SOLUTION: what three digit number made of consecutive digits is 2 less than a cube and 2 more than a square

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Question 253639: what three digit number made of consecutive digits is 2 less than a cube and 2 more than a square
Found 2 solutions by Edwin McCravy, drk:
Answer by Edwin McCravy(20081)   (Show Source): You can put this solution on YOUR website!
Then if x = its hundreds digit then
x+1 = its tens digit and x+2 is its units digit

So the number is 

100x + 10(x+1) + (x+2) =

100x + 10x + 10 + x + 2 =

111x + 12

If you let x=1, you get it immediately, because

111(1) + 12 = 111 + 12 = 123  

and  123 + 2 = 125 = 5^3 and 123 - 2 = 121 = 11^2 

Edwin
Answer by drk(1908)   (Show Source): You can put this solution on YOUR website!
Let htu be a three digit number. Since the numbers are consecutive, we have h_h-1_h-2
in decreasing form or
h_h+1_h+2 in increasing form.
--
decreasing order will not produce any answers.
--
increasing :
This can be expressed as h(100) + (h+1)(10) + h+2. Now
(i) S^2 + 2 = h(100) + (h+1)(10) + h+2
(ii) S^2 = 100h +11h + 10
(iii) S^2 = 111h + 10
If h = 1, then
S^2 = 111 + 10 = 121
S = 11.
The number is 123
--
(v) h(100) + (h+1)(10) + h+2 = s^3 - 2
(vi) 100h +11h + 14 = S^3
(vii) 111h + 14 = S^3
If h = 1, then
S^3 = 111 + 14 = 125
S = 15.
The number is 123.
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