Testing testing 2, 34, 56

Let $2N + 34 = 56k$

Then $N = 28k - 17$

Now,

$\begin{array}{c|c} k & N \\ \hline 0 & -17 \\ 1 & 11 \\ 2 & 39 \\ \ldots & \ldots \\ 36 & 991 \\ 37 & 1019 \\ \hline \end{array}$

Since $1 \leq N \leq 1000 \Rightarrow 1 \leq k \leq 36$

Therefore, there are only $\boxed{36}$ possible values for N

Note that $2N+34$ is even and every multiple of 56 is also even.This means that every multiple of 56 will give us an integer value of N.We see that for $N=1000 , 2N+34 =2034$ and $2034\equiv 18 \pmod{56}$ .

$\Rightarrow$ $2034-18=2016$ gives us an integer value of N(which is 991).Also $2016=56 \times {36}$ .Which means that every multiple of 56 up to 2016 gives us an integer value of N less than 1000.

Total values of N=36.

Clearly, $n=11$ works. Then, solutions for $N$ must differ by at least $28$ , and $2(11+28n)+34=22+34+56n=56(n+1)$ Therefore, all values $N\equiv 11\pmod 28$ work. There are $\boxed{36}$ values.

The given expression can be factorized as

$2N+34$ $=$ $2(N+17$ )

Since $56$ $=$ $2^{3} \times 7$ , therefore, $N+17$ must be a multiple of $28$ .

Then, the given fact can be rewritten as follows.

$1000$ $=$ $17 \times 56 + 48$

$1000$ $=$ $17 \times 28 \times 2 +48$

$1000$ $=$ $34 \times 28 +20 +28$

$1000$ $=$ $35 \times 28 +20$

Therefore we can say that there are $35$ multiples of $28$ from $1$ to $1000$ (inclusive).

But we also know that $36 \times 28 - 11$ is smaller than $1000$ . Therefore, there are $35 + 1$ $=$ $36$ integers that satisfy the integers $N$ .

Note that $\forall k, \exists N \text{ such that } 2N+34=56k$ . Hence, our solution is simply: $\left \lfloor \frac{2(1000)+34-(2(1)+34)}{56} \right \rfloor + 1=36$

We have that $2N+34 \equiv 0 \pmod {56}$ $N+17 \equiv 0 \pmod {28}$ $N \equiv 9 \pmod {28}$ This means we can write $N=28k+9$ for integers k. We have that $28k+9 \leq 1000$ , so $28k \leq 991$ , so $k \leq 35$ . $k$ can be anything from 0 to 35, so there are 36 possible values of $k$ , and thus $\boxed{36}$ possible values of $N$ .

When $N=1000, 2N+34=2034$ . Therefore, $\lfloor \frac{2034}{56}\rfloor = \boxed{36}$ Why is this so? The floor function shows that within $2034$ , $36$ values are divisible by $56$ .

NOTE when $N=1$ , $2N+34=36$ (not a multiple of $56$ ). It is important to check the range of $N$ .

We know that we can form any integer that is a multiple of 56 with that given, because 2N has to be even, and you are adding 34, another even number

The largest number we can form where N is less than 1000 that is a multiple of 56 is 2016.

We then divide 2016 by 56 to find the amount of multiples of 56 that are less than 2016

2016/56=36

There are 36 possibilities for N

First, I solved the equation for 1000. This equaled 2034. Then I calculated all the multiples of 56 under 2034. There were 36 of them. I knew that if I plugged the 36 multiples of 56 I found as the answer to the equation, and I solved for N for all the different equations, N could equal 36 different things. So, 36 is the answer.

Let 2N+34=56a.

N=28a-17. Since N is between 1 and 1000 when a=1 until a=36, there are 36 a which satisfies the condition.

There $2N + 34 = 56 \implies N = 11$

Then $2N + 34 = 56 \times 2 = 112 \implies N = 39$

$2N + 34 = 56 \times 3 = 168 \implies N = 67$

$2N + 34 = 56 \times 4 = 224 \implies N = 95$

So, We get a sequence $11, 39, 67, 95, 123 ...$ [Which is increasing by 28]

And there is $1000 / 28 = 35$ numbers in the sequence.

And the 1st one is $11$

So, There is $35 + 1 = \boxed{36}$ integer value of $N$

Using modular arithmetic, we want to find $N$ between $1$ and $1000$ , such that $2N+34 \ \text{mod}\ 56=0$ . The first integer satisfying this is $11$ , and there are then $1000-11=989$ integers left. Since $2\cdot 28=56$ , the next $N$ 's to satisfy the equation will be $11$ plus a multiple of $28$ . The number of such $N\leq 1000$ is found by $\lfloor \displaystyle\frac{989}{28}\rfloor=35.$ So the total number of $N$ is the $11$ and the multiples of $28$ : $1+35=36$

$2N+34=56$ so $N+17=28 \Rightarrow N \equiv 11 mod 28$ that it's equal to $28n-11=N$

$28n-11=1000 \Rightarrow n= 36$

2N+34=56a so 2N=56a-34 so N={56a-34}/2 since N is <or =1000 so, 56a-34< or = 2000 a is ,or = 36.32 Clearly a= \boxed {36}

2N + 34 = 56a

N = 28a - 17

we also know that N is between 1 to 1000 ( inclusive)

therefore to keep N in that range a can only take values from 1 to 36. hence giving us a total of 36 values

GIVEN,N=1-1000 THEREFORE,2N+34=36-2034 BUT,MULTIPLES OF 56 ARE-56,112,168... THEREFORE,MAXIMUM MULTIPLE OF 56 WILL BE LESS THAN OR EQUAL TO 2034 THEREFORE,t<2034 56+[n-1]56<2034 56n<2034 n<2034/56 n<36....THEREFORE n=36

I programmed it:

    int y = 0;

   for (int x = 1; x < 1001; x++) {
       int z = 2*x + 34;
       if (z % 56 == 0) {
           y += 1;
       }
    }
   System.out.println("The number of multiples is: " + y);

Yeah, so I cheated a little bit. ;)