Web1 de nov. de 2024 · Answer: Maximum power of 6 in 120! is given by Highest power of 3 in 120! Hence the maximum power of 6 which can divide 120! without leaving a remainder … WebFind the prime factors of 156, and state the highest power of 2 in it. Easy Solution Verified by Toppr Correct option is A) The prime factors are 156=2×2×13×3 156=2 2×3×13 Hence, the highest power of 2 is 2. Was this answer helpful? 0 0 Similar questions Find H.C.F. 1836,810,1296 and 702. Medium View solution > State true or false:

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Web29 de out. de 2014 · You must find the greatest power of 3 that divides 73!, since there is much more 2 in 73! rather than 3. For finding the greatest power of 3 in 73! you can refer to Highest power of a prime p dividing N! So it is equal to ⌊ 73 3 ⌋ + ⌊ 73 9 ⌋ + ⌊ 73 27 ⌋ = 24 + 8 + 2 = 34. Hence the greatest power of 6, that divides 73! is 34. grangemouth port ais

What is the highest power of 12 that divides 54!?

Web12 de jan. de 2015 · All but the first four terms are eliminated, since $4$ divides each of them (as each has $36$ as a factor). However, the first four terms, mod $4$ are $1,\,1,\,2,\,2$, which sums to $2$ mod $4$. Ergo, $2^{33}$ does not divide the original expression, and $2^{32}$ is thus the maximum power of two dividing the expression. WebSince there were 100 even numbers so u must have extracted 2 exactly 100 times so those even numbers will now look like (2^100 ) ×100! And now we can write. 200! = (2^100) × … Web10 de dez. de 2024 · The process is as follows: we get the prime factorizations and multiply the highest power of all factors present. In the example above, the LCM would be 2 × 2 … grangemouth plant