LCM = product/HCF; HCF is 6 so LCM = 12 x 42 = 504
C (251519) and B (3112331) are both prime. A (2359) is a multiple of 7 and D (7172949) is a multiple of 3. But, without the number for which they could be a factor, it is only possible to identify the prime numbers, not the one that is a prime factor.
What is the h c f of 51 63 and 21
To have a HCF, you need at least 2 numbers. There is no HCF in just 1 number.
If two numbers are expressed as ab and cb this is easier to work out. Assume that a and c have no common prime factors. Thus, the HCF of the two numbers will be b. The LCM is the two numbers multiplied by each other, divided by the HCF. So the LCM will be abc. b is a factor of abc, and so the HCF will always be a factor of their LCM.
No. The LCM MUST be a multiple of the HCF.
multiple choice: a a b a c c a d b b a c a b d d d b c a Thats ill i got
HCF - Highest Common Factor. LCM - Lowest Common Multiple.
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The SSD 3 module 1 exam has up to 27 multiple choice questions. The answers starting at 1 and ending at 27 are as follows: D, B, A, A, C, B, D, D, A, C, A, B, C, C, A, D, C, A, D, A, A, D, D, B, B, D, C.
The SSD 3 module 1 exam has up to 27 multiple choice questions. The answers starting at 1 and ending at 27 are as follows: D, B, A, A, C, B, D, D, A, C, A, B, C, C, A, D, C, A, D, A, A, D, D, B, B, D, C.
The SSD 3 module 1 exam has up to 27 multiple choice questions. The answers starting at 1 and ending at 27 are as follows: D, B, A, A, C, B, D, D, A, C, A, B, C, C, A, D, C, A, D, A, A, D, D, B, B, D, C.
The SSD 3 module 1 exam has up to 27 multiple choice questions. The answers starting at 1 and ending at 27 are as follows: D, B, A, A, C, B, D, D, A, C, A, B, C, C, A, D, C, A, D, A, A, D, D, B, B, D, C.
The SSD 3 module 1 exam has up to 27 multiple choice questions. The answers starting at 1 and ending at 27 are as follows: D, B, A, A, C, B, D, D, A, C, A, B, C, C, A, D, C, A, D, A, A, D, D, B, B, D, C.
c) 882
882
You need at least two numbers to find either of those.