p + 21.
If you replace some number for "p", you can calculate the sum. Otherwise, just write it as "p+9" - you can't simplify this.If you replace some number for "p", you can calculate the sum. Otherwise, just write it as "p+9" - you can't simplify this.If you replace some number for "p", you can calculate the sum. Otherwise, just write it as "p+9" - you can't simplify this.If you replace some number for "p", you can calculate the sum. Otherwise, just write it as "p+9" - you can't simplify this.
6 + n = 21
a+21
No, it is not.Incidentally, is a number p is prime then its 2 factors are 1 and p. Their sum is p+1.If p is a perfect number then this sum must be 2p.That is, p+1 = 2p or p = 1.But 1 is not a prime. So no prime can be a perfect number and since 13 is prime, it cannot be a perfect number.
sum of 14th square number and 10th square number
-8 × (number + 21) = 96 → number + 21 = 96 ÷ -8 → number = 96 ÷ -8 - 21 = -12 - 21 = -33
17
please clarify the question
19 and 2
There can be no number with an odd sum of 8 because 8 is not odd!
p/q form
Nikhil Aggarwalstrong_number: The sum of factorials of digits of a number is equal to the original number.void strong_number(){int num,i,p,r,sum=0,save_num;printf("\nEnter a number");scanf("%d",&num);save_num=num;while(num){i=1,p=1;r=num%10;while(i