Adrian said that any 2 digit number with 2 identical,digits is divisible by 11. Do you agree? Explain.
Multiples of 11:
11, 22, 33, 44, 55, 66, 77, 88, 99
Yes I agree because it can any number such as 11,22,33,44,55,66,77 so on
Yes, I agree with Adrian's statement. Any two-digit number with two identical digits is divisible by 11.
To explain why this is true, let's consider a two-digit number of the form "AA", where A represents any digit from 0 to 9.
We can represent this number as 10A + A, which simplifies to 11A. Since 11 is a prime number, any multiple of 11 must also be divisible by 11. Hence, any two-digit number with two identical digits is divisible by 11.
Yes, I agree with Adrian's statement. Let me explain why.
To determine if a number is divisible by 11, we can use a rule called the "divisibility rule for 11." According to this rule, a number is divisible by 11 if the difference between the sum of its digits in the odd positions and the sum of its digits in the even positions is either 0 or a multiple of 11.
In the case of a 2-digit number with 2 identical digits, we can represent it as AB, where A and B are the same digit.
Let's evaluate the sum of the digits in the odd and even positions for this number. Since A and B are the same digit, the sum of the digits in the odd positions is A + A = 2A, and the sum of the digits in the even positions is B + B = 2B.
Considering the difference between these sums, we have (2A - 2B).
Since the digits A and B are the same, this difference simplifies to 2(A - B). Now, if we consider two identical digits, the difference (A-B) will always be 0.
Therefore, the entire difference (2(A - B)) will also be 0, which means the number is divisible by 11.
In conclusion, any 2-digit number with 2 identical digits is indeed divisible by 11, as the difference between the sums of its odd and even positions is always 0.