Create a stimulating image depicting an abstract representation of the mathematics concept in question but without any text. Display multiple two-digit positive numbers scattered across, some highlighted to imply importance. Then, show a few examples of squares with a spotlight on the last two digits marked '01'. Make sure to denote a sense of calculation or summing.

What is the sum of all two digit positive integers whose squares

end with the digits 01?

the only number whose square ends in a 1 is a number that ends in 9 or 1

so a quick check on my trusty calculator and squaring numbers like 11,21,31,..91,19,29,39,..99

showed me
51^2 = 2601
49^2 = 2401.
99^2 = 9801

the sum of these 3 is 199

(sorry I used some LaTex code just now)Solution:

If  is a two-digit number, then we can write  in the form 10a+b , where  a and b are digits. Then the last digit of n^2 is the same as the last digit of b^2.

The last digit of n^2 is 1. We know that b is a digit from 0 to 9. Checking these digits, we find that the units digit of b^2 is 1 only for  b=1 and b=9 .

If b=1, then n=10a+1, so n^2=100a^2+20a+1.

The last two digits of 100a^2 are 00, so we want the last two digits of 20a to be 00. This occurs only for the digits a=0 and a=5, but we reject a=0 because we want a two-digit number. This leads to the solution n=51 .

If b=9, then n=10a+9, so

n^2=100a^2+180a+81. The last two digits of 100a^2+100a are 00, so we want the last two digits of 80a+81 to be 01. In other words, we want the last digit of 8a+8 to be 0. This only occurs for the digits a=4 and a=9 . This leads to the solutions n=49 and n=99.

Therefore, the sum of all two-digit positive integers whose squares end with the digits 01 is 51+49+99=199.

Well, that's a "square" question! To find the sum of all two-digit positive integers whose squares end with the digits 01, we can start by listing them out:

11^2 = 121
21^2 = 441
31^2 = 961
41^2 = 1681
51^2 = 2601
61^2 = 3721
71^2 = 5041
81^2 = 6561
91^2 = 8281

Now, let's add them up:

121 + 441 + 961 + 1681 + 2601 + 3721 + 5041 + 6561 + 8281 = 32849

Et voila! The sum of all two-digit positive integers whose squares end with the digits 01 is 32849.

Solution:

If  is a two-digit number, then we can write  in the form , where  and  are digits. Then the last digit of  is the same as the last digit of .

The last digit of  is 1. We know that  is a digit from 0 to 9. Checking these digits, we find that the units digit of  is 1 only for  and .

If , then , so

The last two digits of  are 00, so we want the last two digits of  to be 00. This occurs only for the digits  and , but we reject  because we want a two-digit number. This leads to the solution .

If , then , so

The last two digits of  are 00, so we want the last two digits of  to be 01. In other words, we want the last digit of  to be 0. This only occurs for the digits  and . This leads to the solutions  and .

Therefore, the sum of all two-digit positive integers whose squares end with the digits 01 is .

To find the sum of all two-digit positive integers whose squares end with the digits 01, we need to consider all the possibilities and add them together.

First, let's list all the two-digit positive integers. These range from 10 to 99.

Next, we need to find the square of each of these integers and check the last two digits. We can use the following formula to calculate the square of a number:

Square of a number = Number * Number

For example, to find the square of 10, we multiply 10 by itself: 10 * 10 = 100. The last two digits of this square are "00", which do not end in "01", so we exclude 10 from our sum.

We repeat this process for all the two-digit positive integers and add together the numbers whose squares end in "01". For example, for 11, 11 * 11 = 121. The last two digits are "21", which do end in "01", so we include 11 in our sum.

Finally, we add up all the numbers that passed the test. Let's go through each possibility:

11 * 11 = 121 (Include 11)
12 * 12 = 144 (Exclude 12)
13 * 13 = 169 (Exclude 13)
... (continue this process for all two-digit positive integers)

After checking all the possibilities, we add together the numbers that passed the test:

11 + ... (continue this process for all two-digit positive integers)

By doing this, we would get the sum of all two-digit positive integers whose squares end with the digits "01".