Help #56, #57, #59
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Hey fitzrobin.
Can you show your working?
Hint for 56 - You know that you have two digits [so number = 10*a + b where a and b are between 0 and 9 inclusive] and number = 5 * some other number.
If you can list the numbers and get a distribution [probability distribution] then you will be able to answer this quite easily.
Have you studied probability before?
You also have #58 wrong. The instructions are to "cross off each prime number as well as all the numbers in the diagonal extending up and to the right from that prime number". You have correctly crossed off the prime numbers but appear to have only crossed of the number immediately up and to the right but not all the numbers in that diagonal. For example, 19 is a prime number and the numbers in the diagonal up and to the right are 14, 9, and 4. You crossed off 14 but neither 9 nor 4 and included 9 and 4 in the sum when they should not have been.
#56 says "Mia Zhang wrote down all the two digit multiples of 5. What is the probability that one of these numbers, chosen at random, has exactly two prime numbers that are factors?
Okay, the "two digit multiples of 5" are 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95. A total of 18 numbers. How many of those number is a product of exactly two primes? You can answer that by looking at the prime factorizations of those numbers: 10= 2*5, 15= 3*5, 20= 4*5= 2*2*5, 25= 5*5, 30= 2*15= 2*3*5, 35= 5*7, 40= 4*10= 2*2*2*5, 45= 5*9= 5*3*3, 60= 2*25= 2*5*5, 65= 5*13, 70= 2*35= 2*5*7, 75= 3*25= 3*5*5, 80= 5*16= 2*2*2*2*5, 85= 5*17, 90= 9*10= 2*3*3*5, 95= 5*19.
How many of those 18 numbers have exactly two prime factors?
For 57, there are 25 numbers. The "median" is the "middle" number, the 13th number, 24. Of the 12 numbers less than that the "median" is half way between the 6th and 7th numbers. The 6th and 7th numbers are both 11 so the "lower quartile" is 11. Of the 13 numbers larger than 24, the 6th number and 7th numbers are both 37 so the "upper quartile" is 37. The "absolute difference" between them is 37- 11= 26.
59 asks for 1,000,000!- 999,999!. Do know what that "!" means? For any positive integer n, n! is defined as n times all the positive integers less than n. For example, 5!= 5(4)(3)(2)(1)= 120. In particular, for any positive integer n, "the product of all positive integers less than n" is (n-1)! so n!= n(n- 1)!. So 1,000,000!= (1,000,000)(999,999!) so 1,000,000!- 999,999!= (1,000,000)(999,999!)- (999,999!)= (1,000,000- 1)(999,999)= (999,999)(999,999!). I don't think that can be written in any simpler form.
I don't see any calculations here that would be difficult for a sixth grader (or for a sixth grader's mother) so I suspect the difficulty is with definitions. When you are helping your child, I recommend by specifically asking for the definition of any things you are not clear about. If your child does not know those definitions, then you know what to do- look up those definitions.
Actually, I think Halls misread problem 59.
The problem is
$1,000,000! \div 999,999! = \dfrac{1,000,000!}{999,999!}.$
But as Halls explained $n > 1 \implies n! = n * (n - 1)!.$
$\text {So } \dfrac{1,000,000!}{999,999!} = \dfrac{1,000,000 * \cancel {999,999!}}{\cancel {999,999!}} = 1,000,000.$