warm up problem 17: sum of divisible numbers

Number Theory Level 3

Find the sum of all x such that $300<x<3000 \quad and \quad x\equiv 0 \pmod{11}$

The answer is 404250.

2 solutions

Shubhendra Singh
Oct 31, 2014

According to the given condition $x$ needs to be a multiple of 11 lying between 300 and 3000.

The values will be 308, 319, 330 ..............2992

It will be an AP with $a=308 \ d=11 \ and \ t_{n}=2992$ . By the formula of $n^{th}$ term we can find $n$

$t_{n} =a+(n-1)d$ from here we will get $n=245$ .

Sum of this AP will be

$S_{n}= \frac{n}{2}(a+t_{n})$

$=\frac{245}{2}(308+2992)$ $= \boxed{404250}$

Simple NCERT question!

Kartik Sharma - 6 years, 7 months ago

Has this appeared before, like, the exact problem?

Trevor Arashiro - 6 years, 7 months ago

No, I don't know but it usually is one of the problems in the indian school books.

Kartik Sharma - 6 years, 7 months ago

Not exactly this problem, but this type of problem! Also, in the NCERT books, these type of problems are written in words and not using the notations of modular arithmetic. On a side note, this problem shouldn't be level 2, it should be level 1.

Prasun Biswas - 6 years, 5 months ago

Dang, I figured it from 0. I need to read the problems a little more closely.

Daniel Barnhurst - 6 years, 7 months ago

Ratnadip Kuri
Jan 10, 2015

Using C++ it is very simple:

#include<stdio.h>

int main()
{
    int sum=0;

    for(int i=301;i<3000;i++)
    {
        if(i%11==0)
            sum+=i;
    }

    printf("Sum is: %d",sum);

    return 0;
}

warm up problem 17: sum of divisible numbers

The answer is 404250.

2 solutions

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