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
What is the sum of all two digit positive integers whose squares
end with the digits 01?
3 answers
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 .
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 .
(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.
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.
1 answer