SOLUTION: i have a math problem that say " I am thinking of a number that when multiplied by itself gives me a 3-digit product where the hundreds digit and the ones digit are the same and th

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Question 661703: i have a math problem that say " I am thinking of a number that when multiplied by itself gives me a 3-digit product where the hundreds digit and the ones digit are the same and the tens digit is the sum of the other two digits. what number am i?
Found 2 solutions by htmentor, ReadingBoosters:
Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = the number
Let n = the hundreds digit = the ones digit
Then the tens digit = 2n
The 3-digit product can be written as
x^2 = 100n + 10(2n) + n = 121n
This gives x = 11sqrt%28n%29
The product is three digits, and sqrt%28n%29 must be an integer
If n = 1, x = 11 -> x^2 = 121
If n = 4, x = 22 -> x^2 = 484
It seems both of these meet the conditions of the problem, so the possible answers are 11 and 22

Answer by ReadingBoosters(3246) About Me  (Show Source):
You can put this solution on YOUR website!
product where
x = hundreds
y = tens
z = ones
x=z
y=x+z = z+z=2z
Substituting into expanded form of a 3 digit number
100(x) + 10(y) + 1(z) = 3 digit number
100(z)+10(2z)+z
100z + 20z + z = 121z
recognizing that 121 fits the description
n^2 = 121
n = 11