 given numbers will be the largest of the two or larger than the largest number.
Keeping this in mind, we can notice immediately that the LCM has to be 45 or bigger than 45. As 15x3 = 45, and 45 x1 = 45, we can safely conclude that 45 is the smallest value of number n
If m is the LCM of a and b , then m will be the minimum number divisible by a and b .
LCM =45 =3*3*5
one number =15 =3*5
45÷15=3 , 3 is one quotient , so other quotient will be 5 , so other number will be 3*3=9
hence the answer
If m be the LCM of a and b , then m will be the minimum number divisible by a and b ,and the quotients are coprimes .
LCM =45 =3×3×5
one number =15 ,
45÷15=3 , 3 is one quotient , so other quotient will be 5 , so other number will be 3×3=9
LCM of 45
45=5×3×3
Prime factorisation of
15= 1×3×5
9 =3×3
LCM of 15 and 9
3×3×5=45
Smallest number is 9
Using the J programming language:
n#~45=15*.n=.i.30
9
The answer is 9
<<<>>>
Check:
15 *. 9
45
The possible values of n are factors of 45. The factors of 45 are 1, 3, 5, 9, 15, 45. So we see that n can be 45 or 9.
Step-by-step explanation:
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