Question 612808
.
A two digit number is 5 times the sum of its digits and is also equal to 5 more than the product of its digits. 
Find the number.
~~~~~~~~~~~~~~~~~~~~~~~~~~~



            In this my post,  I will  PROVE  that the analysis and the answer by @Theo is  INCORRECT.


            I also will prove that the problem is posed incorrectly and that such a number as described in the post does not exist.



<pre>
Let N be a two-digit integer number equal to 5 times the sum of its digits.

Let's write N as a two-digit number (ab), where 'a' is a tens digit and 'b' is the ones digit.



Then from the problem, 'b' is either 0 or 5.

     We will consider these cases below, separately.



Also from the problem we can write  N = 10a + b = 5(a+b).


From this equation, we have

    10a + b = 5(a+b),

    10a + b = 5a + 5b,

    5a = 4b.    (*)


Now,  from this last equality  (*),   if b = 0,  then a = 0, and this case doesn't work, 
producing the number 00, which we are not going to interpret as a two-digit number.



If, on contrary,  b = 5,  then  from (*)  4b = 20  --->  a = 20/5 = 4,

    hence the number  N  is  45.


But then the product of its digits is  4*5 = 20,  and  N = 20+5 = 25, according to the second condition
"N is 5 more than the product of its digits".


But 25 = 45 is the contradiction, which ruins the problem itself into the dust and the analysis by @Theo, also into the dust.
</pre>

Thus the problem is solved completely, &nbsp;i.e. &nbsp;DISPROVED.


I don't know how and from which source such idiotic problems come to the forum.