Arithmetic progressions

Consider all sequences with common difference 0. There must be 70 of these, from (1,1,1...) to (70, 70, 70...). Then consider all INCREASING sequences of common difference 1. The first such sequence starts with 1, and the last such sequence starts with 62 (To find this out, I just counted back from 70). So there are 62 of these, plus an equal number of decreasing sequences. Continuing, there are 54 increasing sequences of difference 2, 46 sequences of difference 3, with 8 fewer sequences every time the difference was increased, concluding with 6 sequences of difference 8. To get the solution, double the numbers of each non-zero difference progression to account for decreasing sequences, then add 70 for the zero-difference progressions. 70 + 2*(62 + 54 + 46 + 38 + 30 + 22 + 14 + 6) = 614

But then I started wondering "WHY are there 8 fewer sequences every time the difference is increased by 1?" Each progression can be generalized as: [x, x + d, x + 2d, .... x + 8d], which shows that every time you increase d by 1, you increase the final term in the progression by 8. This means you can fit 8 fewer progressions before you hit 70 as the final term.

I drew a graph and used Pick's Theorem.

With $a$ as the horizontal axis and $d$ as the vertical axis, we are looking for the number of lattice points that lie on or inside the parallelogram bounded by: $a=1$ $a=70$ $a+8d=1$ $a+8d=70$

The corners of this parallelogram are $(1,0), (70, -\frac{69}{8}), (70, 0), (1, \frac{70}{8})$ .

Two of these are not lattice points , so we will chop a little triangle off those corners, giving the hexagon $(1,0), (65, -8), (70, -8), (70, 0), (6, 8), (1, 8)$ . The chopped-off areas contain no interior points because their $d$ -dimension was less than $1$ .

The number of boundary points on this figure can be found by checking each side, again in the order given above (and counting the first corner but not the last for each edge): $B = 8 + 5 + 8 + 8 + 5 + 8 = 42$ .

The area of the figure is the parallelogram area $69 \cdot \frac{69}{8}$ minus the two small triangles, each $\frac{1}{2}\cdot 5 \cdot \frac{5}{8}$ , giving $A = 592$

By Pick's Theorem, $I = 592 - \frac{1}{2}42 + 1 = 572$ , and then the total number of points is $I + B = 572 + 42 = \boxed{614}$ .

d = 0 gives 70 solutions d = 1 gives 62 solutions (How? Try yourself) d =2 gives 54 solutions ..... .... d = 8 gives 6 solutions d =9 and onwards no solution.

However since series can be decreasing; so number of solutions obtained for d =1 to d = 8 need to be multiplied by 2. So, total no. of solutions =[ 6 + 14+...............62 ]* 2 + 70 = 614

With a difference of 0, the maximum starting digit is 70. With a difference of 1, the maximum starting number is 62. With a difference of 2, the maximum starting number is 54. Therefore:

$n = 70 - 8d$

Where there are $n$ possible sequences with a difference of $d$ .

To calculate the number of possible sequences with a positive difference we use:

$\displaystyle \sum_{i=1}^8 70 - 8i = 272$

To account for a negative difference, we times by 2, and for a difference of 0, we add 70. So the final answer is:

$272 \cdot 2 + 70 = \fbox {614}$

I have found generalization of this explained here How Many A.Ps?Theory

The question 'How many arithmetic progressions of length $t$ are there, such that the terms are integers from $1$ to $n$ ?'

The formula is $n+(p+1)[2(n-t+1)-p(t-1)]$ [where $[\frac {n}{t}]=p$ ]

Thus here the answer is $70+(7+1)[2(70-9+1)-7(9-1)]=70+8*[2*62-56]= \boxed{614}$

Firstly, let us note that if the last term of the progression belongs in the interval $[1,70]$ and is an integer, then it is absolutely true that all of the terms of the progression belong to the interval (and are integers). With this in mind, we generally have to find all the $9_{th}$ terms of the progressions in order to find all of them.

For illustrative purposes, let us denote $d$ as the difference between $2$ consecutive terms in the arithmetic progression we are taking into consideration at the given time.

Also, to avoid repetition, bear in mind that we will need to double the number of progressions with a positive $d$ , due to the fact that $d$ can be negative as well, giving the same progressions with reversed terms.

With inspection, we may notice that $d$ can not be greater than $8$ (or lower than $-8$ for that matter), due to the fact that if $d=9$ , the best case scenario we could get would be the progression with a last term $a_{9}=9\cdot8+1=73$ , meaning that it does not belong to the interval, whereas for $d=8$ the last term would be $a_{9}=8\cdot8+1=65$ , which is in the interval.

Therefore, we have $9$ possible cases for $d$ :

Case 1

$d=0$

In this case we have $70$ different progressions since all the terms are the same and are integer values in the interval $[1,70]$ , which are exactly $70$ . In this case, we should not double the number of progressions because we would get the exact same ones all over again $((1,1,1,1,1,1,1,1,1)=(1,1,1,1,1,1,1,1,1))$ .

Case 2:

$d=1$

The last term of the very first progression formed this way would be $8\cdot1+1=9$ , whereas in the last it would be $70$ . Since $d$ is constant, the only thing changing in the progressions in-between are the first terms, which change by $1$ . Therefore, the only possible last terms with $d=1$ are the last terms in the interval $[9,70]$ (considering only the integer value of it of course), which gives us $62$ last terms and $62$ progressions. Thus, the total number of progressions with $d=\pm 1$ is $124$ .

Case 3:

$d=2$

Similarly to the previous case, we get the last terms of the first and last progressions, which are $17$ and $70$ . Thus, there are as many progressions as the number of integers in the interval $[17,70]$ is. Thus, for $d=\pm2$ we have a total of $108$ progressions.

$\vdots$

To avoid repeating myself, we use the same method to find the number of progressions for $d=\pm3,\pm4,\pm5,\pm6,\pm7,\pm8$ .

Finally, that gives us a total of $124+108+92+76+60+44+28+12+70=614$ arithmetic progressions with terms that are integers from $1$ to $70$ .

Difference:0

1~1 ,2~2......70~70(70 groups)

Difference:1

1~9......62~70(62 groups)

Difference:2

1~17......54~70(54 groups)

Difference:3

1~25......46~70(46 groups)......

There is a pattern:If the difference pluses 1,the number of the progressions will minus 8.Sequence:70,62,54,46,38,30,22,14,6 (difference of 0,1,2,3,4,5,6,7,8).

The sequence of the progression also can be reversed,except the groups of difference of 0.Hence,the number of progressions are

70+2(62+54+46+38+30+22+14+6)=70+544=614

Ans=614

So, I'm not sure why this works, but it does. Let x represent (length given - 1). Let n represent the range of integers given.

x = 8, n = 70

What I did was this: 70 + 2(70-x + 70-2x + .... 70 - 8x)

So basically, the last term has to be greater than 0, so you can't have 70 - 9x because that would result in a negative number.

I multiplied everything but the first term by 2, because it could be descending or ascending: (1,2,3,4) vs (4,3,2,1) are two different sequences. However, the first term (70) represents any sequences with 0 being the constant difference.

Then add it up and it'll equal 614.