p+n=x
If ( p ) is an integer between 1000 and 1030, it can be expressed as ( p = 1000 + n ), where ( n ) ranges from 0 to 30. The sum of the digits of ( p ) is given by ( 1 + \text{(sum of the digits of } n) ). Since 1 is odd, for the total sum of the digits to be odd, the sum of the digits of ( n ) must be even. As a result, if ( p ) is odd, ( n ) must be odd (e.g., 1, 3, 5, etc.), confirming that ( p ) is indeed odd. Thus, the statement is true: if the sum of the digits of ( p ) is odd, then ( p ) must be odd.
6+n = 2n+5
if p represents your positive number, and n represents all of your negative numbers, then: |∑n| < p
The sum of n and 5 is n+ 5
void main () { int no1, sum, n; clrscr() sum = 0; n = 1; printf("\n enter the number to which sum is to be generated"); scanf("%d",&no1); while(n<=no1) { sum=sum+n; n=n+1 } printf("\n the sum of %d = %d, no1, sum"); getch (); }
18 + n
void main() { int n=0,i,*p; clrscr(); printf("enter the value of n"); scanf("%d",&n); p=&n; for(i=0;i<=n;i++) { *p=*p+i; } printf("value of the sum is %d",*p); getch(); }
If ( p ) is an integer between 1000 and 1030, it can be expressed as ( p = 1000 + n ), where ( n ) ranges from 0 to 30. The sum of the digits of ( p ) is given by ( 1 + \text{(sum of the digits of } n) ). Since 1 is odd, for the total sum of the digits to be odd, the sum of the digits of ( n ) must be even. As a result, if ( p ) is odd, ( n ) must be odd (e.g., 1, 3, 5, etc.), confirming that ( p ) is indeed odd. Thus, the statement is true: if the sum of the digits of ( p ) is odd, then ( p ) must be odd.
2(n+3)
6+n = 2n+5
The quotient of two and the sum of a number and twenty
The sum of (base current) plus (collector current).
3(n-9)
if p represents your positive number, and n represents all of your negative numbers, then: |∑n| < p
if p represents your positive number, and n represents all of your negative numbers, then: |∑n| < p
N can have any of four values: 125, 161, 209, and 221. Factors of 125 are 1, 5, 25, and 125. Factors of 161are 1, 7, 23, and 161. Factors of 209 are 1, 11, 19, and 209. Factors of 221 are 1, 13, 17, and 221.
/* Discrete Fourier Transform and Power Spectrum Calculates Power Spectrum from a Time Series Copyright 1985 Nicholas B. Tufillaro */ #include #include #define PI (3.1415926536) #define SIZE 512 double ts[SIZE], A[SIZE], B[SIZE], P[SIZE]; main() { int i, k, p, N, L; double avg, y, sum, psmax; /* read in and scale data points */ i = 0; while(scanf("%lf", &y) != EOF) { ts[i] = y/1000.0; i += 1; } /* get rid of last point and make sure # of data points is even */ if((i%2) == 0) i -= 2; else i -= 1; L = i; N = L/2; /* subtract out dc component from time series */ for(i = 0, avg = 0; i < L; ++i) { avg += ts[i]; } avg = avg/L; /* now subtract out the mean value from the time series */ for(i = 0; i < L; ++i) { ts[i] = ts[i] - avg; } /* o.k. guys, ready to do Fourier transform */ /* first do cosine series */ for(k = 0; k <= N; ++k) { for(p = 0, sum = 0; p < 2*N; ++p) { sum += ts[p]*cos(PI*k*p/N); } A[k] = sum/N; } /* now do sine series */ for(k = 0; k < N; ++k) { for(p = 0, sum = 0; p < 2*N; ++p) { sum += ts[p]*sin(PI*k*p/N); } B[k] = sum/N; } /* lastly, calculate the power spectrum */ for(i = 0; i <= N; ++i) { P[i] = sqrt(A[i]*A[i]+B[i]*B[i]); } /* find the maximum of the power spectrum to normalize */ for(i = 0, psmax = 0; i <= N; ++i) { if(P[i] > psmax) psmax = P[i]; } for(i = 0; i <= N; ++i) { P[i] = P[i]/psmax; } /* o.k., print out the results: k, P(k) */ for(k = 0; k <= N; ++k) { printf("%d %g\n", k, P[k]); } }