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 digit find the number
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the number, as far as i can tell, is 50.
the first digit is 5.
the second digit is 0.
it can't be 05 because that's a one digit number of 5.
when the most significant digit is 5 and the least significant digit is 0, you get:
5+0=5
5*0+5=5
the sum of the digits is equal to the product of the digits plus 5.
the equation is generated and solved for as follows:
let x equal 1 of the digits and y equal the other digit.
you get:
xy+5=x+y
subtract x and subtract 5 from both sides of this equation to get:
xy-x=y-5
factor out the x on the left side of the equation to get:
x(y-1) = (y-5)
divide both sides of the equation by (y-1) to get:
x = (y-5)/(y-1)
since x or y have to be single digits from 0 to 9, it's then a matter of substituting a single digit from 0 to 9 for y and seeing what the value of x becomes.
you get:
y x
0 5
1 undefined
2 -3
3 -1
4 -.3333...
5 0
6 .2
7 .3333...
8 .42857...
9 .5
you can see that the only digits that come out to be intergers in the interval of from 0 to 9 are when x = 0 and y = 5 or when x = 5 and y = 0.
if x is the most significant digit, then the solution is x = 5 and y = 0.
if x is the least significant digit, then the solution is x = 0 and y = 5.
Answer by ikleyn(52914)  (Show Source):
You can put this solution on YOUR website! .
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.
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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.
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.
Thus the problem is solved completely, i.e. DISPROVED.
I don't know how and from which source such idiotic problems come to the forum.
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