# CMI2014-A-03

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In the code fragment on the right, start and end are integer values and $\text{prime}(x)$ is a function that returns true if $x$ is a prime number and $\text{false}$ otherwise. At the end of the loop:

i := 0; j := 0; k := 0;
for (m := start; m <= end; m := m+1){
k := k + m;
if (prime(m)){
i := i + m;
}else{
j := j + m;
}
}
1. $k < i+j$
2. $k = i+j$
3. $k > i+j$
4. Depends on $\text{start}$ and $\text{end}$

edited

Explanation:

i := 0; j := 0; k := 0;
for (m := start; m <= end; m := m+1){
k := k + m; // "k" will have the accumulated sum from start to end.
if (prime(m)){
i := i + m; //prime values are added to "m".
}else{
j := j + m; //Values which are NOT prime are added to "j".
}
}

Here,

$\large \color {red}{{\mathrm {\large k}}}$ will contain the sum of all the values from start to end.

$\color {red} {\mathrm {\large i}}$ will contain the sum of values which are prime.

$\color {red} {\mathrm {\large j}}$ will contain the sum of values which are NOT prime.

Hence, finally the values in k is the sum of values in i and j.

$\therefore$ The relation is: $\color {blue}{\mathrm {\large k = i + j} }$

Hence, is the right option.

edited by
1
Nicely explained. π
1
Thanks.
0
I think the below line:-

i := i + m; //prime values are added to "m".

should be changed to:-

i := i + m; //prime values are added to "i".

B. $k = i + j$

Whatever be the value of m, the value added to k is also added to exactly one of i and j.

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