I am less than 500. All three digits are odd. All three digits are different. The sum of the digits is thirteen. The product of my digits is greater than 30. I am not divisible by 5. Who am I?
+2498 \(\text{ let s say x y z is the digits of your 3 digit number } \\ xyz<500 \\x+y+z=13 \\x*y*z>30 \\\text{i am not divisible by five mean that your number cannot end with 0 or 5} \\ \text{so z cannot be 0 and 5 and your number less than 500 so x<5 x can be (3,1): } \\\text{there is only one number:} \text{ x=1 y=5 z=7 (157)}\)
.+2498 \(\text{ let s say x y z is the digits of your 3 digit number } \\ xyz<500 \\x+y+z=13 \\x*y*z>30 \\\text{i am not divisible by five mean that your number cannot end with 0 or 5} \\ \text{so z cannot be 0 and 5 and your number less than 500 so x<5 x can be (3,1): } \\\text{there is only one number:} \text{ x=1 y=5 z=7 (157)}\)
+298 397 can't be also the answer as 3+9+7 is 19, not 13, as again mentioned above.
+298 You are a human being, duh.
Now don't expect me to write your name.
Still, I think the answer is 157.
+298 But SolveIt, 3x9x1 is 27, which is not greater than 30, as mentioned above.
+298 What? I'm just telling the truth, but wait.....there may be 0.00000001% chances he might be an extra terrestial body, plus I gave him the answer too.
+129852 Nice work, OneDirection and Solveit.....points from me.....!!!
Here's a way to confirm this......
x + y + z = 13
x * y * z > 30
Now ....assume z = 9.....but this means that x and y could only be 1 and 3 since 1 + 3 + 9 = 13......but 1 * 3 * 9 = 27
And that's < 30
So......let's assume that z = 7
So z = 13 - x - y → 7 = 13 - x - y → x + y = 13 - 7 → x + y = 6
Thus.....either x = 1 and y = 5 or vice-versa
So......if we write the number as xyz, the only possibility is xyz = 157
+26393 I am less than 500. All three digits are odd. All three digits are different. The sum of the digits is thirteen. The product of my digits is greater than 30. I am not divisible by 5. Who am I?
\(\begin{array}{lrcll} (1) & x + y + z &=& 13 \\ & z &=& 13-(x+y) \\ (2) & x\cdot 100 + y\cdot 10 + z &<& 500 \\ & x\cdot 100 + y\cdot 10 + 13-(x+y) &<& 500 \\ & 99x + 9y + 13 &<& 500 \\ & 9y &<& - 99x + 500 - 13 \\ & 9y &<& - 99x + 487 \\ && & \boxed{~ \mathbf{ y } ~ \mathbf{<}~ \mathbf{- 11x + \frac{487}{9} } \\ ~}\\ (3) & x\cdot y \cdot z &>& 30\\ & x\cdot y &>& \frac{30}{z}\\ &&& \boxed{~ \mathbf{x\cdot y } ~\mathbf{>}~ \mathbf{\frac{30}{13-(x+y)} }\\ ~} \end{array} \)
I have put \(y ~<~ - 11x + \frac{487}{9} \) and \(x\cdot y ~>~ \frac{30}{13-(x+y)} \) in https://www.desmos.com/calculator
se we have a result of 5 Points. see: 
Now let us see:
\(\begin{array}{|r|r|r|r|} \hline x & y & z=13-(x+y) & \text{result}\\ \hline 1 & 5 & 13 - (1+5) = 7 & {\color{red}\text{yes}}\\ 1 & 7 & 13 - (1+7) = 5 & \text{no, because }~ 175 ~\text{ is divisible by }~ 5\\ 3 & 3 & 13 - (3+3) = 7 & \text{no, because }~ x = y \\ 3 & 5 & 13 - (3+5 )= 5 & \text{no, because }~ y = z \\ 3 & 7 & 13 - (3+7 )= 3 & \text{no, because }~ x = z \\ \hline \end{array} \)
There is only one number 157
