Factorial – is the product of natural numbers, for example:

Mathematics

1 answer:

svp [43]3 years ago

8 0

Answer:

Step-by-step explanation:

11! = 11*10*9*8*7*6*5*4*3*2*1 = 3991600

15! = 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1=1.307674368 * $10^{12}$

also looks like this 1,307,674,368,000

20! = 20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2 = 2.432902008 * $10^{18}$

also looks like this 2,432,902,008,000,000,000

8!(7-3)! = 8!4! = (8*7*6*5*4*3*2)(4*3*2) = 967,680

10!(9-5)!13! = 10!4!13! =3,628,800*24*6,227,020,800 = 5.423187139* $10^{17}$

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Jobisdone [24]

The answer to this question would be: p+q+r = 2 + 17 + 39= 58

In this question, p q r is a prime number. Most of the prime number is an odd number. If p q r all odd number, it wouldn't be possible to get 73 since
odd x odd + odd= odd + odd = even
Since 73 is an odd number, it is clear that one of the p q r needs to be an even number.

There is only one odd prime number which is 2. If you put 2 in the r the result would be:
pq+2= 73
pq= 71
There will be no solution for pq since 71 is prime number. That mean 2 must be either p or q. Let say that 2 is p, then the equation would be: 2q + r= 73

The least possible value of p+q+r would be achieved by founding the highest q since its coefficient is 2 times r. Maximum q would be 73/2= 36.5 so you can try backward from that. Since q= 31, q=29, q=23 and q=19 wouldn't result in a prime number r, the least result would be q=17
r= 73-2q
r= 73- 2(17)
r= 73-34=39

p+q+r = 2 + 17 + 39= 58

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