Question 1190791: Split 69 into three parts such that they are in A. P. and the product of the two smaller parts is 483
Found 3 solutions by math_helper, ankor@dixie-net.com, greenestamps: Answer by math_helper(2461) (Show Source):
You can put this solution on YOUR website!
a1 + a2 + a3 = 69
Since we are given a1,a2,a3 form an AP, we can write:
a1 + (a1+k) + (a1+2k) = 69 where k is the common difference
3a1 + 3k = 69
a1 + k = 23
Trying a1 = 21, a1+k = 23, and a1+2k = 25, we see this AP meets the requiremnt that the first two terms multiply out to 483, thus {21,23,25} satisfies the condition.
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Alternate solution:
a1 * (a1+k) = 483


Expanding it back to a quadratic:
so we see a1=21, k=2 works, resulting in {21,23,25} as above
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Split 69 into three parts
a + b + c = 69
such that they are in A. P.
Assume this means Arithmetic Progression. (dif between two Nos. is the same)
let d = difference between consecutive numbers
Then
b = a + d
c = a + 2d
the product of the two smaller parts is 483
ab = 483
replace b with a+d
a(a+d) = 483
a^2 + ad = 483
:
in the first equation write it
a + (a+d) + (a+2d) = 69
3a + 3d = 69
simplify, divide by 3
a + d = 23
d = (23-a)
:
Back to a^2 + ad = 83, replace d with (23-a)
a^2 + a(23-a) = 483
a^2 + 23a - a^2 = 483
a^2 drops out
23a = 483
a = 21
Find d
23-21 = 2 is the difference
The numbers then are:
21, 23, 25
:
Check:
21 * 23 = 483
and
21 + 23 + 25 = 69
Answer by greenestamps(13200) (Show Source):
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