POTW 23 is below:
Thursday, February 18, 2016
POTW #23 -
The final answer for POTW 22 saw some discrepancy. It was interesting to see that how many decimal numbers you hold for Pi (e.g. 3.14 vs. 3.14159) would change your decimal answer a tad. The answer was indeed 82.3cm squared (or 82.27 or some other variety, of 82.26, depending on your digits of Pi).
POTW 23 is below:
POTW 23 is below:
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To start off, i will draft all the prime years between 5 - 18. 5, 7, 11, 13, 17.
ReplyDeletethen using common sense, simply to get to 40, i would need 3 double digit numbers that add up to 40. there is only 3 double digit numbers. 17, 13 and 11. 17 + 13 = 30, 30 + 11 = 41. the children's 3 ages are 11, 13 and 17 years old.
First of all to name some prime numbers:
ReplyDelete3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41
To find a number that has a 16 difference
3 + 19 = 22 41-22= 19
*7 + 23 = 30* 41-30 = 11
13 + 29 = 32 41-32 = 9
The 7, and 23 combination is the only one that leaves a prime number for the 3rd child. That gives me the conclusion that the child's ages are 7, 11, and 23.
To answer this question I first find the prime numbers up to fifty which are:
ReplyDelete2,3,4,7,11,13,17,19,23,29,31,37,41,43,47
Looking at the question their ages have to sum up to 41. Inferring that the 16 year difference is between the youngest child and the oldest child I found 2 numbers with the difference of 16. I got a total of 2 pairs. They are:
3,19 and 7,23
I added both up to get:
22 and 30
I subtracted both numbers from 41 to get:
19 and 11.
Because I already used 19 as my oldest age for the first pair 11 is the valid number and is the middle of 7 and 23. Thus the first child's age is 7 years old, the second is 11 years old and the oldest is 23 years old.
Since there were 3 prime ages to add up to 41, I divided 41 by 3 and the closest prime number was 13.
ReplyDeletea+b+c=41
a+16=b
a+a+16+c=41 (c=13 in this case)
a+a+16+13=41
2a=12
a=6 (not prime so it's wrong)
This didn't work so I tried the same equation with the closest prime number which was 11.
a+a+16+11=41
2a=14
a=7 (correct answer because 7 is a prime number)
Therefore, the 3 ages are 7, 23(7+16), and 11.
Prime numbers in order from least to greatest...
ReplyDelete3,5,7,11,13,19,23,29,31,37 <- Must be less than 41 <- Sum of ages
Child 1 = 7 + 16 = Child 2 = 23
23 + 7 = 30
41 - 30 = 11 <- Child 3
So...
Child 1 = 7
Child 2 = 23
Child 3 = 11
Check...
7, 23, 11 are all prime numbers
23 - 7 = 16 <- 16 year difference between 2 of the children
7 + 23 + 11 = 41 <- Their ages add up to 41
Final answer:
In the family of 3 children, one is 7, the other is 23, and the last one is 11 years old.
i started to list sets of numbers with a difference of 16 starting with the lowest prime numbers.
ReplyDelete2 and 18 (doesn't work 18 is composite)
3 and 19 (works - that was pretty fast)
I found 3 and 19 which are both prime and a have a difference of 16.
the sum of the ages had to be 41.
3+19=22
41-22=19
19 as established is a prime number
so the ages of the children are 3,19, 19
The ages of the children are 7, 11, 23, as neither are composite numbers, all are different, 7 + 16 = 23, and 7 + 11 + 23 = 41.
ReplyDeleteHard Copy
ReplyDeleteFirst I looked at what prime numbers there are.
ReplyDelete2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41.
Then I took 16 and prime numbers and added it together to get another prime number.
16 + 7 = 23.
This means that 2 of the prime numbers are 7 and 23.
Now I do 23+7=30.
41-30=11
So all the numbers are 7, 23, 11.
11+23+7=41
We can start by listing all of the prime numbers less than 41. The possible
ReplyDeleteprimes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, and 37. We could actually
eliminate some of the higher primes from this list since there are three different
primes in the sum.
Now we will look for all prime pairs from this list that differ by 16. The pairs
include 3 and 19, 7 and 23, and 13 and 29. We will look at each of these pairs
and determine the third number so that the sum of the three ages is 41.
For the pair 3 and 19, the sum is 22. The third age would be 41 - 22 = 19. The
ages of the three children would be 3, 19, and 19 but all three children have a
different age. Therefore this solution is not acceptable.
For the pair 7 and 23, the sum is 30. The third age would be 41 - 30 = 11. The
ages of the three children would be 7, 11, and 23. Each of these numbers is a
different positive prime. Therefore this is a possible solution. But are there other
solutions?
For the pair 13 and 29, the sum is 42. The sum is already over 41 so this is not a possible solution.
Therefore, the only possible ages for the children are 7, 11 and 23.
Oh, my answer didn't post so let me just say how I answered it here:
ReplyDeleteI listed the prime numbers that are less than 41:
2,3,5,7,11,13,17,19,23,29,37,
then found the pairs that have a difference of 16:
3 &19, 7 &23, and 13 &19
13 and 19 isn't possible because the sum is over 41 (42) and for 3 and 19 the second number would be 19 but there are all a different age so it isn't possible, and for 7 and 23 the last number would be 11 and each are positive prime numbers.
Meaning that the ages of the children are 7 ,11, and 23