SOLUTION: find four natural number in AP. such that their sum is 24 and their product is 945.

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Question 1103964: find four natural number in AP. such that their sum is 24 and their product is 945.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Different students would solve it different ways.
Different teachers would expect (maybe even require) different solutions.
Figure out what your teachers expects,
how you would solve the problem,
and how convenient it would be to please the teacher.
Some of the many possibilities are listed below.

THE FIFTH-GRADER WAY:
They must be easy small numbers to have a sum of 24=6%2B6%2B6%2B6 .
As the product, 945, is odd,
all four numbers must be odd
1%2B3%2B5%2B7=16 ,
3%2B5%2B7%2B9=24 ,
1%2B5%2B9%2B11=26 ,
and any other sum of four odd numbers would be larger than 24 .
So, is 3%2A5%2A7%2A9 equal to 945 ?
It is obviously a multiple of 3 , of 5 and of 9 ,
and 945=700%2B245=700%2B210%2B35 is also a multiple of 7 .
The four numbers must be 3 , 5 , 7 and 9 .
Let me check the product.
3%2A5%2A7%2A9=3%2A7%2A5%2A9=21%2A45=900%2B45=945
It checks, so the answer is highlight%283%29 , highlight%285%29 , highlight%287%29 , and highlight%289%29 .
MY WAY (I like primes and prime factors):
Dividing by prime numbers in order,
we can find the prime factorization of 945 to be
945=3%2A3%2A3%2A5%2A7 .
Four numbers in arithmetic progression (arithmetic sequence in the USA)
with an average of 24%2F4=6 will have 6 as their mean and median,
meaning that two of the four numbers will be less than 6,
and two will be more than 6.
The only integer factors that can be less than 6 are 3 and 5 ,
so re-writing 945 as a product of 4 integers,
945=3%2A3%2A3%2A5%2A7=+3%2A5%2A7%2A%283%2A3%29=3%2A5%2A7%2A9 ,
tells us the four natural numbers are
highlight%283%29 , highlight%285%29 , highlight%287%29 , and highlight%289%29 .

THE ALGEBRA-AS-USUAL WAY:
An AP could be 6, 6, 6, 6, ....,
but in this case the product of four consecutive terms would be 6%5E4=1296%3C%3E945 ,
so the numbers are not all the same;
one of them is less than the others.
Let a be the least of the natural numbers, and
d be the positive common difference
(the difference between the least and the closest of the other 3 numbers).
That makes the numbers
a , a%2Bd , a%2B2d , and a%2B3d .
The sum of the four numbers is
a%2Ba%2Bd%2Ba%2B2d%2Ba%2B3d=4a%2B6d .
The sum of the four numbers is
4a%2B6d=24 <--> 2a%2B3d=12
The product of the four numbers is
a%28a%2Bd%29%28a%2B2d%29%28a%2B3d%29=945 --> a%5E4%2B6a%5E3d%2B11a%5E2d%5E2%2B6ad%5E3=945 .
So the answer is a solution of
system%282a%2B3d=12%2Ca%5E4%2B6a%5E3d%2B11a%5E2d%5E2%2B6ad%5E3=945%29 ,
with a and d being integers,
and a%3E0 , so that it will be a natural number.

THE ALGEBRA-WITH-COMMON-SENSE WAY:
As above, we get to
2a%2B3d=12 ,
and common sense tells us that
a and d must be really common small numbers.

In more mathematically rigorous words,
as a%2Bd and a are both natural numbers, with d%3E0 ,
%28a%2Bd%29-a=d is a positive integer (a natural number).
As a is a natural number,
2a is an even natural number.
Being the difference of two even numbers,
12-2a=3d must be an even number,
and as 3 is odd, d must be even.
Even numbers are 2, 4, 6, ...
d%3E=4 makes 3d%3E=12 , 2a=12-3d%3C=0 , and a%3C=0 .
The only solution is highlight%28d=2%29 ,
which makes 2a=12-3d=12-6=6 , and highlight%28a=3%29 .

So, the four natural numbers are
a=highlight%283%29 , a%2Bd=3%2B2=highlight%285%29 , a%2B2d=highlight%287%29 , and a%2B3d=highlight%289%29 .

THE ALGEBRA-WITH-A-TWIST (BUT NO COMMON-SENSE) WAY:
In an AP, the mean (average) is also the median,
so 24%2F4=6 is the median.
Let 2x be the common difference.
The four numbers are
6-3x , 6-x , 6%2Bx , and 6%2B3x .
Their product is
%286-3x%29%286%2B3x%29%286-x%29%286%2Bx%29=945
%28%286-3x%29%286%2B3x%29%29%2A%28%286-x%29%286%2Bx%29%29=945
%2836-9x%5E2%29%2836-x%5E2%29=945
36%5E2-36x%5E2-324x%5E2%2B9x%5E2=945
1296-36x%5E2-324x%5E2%2B9x%5E2-945=0
9x%5E2-360x%2B351=0
x%5E2-40x%2B39=0 --> system%28x=-39%2C%22or%22%2Cx=1%29 .
x=-39 would give us negative numbers (not natural numbers) for the AP.
x=1 gives us
6-3x=6-3=3 , 6-x=6-1=5 , 6%2Bx=6%2B1=7 , and 6%2B3x=6%2B3=9 .
So, the four natural numbers are
highlight%283%29 , highlight%285%29 , highlight%287%29 , and highlight%289%29 .