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Martin Bashir
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How can i prove that square of an integer ends with 0,1,4,5,6,9 ?
Martin Bashir said:How can i prove that square of an integer ends with 0,1,4,5,6,9 ?
To prove this, we can use the concept of modular arithmetic. We know that the square of any number can be written as (n^2) = (10k + r)^2, where k is an integer and r is the remainder when n is divided by 10. By expanding the expression, we get (n^2) = 100k^2 + 20kr + r^2. Since the first two terms are divisible by 10, the remainder will only depend on r^2. We can observe that the only possible remainders when r^2 is divided by 10 are 0, 1, 4, 5, 6, and 9. Therefore, the square of any integer will always end in one of these digits.
Yes, there is a repetitive pattern to the last digit of squares of integers. The last digit will always be 0, 1, 4, 5, 6, or 9, in that order. For example, the squares of 1, 11, 21, and so on, will always end in 1. Similarly, the squares of 2, 12, 22, and so on, will always end in 4.
Sure, let's take the number 7. When we square 7, we get 49, which ends in 9. Using the formula (n^2) = (10k + r)^2, we can write 7^2 = (10*4 + 9)^2. Here, k = 4 and r = 9. The remainder when 7 is divided by 10 is 7, which is represented by r. And we know that the square of 7 will always end in the same remainder, which is 9 in this case.
No, this concept only applies to integers. When we square a decimal number, the resulting number will not follow the same pattern and can end in any digit. For example, the square of 1.5 is 2.25, which does not end in 0, 1, 4, 5, 6, or 9.
Yes, this proof is valid for negative integers as well. When we square a negative integer, the result will always be a positive number, and the last digit will follow the same pattern as positive integers. For example, the square of -3 is 9, which ends in 9. The square of -13 is 169, which ends in 9 as well.