Consider all integer combinations of ab for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:
22=4, 23=8, 24=16, 25=32
32=9, 33=27, 34=81, 35=243
42=16, 43=64, 44=256, 45=1024
52=25, 53=125, 54=625, 55=3125
If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?
SOLUTION
#include<stdio.h>main()
{
int res=9801;
int a,b,x;
for(a=2;a<=100;a++)
{
float p=2;
int i=2;
while(p>=2)
{
p=pow(a,1.0/i);
int k=p;
if(k==p){res=res-(98/i);}
i++;
}
}
res=res+2;//As there will be two numbers which wont be there
printf("%d",res);
}
TIME TAKEN:
9ms
ANSWER:
9183
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