Think of a 2-digit number, Reverse the digits and add it to your original number that is a multiple of 11, why? (with proof)
You can write original number as :
x = 10 a + b
Reverse number you can write as :
y = 10 b + a
x + y = 10 a + b + 10 b + a = 11 a + 11 b = 11 ( a + b )
Example :
Original number:
x = 23
a = 2 , b = 3
x = 10 a + b = 10 * 2 + 3 = 20 + 3 = 23
Reverse number:
y = 10 b + a = 10 * 3 + 2 = 30 + 2 = 32
x + y = 23 + 32 = 55
55 = 11 * 5
To understand why adding a reversed 2-digit number to its original number results in a multiple of 11, let's break it down step by step.
Let's assume the 2-digit number is represented as AB, where A represents the tens digit and B represents the units digit. So, the original number can be expressed as 10A + B.
Now, if we reverse the digits, we get BA, which can be expressed as 10B + A.
Adding the reversed number to the original number gives us:
(10A + B) + (10B + A) = 11A + 11B = 11(A + B).
As we can see, the result is a multiple of 11 because it can be factored as 11 multiplied by the sum of the digits A and B.
To further prove this, we can verify with some examples:
Let's say we choose the number 32. Reversing the digits gives us 23. Adding 32 and 23 gives us 55, which is indeed a multiple of 11 (5 x 11).
Similarly, if we choose the number 87, reversing the digits gives us 78. Adding 87 and 78 gives us 165, which is also a multiple of 11 (15 x 11).
Therefore, adding a reversed 2-digit number to its original number always results in a multiple of 11.