A prime sequence

take mod 5 $\begin{array}{c}a p\equiv p\pmod 5\\ p+6\equiv p+1\pmod 5\\ p+12\equiv p+2\pmod 5\\ p+18\equiv p+3\pmod 5\\ p+24\equiv p+4\pmod 5\\ \end{array}$ notice that this are 5 consecutive numbers.5 must divide one of them. so we need the number that 5 divides to be prime.

case 1: $p+3=5\Longrightarrow p=2$ . does not satisfy p+6=8

case 2: p=5. satisfies

consider cases

• p=5k, k must be 1, can check all others are prime
• p=5k+1, p+24=5k+25 not prime
• p=5k+2, p+18=5k+20 not prime
• p=5k+3, p+12=5k+15 not prime
• p=5k+4, p+6=5k+10 not prime

therefore only p that satisfies is p=5

Between the five numbers, one of them must be an exact multiple of 5. The only exact multiple of 5 that is prime is 5 itself. Since the numbers are all positive integers, 5 must be the smallest of them. Thus the sequence is 5, 11, 17, 23, 29.

Just put value 1,2,3,4,5
You will get your answer-------5

I tried to see which ones digit makes the sum to be multiple of 5. P + 6 = multiple of 5 on the numbers of ones digit of 4 and 9 P + 12 = multiple of 5 on the numbers of ones digit of 3 and 8 18; can't do 2 & 7 24; can't do 1 & 6 Only ones digits we can use are 0 and 5, and any number with 0 or 5 on ones place is composite number except for 5. Therefore 5 is the only answer.

The easy way is to use numbers in the value of p

First try : let p = 1, 3, 5 ....

Second try : let p = 2, 4,.....

Since the number is positive then the smallest is 5,

Using p = 5 we get:

P=5,

P= 5+6=11

And :: 17,23,29

Hence answer