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 (sod(x))^2 = x+2 (Posted on 2011-01-26)
Determine the probability that for a base ten positive integer x chosen at random from 1 to 9999 inclusively, this relationship is satisfied: (sod(x))2 = x+2, where sod(n) denotes the sum of the digits in the base ten representation of n.

 No Solution Yet Submitted by K Sengupta No Rating

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 a simple proof Comment 3 of 3 |

2 ,23, 62, 119, 194, 287, 398

Since 7+9+9=25 and 25^2<700 our candidate solutions can be derived  examining square numbers below 700, i.e. 26 numbers at most.

The process may me mental, p&p, simple calculator or short excel table. Just to illustrate the point:…..

121==>119==>1+1+9=11  11*11 = 121:  fits<br>

144==>142==>1+4+2=7  7*7 is not 144:<br>

169=>167==>1+6+7=14  14*14 is not 167:<br>

196==>194==>1+9+4=14  14*14 = 167:  fits<br>

….

Analytical solution(not so time-effective):  For 1 or 2 digit number N let N be 10a+b

(a+b)^2=10*a+b+2<br>

Solving for a we get a=5-b+sqrt(3-b)<br>

and              a=5-b-sqrt(3-b)<br><br>

INTEGER SOLUTIONS:<br>

b=2   a=0,6   N=2  or 62<br>

b=3   a=2  N=23<br>

For 3 digit number less than 200 solve:

(a+b+1)^2=10*a+b+102    and so on.

Edited on January 27, 2011, 8:29 am
 Posted by Ady TZIDON on 2011-01-27 08:23:46

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