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 3 solutions by Theo, ikleyn, greenestamps:
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(52915)  (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.
Answer by greenestamps(13216)  (Show Source):
You can put this solution on YOUR website!
Here is a response using different logical reasoning to show that the problem has no solution.
(1) The 2-digit number is 5 times the sum of its digits, so the units digit is either 5 or 0.
(2) The 2-digit number is also 5 more than the product of its digits. If the units digit is 0, then the product of the digits is 0, and 5 more than 0 is 5, which is not a 2-digit number.
(3) From (1) and (2), the units digit of the 2-digit number must be 5.
(4) From (1) and (3), the sum of the digits must be odd. Since the units digit is 5, the tens digit must be even.
That reasoning leaves only these possibilities for the 2-digit number: 25, 45, 65, and 85. Of those, only 45 satisfies the condition that the number is 5 times the sum of its digits; and 45 does not satisfy the condition that it is 5 more than the product of its digits.
ANSWER: No solution
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