where p and q are prime.

Those divisors are 1,2,4,8,16,32,64 ----- If you need more help, or if you have any questions about the problem, feel free to email me at jim_thompson5910@hotmail.comjim Given that is the factorial base expansion of , find the value of . OR. What positive integers has exactly (and prove your result) (a) one positive divisor; (b) two positive divisors; (c) three positive divisors; (d) four positive divisors; (e) five positive divisors. My point being that you $24=4 \times 3 \times 2 \Rightarrow 2^{3} \times 3^2 \times 5^1$$24=6 \times 2 \times 2 \Rightarrow 2^{5} \times 3^1 \times 5^1$$24=3 \times 2 \times 2 \times 2 \Rightarrow 2^{2} \times 3^1 \times 5^1 \times 7^1$But why is the prime factorisation not working?It works in other cases.........like if we do for 60 divisors.I don't understand that if 1 and the number itself are included as divisors ,then will it change the process somehow?Because $2^3 \times 3^2 \times 5^1 <2^2 \times 3^1 \times 5^1 \times 7^1$ which is ultimately because $2^1 \times 3^1 < 7^1$.It's nothing to do with the fact that $1$ and $n$ are included as divisors.Oh i got it! Can you see where does your reasoning fail?$2^{23} $ will have 24 factors as 24=24 as well as 2.2.2.3. Discuss the workings and policies of this site The answer is 64 = 2 (7 − 1) which means seven is one of the most boring possible numbers to choose for this question. Solution. This happened recently with a very similar question, where it was 9 "factors" that were wanted.When someone mentions "factors" of an integer, N, it can be taken to mean that you've split N into a product of smaller integers, and those are the factors being referred to, e,g,:In other words, that N is exactly equal to the product of its factors, and there will usually be different ways to split N into factors.So given this, one possible answer to your question would bebecause that's the smallest N>0 that can be broken into 20 factors (each ≠ 1, because if you allow 1, then every N, including 1, will have an unlimited number of factors).Another answer is obtained by taking the meaning to be, that you want a product of integers > 0, that are all different (so now we can allow 1, because it can only be used once), and then the smallest such N isBut I suspect what you are looking for is the smallest N>0 with 20 different *divisors* -- and that will be found by considering the number of divisors N will have, based on its prime factorization. From here it should be obvious that the smallest such number is 2^6 * … For [math]n \in \mathbb N[/math], let [math]d(n)[/math] denote the number of positive divisors of [math]n[/math]. If one of the parallel sides is longer than the other by 6cm, find the lengths of two parall n = 20! where $a=7$, $b=5$, $c=3$ and $d=2$. Find possible values of the constant b?Divide p(x) by g(x) if, p(x)=x³+31/6*x²-2x+4 and g(x)=1+3x/2The area of a trapezium is 105cm² and its height is 7 cm. Examples: Input : N = 16 Output : 4 9 4 and 9 have exactly three divisors. a) We could have two primes that divide the integer . Find . Then, what is the next one after that (second smallest positive number with exactly ten positive integer divisors). So given that this product is 20, its possible factorizations are[Note that you can tack on an arbitrary number of other prime-to-the-0-power-equals-1 terms, without changing D(N). Step-by-step explanation: The proper divisors of 12 are. So Given a number N, print all numbers in range from 1 to N having exactly 3 divisors. Learn more about Stack Overflow the company The best answers are voted up and rise to the top

If p, q, and r are different primes, thendivisors.

prime factors of $24$ are : $(2,2,2,3)$ Problem 14.


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