But What's in the 1000th Row?

We do some short number theory problems involving triangular arrangements of numbers, pattern recognition, and some basic familiar formulas from elementary number theory. This includes the sum of consecutive positive integers and the sum of consecutive odds. #mathproblems #mathpuzzle #mathskills

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0:00 Intro
0:19 Problem 1
3:56 Problem 2
6:02 Problem 3

[levels]

Sums are also a great way to find the first or last number, if m is the row # then the first one could be solved by (∑^m _n=0 of n) - m + 1 so plug in any number for m and you get the first number in that row. The second one (not a sum but yk whatever) 2^(m) - 1 is the last number of any row. Last but not least the third one (∑^m-1 _n=0 of 2n+1) - 2(m-1) could be used to find the first number in any row.

[MrDannyDetail]

I solved the actual puzzle shown in the thumbnail before pressing play on the video, and thus didn't know you had in fact set the triangle at 45 degrees from the orientation necessary to correctly identify what the rows were. That meant that I perceived 1, 4, 9, 16 as the first row, then 3, 8, 15 as the second row, and so on. Under that perception I believe the first number of the 1000th row would be 250, 502, though I could potentially have gone wrong somewhere in my calculations.

[nehalahmad]

You really teach in an engaging and interesting way and I love the topics you bring up. Keep it up bro you'll definitely get success

[HarryplaysOMG]

r = row

1. What's the first number in row 100?
summation: lowerbound\sum\upperbound { sum of thing
f(r) / Last number formula: n=1\sum\r { n
r=1: 1
r=2: 1+2
r=3: 1+2+3
First number formula: f(r) + 1
FN in row 100 = 4950 + 1 = 4951

2. What's the last number in row 10?
Last number formula: 2^r - 1
LN in row 10 = 2^10 - 1 = 1024 - 1 = 1023

3. What's the first number in row 1000?
First number formula: (r - 1)^2 + 1
FN in row 1000 = 998002

Explanation for 1:
row 4 in the right side is 10. this is 1 + 2 + 3 + 4. so 1 + 2 + .... + row. the first number for the NEXT row is the sum + 1.

Explanation for 2:
the first number in row 3 -> 4 doubles. we can infer this as 2^row. the last number of the previous row is 2^row - 1.

Explanation for 3:
the last number of the row is row^2. the first number in the next row is row^2 + 1. we must do row-1^2 to do the row we want.

[thethinkinlad]

great video. these problems were easy to solve but i was very intrigued by the different ways in which we approached them. math is beautiful!

[amogus_impasta]

in the second problem, the nth row actually contains all the n digits numbers in binary

[diaklau]

You have 10 apples
You buy 10 + 1 eat 2
You buy 10 + 2 eat 4
You buy 10 + 4 eat 8

If you do that kind of logic 1000 times
how many apples do you have?

Bonus: what is (1/2)×59^2?

[THE_HONOURED_ONE_LOL]

1:14 Aint we talking we talking triangle numbers?

[maxvangulik]

green:
T(n)=sum[k=0, n](k)=n(n+1)/2
T(0, 1, 2, 3, 4, 5)=(0, 1, 3, 6, 10, 15)
T(n)+1=(n^2+n+2)/2
T(0, 1, 2, 3, 4, 5)+1=(1, 2, 4, 7, 11, 16)
T(99)+1=T(100)-99=4951

blue:
f(n)=2^n-1
f(10)=2^10-1=1023

purple:
the last digit in row n is n^2
=> the first digit in row n+1 is n^2+1
f(n)=(n-1)^2+1=n^2-2n+2

=998002