Q: What is the smallest value of n 354N6 35455?

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The smallest value is 5.

After n years of 7 percent growth, the value is (1.07)n times the starting value. The answer to the question is the smallest value of n such that (1.07)n >=2 or nlog(1.07)>= log(2) or n >= log(2)/log(1.07) = 10.2 So the GDP will double during the 11th year.

In Statistics, ra nge does n't mea n divisio n but it is the differe nce of the largest value a nd the smallest value i n the list of numbers give n. Example: 11, 12, 13, 14, 15, 20, 23, 25 Ra nge = 25 - 11 = 14

Let's us say that the pth multiple of 14 is 63n i.e. 14p = 63n.14p = 63n and n = 14p/63....(1)From the relation 1 it is clear that we have to look for the multiple of 14 which is divisible by 63.Method 1:Let's us write the multiples of 14 greater than 63: 70, 84, 98, 112, 126, 140 etc.Is any of these multiples a multiple of 63?126 is the required multiple and it is the second multiple(n=2) of 63.So the smallest value of n is 2.Method 2:The smallest value of n can be found by calculating the LCM of 63 and 14.LCM of x and y gives us the smallest number divisible by x and y.LCM of 63 and 14 is 126. And putting 63n equal to 126 gives:63n = 126n =2

If you denote N as the smallest of them, you get N + (N + 1) + (N + 2) = 72. That gives you:3N + 3 = 72 OR 3N = 69.Divide both sides by 3 and you get N = 23

Related questions

The smallest value is 5.

n=27

27

7

#include #include #include int main(int argc, char *argv[]){int n, smallest, largest, sum, temp;if(argc < 2){printf("Syntax: foo val1[val2 [val3 [...]]]\n");exit(1);}smallest = largest = sum = atoi(argv[1]);for(n = 2; n < argc; n++){temp = atoi(argv[n]);if(temp < smallest) smallest = temp;if(temp > largest) largest = temp;sum += temp;}printf("Smallest: %i\nLargest: %i\nAverage: %i\n", smallest, largest, sum / (argc - 1));return 0;}

5

273

void main() { int a,b,c; clrscr(); printf("Enter the value of a:"); scanf("%d",&a); printf("\nEnter the value of b:"); scanf("%d",&b); printf("\nEnter the value of c:"); scanf("%d",&c); if(a>b) { if(a>c) { if(b>c) { printf("c is smallest\n"); printf("b is middle\n"); printf("a is largest\n"); } else { printf("b is smallest\n"); printf("c is middle\n"); printf("a is largest\n"); } } else { printf("b is smallest\n"); printf("a is middle\n"); printf("c is largest\n"); } } else if(b>c) { if(a>c) { printf("c is smallest\n"); printf("a is middle\n"); printf("b is largest\n"); } else { printf("a is smallest\n"); printf("c is middle\n"); printf("b is largest\n"); } } else { printf("a is smallest\n"); printf("b is middle\n"); printf("c is largest\n"); } getch(); }

The operation you describe is not a recursive one, it is iterative (linear in nature). It's also better to return the index of the smallest value rather than just the value itself. In this way we know the position of the smallest value as well as the value itself. To find the smallest element in an array, we start with the first element. Being the only value encountered so far, it must be the smallest, so we store its index. We then compare the value at the stored index to that of each of the remaining elements, one by one. When we find a smaller value, we update the stored index, because that value is the smallest encountered so far. When we reach the end of the array, the stored index will tell us which element was the smallest, so we simply return that index. The algorithm can (and should) be generalised to cater for arrays of any length. // Returns the index of the smallest value in array a of length n int smallest (const int* const a, int n) { if (n<1) return -1; // sanity check: the array must have 1 or more elements to be valid int i, j; i = 0; // assume first element holds smallest value until proven otherwise for (j=1; j<n; ++j) if (a[j] < a[i]) i = j; // update i each time a smaller value is found at index j return i; // a[i] holds the smallest value in the array } int main (void) { const int max = 10; int a[max] = {5, 4, 6, 3, 0, 7, 2, 8, 1, 9}; int i; i = smallest (a, max); assert (i==4); // a[4] holds the smallest value return 0; }

#include<iostream> #include<sstream> unsigned smallest_digit (double value) { unsigned long long integral = static_cast<unsigned long long>(value); while (value > integral) { value*=10; integral = static_cast<unsigned long long>(value); } unsigned smallest = integral % 10; while ((integral /= 10) != 0) { unsigned digit = integral % 10; if (digit < smallest) smallest = digit; } return smallest; } int main() { double d; while (true) { std::cout << "Enter a number: "; std::string s; std::getline (std::cin, s, '\n'); std::stringstream ss; ss << s; if (ss >> d) break; std::cerr << s << " is not a valid number.\n"; } std::cout << "The smallest digit in " << d << " is " << smallest_digit (d) << '\n'; }

After n years of 7 percent growth, the value is (1.07)n times the starting value. The answer to the question is the smallest value of n such that (1.07)n >=2 or nlog(1.07)>= log(2) or n >= log(2)/log(1.07) = 10.2 So the GDP will double during the 11th year.

In Statistics, ra nge does n't mea n divisio n but it is the differe nce of the largest value a nd the smallest value i n the list of numbers give n. Example: 11, 12, 13, 14, 15, 20, 23, 25 Ra nge = 25 - 11 = 14