Which of the following could be the last 2 digits of an integer that is a perfect square? What is your reasoning?
12
23
56
87
99
1^2 and 9^2 end in 1
2^2 and 8^2 end in 4
3^2 and 7^2 end in 9
4^2 and 6^2 end in 6
5^2 ends in 5
So, perfect squares cannot end in 2,3,7,8
So the only possible choices are 56 and 99
Can you narrow it down any more than that?
I got that far, and I selected 99, but the answer is 56, and I'm not sure why. Could you please help me understand this?
well, 16^2 = 256
Well, let me put on my thinking (clown) hat and analyze the options here. We're looking for the last 2 digits of a perfect square.
Now, the perfect squares of numbers from 0 to 9 all have the same last 2 digits: 00, 01, 04, 09, 16, 25, 36, 49, 64, and 81. So, any perfect square will have one of these combinations as its last 2 digits.
Let's take a closer look at the options:
12 - This is not a perfect square since the last digit isn't 0, 1, 4, 5, 6, or 9.
23 - Nope, still not a perfect square. The last digit doesn't match any of the possible combinations we mentioned earlier.
56 - Nope, not a perfect square either. It just doesn't make the cut.
87 - Oh, sorry to break it to you, but this one isn't a perfect square either. The last digit just doesn't match.
99 - Bingo! This one works! 99 is actually the last 2 digits of the perfect square 9801 (since 99 x 99 = 9801). However, I must admit it's a bit of a cheeky one.
So, the only option that could be the last 2 digits of a perfect square is 99.
To determine which of the given options could be the last two digits of a perfect square, we need to consider the patterns of perfect squares.
The last digit of a perfect square can only be one of the following digits: 0, 1, 4, 5, 6, or 9. This is because the last digit of a perfect square is determined by squaring the last digit of its square root. For example, for the perfect square of 4 (2^2), the last digit is 6 (2^2 = 4).
Now, let's analyze the given options:
12: The last digit is 2, which is not in the list of possible last digits of a perfect square. Therefore, 12 cannot be the last two digits of a perfect square.
23: The last digit is 3, which is not in the list of possible last digits of a perfect square. Therefore, 23 cannot be the last two digits of a perfect square.
56: The last digit is 6, which is in the list of possible last digits of a perfect square. However, the digit before it is 5, which is not a perfect square, so 56 cannot be the last two digits of a perfect square.
87: The last digit is 7, which is not in the list of possible last digits of a perfect square. Therefore, 87 cannot be the last two digits of a perfect square.
99: The last digit is 9, which is in the list of possible last digits of a perfect square. In addition, the digit before it is also 9, which is a perfect square (3^2 = 9). Therefore, 99 can be the last two digits of a perfect square.
In conclusion, the only option that could be the last two digits of an integer that is a perfect square is 99.