The 3 digit number in which the ones digit is three times the tens digit and the tens digit is three times the hundreds digit is?
1*3 = 3
3*3 = 9
139
think a minute what can you multiply twice by 3 and still be less than 10?
Let's solve this step-by-step:
Step 1: Let's assign variables to each digit: hundreds digit = x, tens digit = y, ones digit = z.
Step 2: From the given conditions, we can write the equations:
z = 3y (the ones digit is three times the tens digit)
y = 3x (the tens digit is three times the hundreds digit)
Step 3: Now, we can solve for the digits. Let's start with the last equation:
y = 3x
Step 4: Substituting y in the first equation, we get:
z = 3(3x) = 9x
Step 5: Finally, we can write the 3-digit number using the digits we found:
xyz = 100x + 10y + z = 100x + 10(3x) + 9x
Step 6: Simplifying the equation, we get:
xyz = 100x + 30x + 9x = 139x
Therefore, the 3-digit number in which the ones digit is three times the tens digit and the tens digit is three times the hundreds digit is 139x. Note that the value of x was not given, so you will need additional information to determine the specific number.
To find the 3-digit number where the ones digit is three times the tens digit and the tens digit is three times the hundreds digit, we can use algebraic representation.
Let's assign variables to the digits. We can call the hundreds digit "h," the tens digit "t," and the ones digit "o."
We know from the problem that the ones digit is three times the tens digit, so we can write the equation: o = 3t.
Similarly, the tens digit is three times the hundreds digit, so we can write the equation: t = 3h.
Now, we have a system of two equations with two unknowns:
o = 3t,
t = 3h.
To find the 3-digit number, we need to solve these equations simultaneously.
Substituting the second equation into the first equation, we get:
o = 3(3h),
o = 9h.
Since o represents the ones digit, it must be less than 10. Therefore, h can only be 1.
Substituting h = 1 into the second equation, we get:
t = 3(1),
t = 3.
Finally, substituting h = 1 and t = 3 into the original equation, we get:
o = 3(3),
o = 9.
So, the 3-digit number where the ones digit is three times the tens digit and the tens digit is three times the hundreds digit is 139.