E
The triangular numbers are of the form T n = 2 n ( n + 1 ) , where n is a positive integer. How many possibilities are there for the last digit of T n ?
The answer is 6.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
4 solutions
Moderator note:
How would you show that the last digits must be periodic?
That's much better than my method. Good job.
WHY TO FIND MODULO 10 ?
Log in to reply
Do you mean "Why did you use modulo 10" or "How do you find modulo 10." The latter is very simple, it's just the remainder when divided by ten, or the last digit. As for the former, I used mod 10 because it's a lot more convenient as the numbers increase since all the really matters is the last digit.
nice one
right
Can you explain why "This is because T 0 ≡ 0 , T 1 0 ≡ 5 ( m o d 1 0 ) is an explanation for "after every twenty triangular numbers, the residues modulo 10 cycle? How do you know that T 2 3 ≡ T 3 ( m o d 1 0 ) ?
Log in to reply
Suppose n ≡ a ( m o d 2 0 ) and ( n + 1 ) ≡ b ( m o d 2 0 )
Let n = 2 0 m + a and n + 1 = 2 0 n + b T n = 2 ( n ) ( n + 1 = 2 ( 2 0 m + a ) ( 2 0 n + b ) = 2 4 0 0 m n + 2 0 m b + 2 0 n a + a b = 2 1 0 ( 4 0 m n + 2 m b + 2 n a ) + a b = 1 0 ( 2 0 m n + m b + m a ) + 2 a b ≡ 2 a b = T a ( m o d 1 0 )
Log in to reply
No. For example, 1 1 ≡ 1 ( m o d 1 0 ) , 1 2 ≡ 2 ( m o d 1 0 ) , but we have 1 ≡ 2 1 × 2 ( m o d 1 0 ) while 6 ≡ 6 6 ≡ 2 1 1 × 1 2 ( m o d 1 0 ) .
You can't simply divide in a modulus statement.
Note also that the period is 20, and not 10.
Log in to reply
@Calvin Lin – Oh I see where I got wrong. I typed 20 into 10 :) Am I right now?
Ok, so first note that in mod 10, we are essentially adding, 1, 2, 3, ..., 9, 0, 1, 2, ..., 9, 0, etc. and that T 0 = 0 . So if we find a T n ≡ 0 ( m o d 1 0 ) s.t., n ≡ 0 ( m o d 1 0 ) , then it would begin a periodic cycle, starting again with the addition of 1, 2, 3, ...9, 0, 1, 2, ..., etc. Now 1 + 2 + ⋯ + 9 + 0 ≡ 5 ( m o d 1 0 ) , so we require two sums of 0 to 9 to reach the start of a new cycle ( 5 + 5 ≡ 0 ( m o d 1 0 ) ).
1st term: 1
2nd term: adding 2 to get three
3rd term: adding 3 to get 6
And so on.
The last digits follow a pattern as such: 1 , 3 , 6 , 0 , 5 , 1 , 8 , 6 , 5 , 5 , 6 , 8 , 1 , 5 , 0 , 6 , 3 , 1 , 0 , 0
After which you will be adding 1 to get the same pattern.
0 , 1 , 3 , 5 , 6 , 8 is 6 numbers.
How would you demonstrate that such a pattern must exist?
how to find the mod? i am unable to understand the problem actually.
nice one
If you write out the first 20 terms of the triangular numbers, you realise that the last digit would form the cycle of: 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0...... This would mean that the cycle actually repeats and thus 6 different numbers appear.
How would you demonstrate that such a pattern must exist?
Since n(n+1) is a product of two consecutive numbers then the last digit of n must be 1 and the last digit of n+1 must be 2 or the last digit of n must be 2 and the last digit of n+1 must be 3 and so on................the last digit of must be 9 and the last digit of n+1 must be 0. Let's multiply: 1 2=2 2 3=6 3 4=12 4 5=20 5 6=30 6 7=42 7 8=56 8 9=72 9 10=90 10 11=110 11 12=132............ the last digit of a product is the last digit of the product of the last digits of these two consecutive numbers for example 23 24=552 beacuse the last digit of 3*4=12 is 2.Thats why the last digit of n(n+1) can be 2,6 or 0. since n(n+1) is even then 2 is a divisor of this number. lets consider when n(n+1) ends with 2 then n(n+1)/2 must end with 1 or 6 . lets consider when n(n+1) ends with 6 then n(n+1)/2 must end with 3 or 8. lets consider when n(n+1)ends with 0 then n(n+1)/2 must end with 0 or 5 so we have 6 possibilities for the last digit of Tn.
thank you lorence
0,1,3,5,6,8 are the only last digits in a triangular number
Notice that after every twenty triangular numbers, the residues modulo 10 cycle. This is because T 0 = 0 , and T 1 0 ≡ 5 ( m o d 1 0 ) . Afterwards, we again adding 1 , 2 , . . . , 0 ( m o d 1 0 ) , but start at 5 instead of 0. Thus, T n has a cycle of 2 0 ( m o d 1 0 ) . Listing them out, 0 , 1 , 3 , 6 , 0 , 5 , 1 , 8 , 6 , 5 , 5 , 6 , 8 , 1 , 5 , 0 , 6 , 3 , 1 , 0 , giving an answer of 6 possible last digits.