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Chứng minh
b) Thiếu đề với p>3. nhé!. Vì p=3 thì p+100=103 là số nguyên tố
p là số nguyên tố nên có dạng 3k+1, 3k+2, thuộc N
Với p=3k+1 => p+8=3k+9 \(⋮3\)loại vì p+8 là số nguyen tố
Với p=3k+2=> p+100=3k+2+100=3k+102 =3(k+34) chia hết cho 3
=> p+100 là hợp số.
Bài 1:
+Nếu p = 2 ⇒⇒ p + 2 = 4 (loại)
+Nếu p = 3 ⇒⇒ p + 6 = 9 (loại)
+Nếu p = 5 ⇒⇒ p + 2 = 7, p + 6 = 11, p + 8 = 13, p + 12 = 17, p + 14 = 19 (thỏa mãn)
+Nếu p > 5, ta có vì p là số nguyên tố nên ⇒⇒ p không chia hết cho 5 ⇒⇒ p = 5k+1, p = 5k+2, p = 5k+3, p = 5k+4
-Với p = 5k + 1, ta có: p + 14 = 5k + 15 = 5 ( k+3) ⋮⋮ 5 (loại)
-Với p = 5k + 2, ta có: p + 8 = 5k + 10 = 5 ( k+2 ) ⋮⋮ 5 (loại)
-Với p = 5k + 3, ta có: p + 12 = 5k + 15 = 5 ( k+3) ⋮⋮ 5 (loại)
-Với p = 5k + 4, ta có: p + 6 = 5k + 10 = 5 ( k+2) ⋮⋮ 5 (loại)
⇒⇒ không có giá trị nguyên tố p lớn hơn 5 thỏa mãn
Vậy p = 5 là giá trị cần tìm
Bài 2:
ta có: p + 8 là số nguyên tố
=> p > 3
mà p là số nguyên tố
=> p được viết dưới dạng: 3k+1; 3k+2
nếu p = 3k + 1 => p + 8 = 3k + 1 + 8 = 3k + 9 chia hết cho 3 ( vô lí, p + 8 sẽ không là số nguyên tố ( đầu bài cho)) (Loại)
nếu p = 3k + 2 => p + 100 = 3k + 2 + 100 = 3k + 102 chia hết cho 3
=> p + 100 là hợp số (đpcm)
TL:
a)Để P+2;P+6; P+8 là số nguyên tố thì \(P=5\)
hc tốt
hàng số 2
1 số 8 ở hàng 2