"FOR" loop

I have to find the prime numbers from that lead up to 100 and then use the 'for' function to multiply adjecent prime numbers: example:

first prime numbers

2 3 5 7

and i need the 'for' to do the following with the prime numbers

2*3 3*5 5*7 etc...HELP PLEASE

[Dinesh Kannan Natarajan]

% to Generate prime number between the intervals

clear;

clc;

m=input('provide the required start point of prime number: ');

n=input('provide the required end point of prime number: ');

if n<=0

pr='Error';

disp('ERROR: Input argument must be a positive integer')

elseif round(n)~=n||round(m)~=m

pr='Error';

disp('ERROR: Input argument must be positive integer')

elseif n <= m

pr='Error';

disp('ERROR: end point must be greater than start point')

else

k=1;

for i=m:n

c=0;

for j=2:i-1

if rem(i,j)==0

c=1;

break

end

end

if c==0

pr(k)=i;

k=k+1;

end

end

end

pr

Answer 1

[Junaid]

let say your prime vector is P;

p=primes(100);
a = zeros(1,length(p)-1);
for i=1:length(p)-1
 a(i) = p(i) * p(i+1);
end
a

though you can do it also without loop but you asked for a loop so it is loop solution.

[Juan Alvarez]

im sorry but this did not help...

i am at here so far

k=primes(100)

then it list the primes but i cant figure out how to use the for loop at all

Answer 2

[Junaid]

p=primes(100);
A = zeros(1,length(p)-1);
for i=1:length(p)-1
 A(i) = p(i) * p(i+1);
end
A

just copy and paste the code, answers is in vector (A) try this, do let us know if it is what you wanted to do.

[Juan Alvarez]

This helped a lot!! thanks!

Answer 3

[Sean de Wolski]

Loop sure seems like overkill

x = primes(100);
x1p1 = x(1:end-1).*x(2:end);

Answer 4

[Andrei Bobrov]

a = 1:100;
yourprime = a(a - sum(rem(triu(bsxfun(@rdivide,a,a.')),1)>0)<=2);
out = prod(yourprime(bsxfun(@plus,1:numel(yourprime)-1,(0:1)')))

[Paulo Silva]

It takes a while for me to understand your code, nicely done