My tenths digit is twice greater than my thousandths place.My hundredths digit is the sum of my tenths and thousandths digits.Who am i?
To solve this riddle, we need to analyze the clues given. Let's assign variables to each digit to make it easier to understand.
Let's say the thousandths digit is A.
The tenths digit is B.
The hundredths digit is C.
According to the riddle, the tenths digit is twice greater than the thousandths place. In other words, we can write this as B = 2A.
The riddle also states that the hundredths digit is the sum of the tenths and thousandths digits. So, we can write this as C = B + A.
Now, let's solve the system of equations to find the values of A, B, and C.
From equation 1, we have B = 2A.
Substituting this into the second equation, we get C = 2A + A, which simplifies to C = 3A.
Now, we have two equations:
B = 2A (equation 1)
C = 3A (equation 2)
If we substitute the value of C from equation 2 into equation 1, we get:
B = 2(1/3)C.
Simplifying, we find B = 2/3C.
This means that the tenths digit is two-thirds of the hundredths digit.
To determine the values of A, B, and C, we can substitute the possible values for C and solve for their respective values.
Let's start with the smallest possible value for C, which is 0. In this case, B would be 2/3(0) = 0, and A would be 0/2 = 0.
Therefore, in this case, all the digits are 0, which does not have a tenths digit greater than its thousandths place.
Since this case doesn't hold true, we need to try a different value for C. The next smallest possible value for C is 1. In this case, B would be 2/3(1) = 2/3, and A would be (2/3)/2 = 1/3.
Therefore, the number that satisfies the given conditions is 0.1(2/3)1.
To summarize, the answer is 0.123.
thousandth digit --- t
tens digit --------- 2t
hundredth digit ---- t + 2t = 3t
well, digits can only range from 0 to 9
let's rule out t = 0 , since that would not be a sensible number
then tenth digit is obviously an even number
so it could be 2,4,6,8
which means that t can only be 1, 2, 3, or 4
if t = 1, # is 132x , where x is any number from 0 to 9
if t=2, # is 264x
if t = 3, # is 396x
So you have 30 possible numbers