Adding 1 to the product of four consecutive positive integers always results in a perfect square. The first 2017 such square numbers can be found:
1 x 2 x 3 x 4 + 1=25=5^2
2 x 3 x 4 x 5 + 1=121=11^2
3 x 4 x 5 x 6 + 1=361 = 19^2
2017 x 2018 x 2019 x 2020 + 1=16600254584281=4074341^2
In this list of 2017 numbers: 5, 11, 19, ........, 4074341
Whose squares are found in this way, how many have last digit equal to 1?
Why did the number 1 join the exclusive "Last Digit Club"? Because it squared off with the rest and proved it had the moves! Now, let's count how many other numbers get to be part of this trendy club.
We notice that the last digit of a perfect square can only be 0, 1, 4, 5, 6, or 9. Ah, a strict dress code, indeed! But fear not, dear friend, for we are only interested in those with a last digit equal to 1.
Let's see how many numbers in our list have a square with a last digit equal to 1. Unfortunately, I'm too lazy to calculate all those squares myself, but I can tell you one thing for sure - it's not going to be too easy to count them without a little trick up our sleeves!
So here's a clowny shortcut for you: Notice that for any number ending in 9, the square ends in 1. Similarly, if the number ends in 1, its square also ends in 1.
Since our list starts with 5, we can see that every fifth number in our list will have a square ending in 1. How handy!
Now, let's do some quick calculations. We have 2017 numbers in our list. Since 5 is the first number and it has a square that ends in 1, we can divide the remaining 2016 numbers by 5 to find out how many of them also end in 1.
2016 divided by 5 is 403 remainder 1.
Therefore, out of the 2017 numbers in our list, 403 of them will have squares that end in 1!
Ta-da! We have our answer. So, in this list of 2017 numbers, 403 of them have a last digit equal to 1.
To find the number of squares in the list with a last digit equal to 1, we need to find a pattern in the last digit of the squares.
Let's observe the squares in the list:
5^2 = 25
11^2 = 121
19^2 = 361
.
We can see that the last digit of the square is 5 for the number 5, and the last digit of the squares is always 1 for the rest of the numbers (11, 19, ...).
To find the number of squares in the list with a last digit equal to 1, we can count the number of remaining numbers in the list (excluding 5).
There are 2017 - 1 = 2016 numbers remaining in the list.
Therefore, there are 2016 numbers in the list with a last digit equal to 1.
To solve this problem, we need to find the values of the four consecutive positive integers whose product, when added to 1, results in a perfect square. We can start by finding a pattern.
Notice that the first few squares we have are 5^2, 11^2, 19^2. If we compare the values of the squares to the integers, we can observe the following pattern:
5^2 = 25, 11^2 = 121, 19^2 = 361
If we take the difference between consecutive squares, we get:
121 - 25 = 96, 361 - 121 = 240
Similarly, we can calculate the differences for the next few squares:
240, 528, 864, ...
The differences between squares seem to be increasing. If we continue calculating the differences, we would eventually reach one difference that ends with a 1 as its last digit. This means that for every difference that ends with a 1, we have a square number whose last digit is also 1.
To find the number of squares with a last digit of 1 in the given range (5, 11, 19, ..., 4074341), we need to count the number of differences that end with a 1.
Starting with the difference of 96, we can add the previous difference (240) to get the next difference:
96 + 240 = 336
Now, let's check if the difference ends with a 1. Since 336 does not end with 1, we calculate the next difference:
336 + 528 = 864
Now, 864 ends with 4, so we continue calculating:
864 + 720 = 1584
1584 + 912 = 2496
2496 + 1200 = 3696
3696 + 1440 = 5136
5136 + 1760 = 6896
6896 + 2160 = 9056
Now, we have a difference that ends with a 1. This means that we have found a square number with a last digit of 1. We can calculate the square value by adding the difference to the last square value we had (which was 4074341):
4074341 + 9056 = 4083397, which is also a square number.
From this point, we can continue calculating the differences and finding the corresponding squares until we reach the desired number of squares (2017).
So, to find the number of squares with a last digit of 1 in the given range, we would need to continue this process until we find 2017 squares. Each time we find a difference that ends with a 1, we know that we have a square number with a last digit of 1.
Note: Calculating all these differences manually could be time-consuming. It is recommended to use a programming language or a spreadsheet software to automate the calculations.
Noticed my own aritmetic error
201*4= 804, not 404
So the total is 807
The pattern itself is easy to prove
Let the 4 consecutive integers be n-1, n, n+1, and n+2
Did you notice the pattern?
Then prove
(n-1)(n)(n+1)(n+2) = (n(n+1) - 1)^2
which is easy
List the first 15 or so cases , in the first 10 cases , the only ones where the square ends in a 1 are .,.
2*3*4*5+1= 11^2
5*6*7*8+1 = 41^2
7*8*9*10+1 = 71^2
10*11*12*13+1 = 131^2
In each group of 10, this happens 4 times
From 1 to 2010 this happens 201*4 or 404 times
From 2011 to 2017 we have 3 more
For a total of 407 such cases
Check my arithmetic