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A three-digit even number is given such that the hundreds, tens, and unit digits respectively form a decreasing arithmetic sequence. When two is added to the number, the same order of the digits now forms a geometric sequence. What is the original number?

 Jun 12, 2021
 #1
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100a + 10b + c

a, b, c is a decreasing arithmetic sequence. 

a, b, c + 2 is a geometric sequence.

Note that the c can't carry over since that only happens when c is 9 (odd number) or 8, but 8 + 2 = 0 which can't be part of a geometric sequence. 

 

c = c

b = c + x

a = c + 2x

 

b^2 = (c + 2)(a)

(c + x)^2 = (c + 2)(c + 2x)

c^2 + 2cx + x^2 = c^2 + 2cx + 2c + 4x

x^2 = 2c + 4x

 

Let's try when c = 2. 

x^2 = 4 + 4x, x is a weird number. 

 

Let's try when c = 4. 

x^2 = 8 + 4x, x is not a nice number. 

 

Let's try when c = 6.

x^2 = 12 + 4x, x is a nice number. 

(x-2)(x-6) = 0

x = 6, this is not possible. 

 

Let's try when c = 0. 

x^2 = 4x

x = 4. 

 

 

c = 0

b = 4

a = 8

Our number is 840. :))

 

=^._.^=

 Jun 12, 2021

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