Is there a number like this?
Is there a 3-digit number that, when the three digits are multiplied together, equal that number?
Like, if the number was xyz, so x*y*z=xyz?
- No BodyLv 410 years ago
The number xyz has a value of:
x*10^2 + y*10 + z
So the question we're asking is is there x,y,z , 0<=x,y,or z<=9 such that:
x*y*z = x*10^2 + y*10 + z
Dividing both sides by 10^2 gives:
x*y*z / 100 = x + y/10 + z/100
Note that since x, y, and z are all less than 10 and nonnegative, x*y*z / 100 <= x. By observing that
x <= x + y/10 + z/100 we can conclude that:
x*y*z / 100 <= x + y/10 + z/100
We can strengthen <= to < by noting that if any one of x, y, or z are nonzero, through appropriate relabeling we are guaranteed to have x < x + y/10 + z/100. Therefore we get:
x*y*z / 100 < x + y/10 + z/100
From this inequality we can conclude that x*y*z cannot equal the number "xyz".
- AlexLv 610 years ago
xyz = 100x + 10y + z
x*y*z = xyz
⇒ x*y*z = 100x + 10y + z
If x = 1
⇒ the maximum value of x*y*z is when y = z = 9 so it's 81, but 100x + 10y + z = 100 + 10y + z, which is greater than 81
⇒ if the 1st digit is 1, there is no such number
If x = 2
⇒ the maximum value of x*y*z is again when y = z = 9 so it's 162, but 100x + 10y + z = 200 + 10y + z, which is greater than 162
⇒ if the 2nd digit is 2, there is no such number
For x = 3,4, ..., 9
⇒ similarly: x*y*z < 100x + 10y + z
⇒ there is no such a 3-digit number
- TheSicilianSageLv 710 years ago
Diophantine: a + 10b + 100c = a * b * c where a<10, b<10, c<10
- 10 years ago