1 b[4]={5,1,32,4};
2 k=++b[1];
3 l=b[1]++;
4 m=b[k++];
5 print k, l, m
Answers
The output is 3, 2, 32.
Skeleton code:
int b[4]={5,1,32,4};
int k,l,m;
k=++b[1];
l=b[1]++;
m=b[k++];
print k, l, m
Output:
3, 2, 32
Explanation:
k=++b[1] : b[1] = 1. On pre-incrementation, it becomes 1+1=2 in this statement itself. So, b[1]=2 and k=2; where updated array will be b[4]={5,2,32,4}.
l=b[1]++ : As the array is updated, now, b[1]=2. As the post incrementation is implemented here, the updation will be done later in the next statement. So, l=2.
m=b[k++] : By now, b[1] will become 3, and the updated array will be b[4]={5,3,32,4}. As k=2, due to post incrementation, it remains the same for this statement and incremented by the next step. So, b[2] which is 32, will be stored in m. So, m=32.
Now, k becomes 3.
The values to be printed are k, l, m. After all the updations are done, k=3, l=2, and m=32.
So, print k, l, m statement prints 3, 2, 32
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Answer:
Given data:
The first 100 natural numbers are taken into consideration. (1, 2 3,..., 100)
To find:
The Units digit of the sum of the third powers (cubes) of the first 100 natural numbers.
Solution:
We know that :
Starting from 1, the sum of cubes of first n natural numbers is: (\frac{n(n+1)}{2})^2(
2
n(n+1)
)
2
Here, n=100.
The sum of the cubes of first 100 natural numbers i.e.,
1³+2³+3³+...+100³ = (\frac{100(100+1)}{2})^2(
2
100(100+1)
)
2
= (\frac{100*101}{2})^2(
2
100∗101
)
2
= (50 x 101)²
= 5050²
As we need to calculate the units digit only, there is no need to calculate the complete value of 5050².
As the last digit of 5050 is 0, Squaring the number 5050 results the ending or units digit as 0.
Therefore, The Units digit of the sum of the third powers (cubes) of the first 100 natural numbers i.e.,
The units digit of 1³+2³+3³+...+100³ is 0.
Check:
5050² = 25502500
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