Number Theory Question

Nice problem! We have to see a series going on here. It is an AP going on in the ceiling function. So, it is not just a simple series. As d = 0.7, after 10 terms, the term which we started our counting will get added by 7(0.7 * 10). In other words, this will be the series

$\left\lceil 1 \right\rceil + \left\lceil 1.7 \right\rceil + \left\lceil 2.4 \right\rceil + \left\lceil 3.1 \right\rceil + \left\lceil 3.8 \right\rceil + \left\lceil 4.5 \right\rceil + \left\lceil 5.2 \right\rceil + \left\lceil 5.9 \right\rceil + \left\lceil 6.6 \right\rceil + \left\lceil 7.3 \right\rceil + \left\lceil 8 \right\rceil + ..... + \left\lceil 999.9 \right\rceil$

Therefore, 10 A.P.'s with d = 7 go on simultaneously in the series.

Now our last term of the series i.e. $\left\lceil 999.9 \right\rceil$ corresponds to $\left\lceil 5.9 \right\rceil$ and so, it is the last term of the AP with first term as 6.

So, we can find the number of terms in the AP using its general term-

6 + (n-1)7 = 1000

n-1 = 142, n = 143

Now, using the sum formula of AP,

S = $\frac{143(6 + 1000)}{2}$

Similarly, for all the AP's with a = 6, 5, 4, 3, 2, 1; n = 143.

For the AP's with a = 7, 8; n = 142.

Hence, finally adding all the sums

$\frac{143(6 + 1000)}{2} + \frac{143(6 + 1000)}{2} + \frac{143(5 + 999)}{2} + \frac{143(4 + 998)}{2} + \frac{143(4 + 998)}{2} + \frac{143(3 + 997)}{2} + \frac{143(2 + 996)}{2} + \frac{143(1 + 997)}{2} + \frac{142(7 + 994)}{2} + \frac{142(8 + 995)}{2} = \boxed{715285}$

In C++

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#include <iostream>
#include<math.h>
using namespace std;

int main() {
     long double sum=0;

    for(long double x=1.0;x<=999.9;x+=0.7)
     { 
         sum+=ceil(x);
     }
     cout<<sum;

    return 0;
}

Output:

1

715285

I wrote the ceiling values and compared it with the natural number series i.e $1 , 2 , 3 , 4 , 5 , 6$ .

I found out the difference in the values and subtracted it from the sum of the natural number series.

My solution goes as -

CLARIFICATION

I found out the number of terms using AP Formula.

The expression can be written like this: $\left\lceil 1 \right\rceil +\left\lceil 1+0.7 \right\rceil +\left\lceil 1+1.4 \right\rceil +\left\lceil 1+2.1 \right\rceil +...+\left\lceil 1+998.9 \right\rceil$ . Given that $\frac { 998.9 }{ 0.7 } =1427$ , then

$\left\lceil 1 \right\rceil +\left\lceil 1+0.7 \right\rceil +\left\lceil 1+1.4 \right\rceil +\left\lceil 1+2.1 \right\rceil +...+\left\lceil 1+998.9 \right\rceil$ $=\sum _{ k=0 }^{ 1427 }{ \left\lceil 1+0.7k \right\rceil } =\sum _{ k=0 }^{ 1427 }{ (1+\left\lceil 0.7k \right\rceil ) }$

$=1428+\sum _{ k=1 }^{ 1427 }{ \left\lceil 0.7k \right\rceil }$ .

Let´s solve the sum. With the first 10 terms of $\left\lceil 0.7k \right\rceil$ : 1, 2, 3, 3, 4, 5, 5, 6, 7, 7, its sum (43) and $\frac { 1427 }{ 10 } =142.7$ , $\sum _{ k=1 }^{ 1427 }{ \left\lceil 0.7k \right\rceil }$ is modified to

$\underbrace { \sum _{ k=1 }^{ 10 }{ \left\lceil 0.7k \right\rceil } +\sum _{ k=1 }^{ 10 }{ \left\lceil 0.7k+7 \right\rceil } +\sum _{ k=1 }^{ 10 }{ \left\lceil 0.7k+14 \right\rceil } +...+\sum _{ k=1 }^{ 10 }{ \left\lceil 0.7k+7(n-1) \right\rceil } +...+\sum _{ k=1 }^{ 10 }{ \left\lceil 0.7k+987 \right\rceil } }_{ 142\quad sums } +\sum _{ k=1 }^{ 7 }{ \left\lceil 0.7k+994 \right\rceil }$ . In this case $n$ is the number of the sum.

Simplifying the equation above

$142\sum _{ k=1 }^{ 10 }{ \left\lceil 0.7k \right\rceil } +10\sum _{ k=1 }^{ 141 }{ 7k } +\sum _{ k=1 }^{ 7 }{ 994 } +\sum _{ k=1 }^{ 7 }{ \left\lceil 0.7k \right\rceil }$

We know that $\sum _{ k=1 }^{ 10 }{ \left\lceil 0.7k \right\rceil } =43$ , therefore the answer is

$1428+142*43+10*7(\frac { 141*142 }{ 2 } )+7*994+(43-7-7-6)= 1428+ 6106+700770+6958+23= \boxed { 715285 }$

Let's tackle it with modulus! Recall that for any rational number, $\left\lceil \frac{m}{n}\right\rceil=\frac{m}{n} - \frac{\mod(m,\:n)}{n} + 1-\begin{cases} 1&\mod(m,\:n)=0\\ 0&\text{otherwise} \end{cases},\qquad m\in\mathbb{Z},\quad n\in\mathbb{N}$

For simplicity, let $N:=1427$ . Then we can rewrite the given series: \[\begin{align} S&:=\sum_{k=0}^N\left\lceil 1+\frac{7k}{10}\right\rceil=\sum_{k=0}^N\green{1+1}+\frac{7k}{10}-\frac{\mod(7k,\:10)}{10}-\begin{cases} \red{1}&\mod(7k,\:10)=0\\ 0&\text{otherwise} \end{cases}\\

&=\green{2(N+1)}+\frac{7N(N+1)}{10\cdot 2}-\red{ \left\lfloor\frac{N}{10}+1\right\rfloor }-\frac{1}{10}\sum_{k=0}^N\mod(7k,\:10)&(*) \end{align}\] The remaining part of the sum is tricky. Notice that $\mod(7k,\:10)$ has a period of 10: $\begin{array}{|c|rrrrrrrrrrrr|} \hline k&0&1&2&3&4&5&6&7&8&9&10&\ldots\\ \hline \mod(7k,\:10)&0&7&4&1&8&5&2&9&6&3&0&\ldots\\ \hline \end{array}$ $\begin{aligned} \Rightarrow &&\sum_{k=0}^9\mod(7k,\:10)&=\sum_{k=0}^9k=\frac{9(9+1)}{2}=45 \end{aligned}$ As long as we sum over a whole period, we will always get $45$ . Our sum consists of $142$ periods plus a rest, or (to make things easier) it shall consist of $143$ periods minus a rest: $\begin{aligned} \sum_{k=0}^N\mod(7k,\:10)&=143\cdot 45-\mod(7\cdot 1428,10)-\mod(7\cdot 1429,10)\\ &=143\cdot 45-6-3=6426 \end{aligned}$

Pluck that value and $N=1427$ into $(*)$ , to get $S=\boxed{715285}$

Now lets first calculate the arithmetic progression without the ceiling function ( f(x)), then sum what is added on from rounding. Step 1: let ${n}$ = number of terms such that ${n}$ = $\frac{999.9 -1}{0.7}$ +1 = 1428, ${a}$ = 1st term = 1, ${d}$ = common difference = 0.7. Now: $s_{n} = \frac{n}{2} (2a +(n-1)d) = 714642.6 < f(x)$ The amount added on each time repeats in a cycle mod (10) (as 0.7 $\times$ 10 = 7) as follows: (0 + 0.3 + 0.6 + 0.9 + 0.2 + 0.5 + 0.8 + 0.1 + 0.4 + 0.7) = 4.5. It is easy to see that $\frac{1428}{10}$ = 142 r8 $\therefore\ f(x) = 714642.6 + (4.5\times\ 142) + 3.4 = \boxed{715 285}$

Great problem! :D

The pattern $1,2,3,4,4,5,6,6,7,8,8,\ldots$ continues in the sequence. Notice that all terms congruent to $1,4,6$ mod $7$ are counted twice except for $1$ . There are a total of $143$ of the $4,6$ mod $7$ pairs and a total of $142$ of the $1$ mod $7$ pairs, so therefore the answer is $\dfrac{1001(1000)}{2}+142 \left(\dfrac{8+995}{2}\right)+143\left(\dfrac{6+1000}{2}+\dfrac{4+998}{2} \right) = \boxed{715285}.$