It is proved that if P is a prime greater than 5, then p 2 - 1 is a multiple of 24

It is proved that if P is a prime greater than 5, then p 2 - 1 is a multiple of 24

It is proved that positive integers can be divided into six classes: 6K, 6K + 1, 6K + 2, 6K + 3, 6K + 4, 6K + 5. Because P is a prime number greater than 5, P can only belong to 6K + 1, 6K + 5. When p = 6K + 1, P2-1 = 36k2 + 12K = 12K (3K + 1), because there must be an even number in K, 3K + 1, then 24 is (P2 -...)

Can the remainder of a prime divided by 6 be 2 or 3? Why?

No way
Let this number 6N + 2 (n ≥ 0)
Then the number must not be prime, contrary to the hypothesis
Let 6N + 3 = 3 (2n + 1)
Then the number must not be prime, contrary to the hypothesis
It's impossible to divide a prime number by two or six

Prime divided by sum equals (prime or sum), hurry

A prime divided by a composite equals (prime or composite) a fraction is neither a prime nor a composite

Divide 2033 by Prime a, the quotient is a two digit number, the remainder is 35, what is prime a

37

200 divided by a two digit prime, the remainder is 14, what is the two digit prime?

200-14=186=31*6
This two digit prime is 31
This kind of topic must pay attention to small words, such as 186 ≠ 2 × 93, why? Because 2 is not a two digit number

23,29,31,37 prime divided by 6, what are the characteristics of the remainder

Remainder = 5 or 1
First of all, prime numbers greater than 2 are odd numbers, so the remainder of dividing by 6 can only be 1, 3, 5. If the remainder of being divided by 6 is 3, it can certainly be divided by 3. Except 3, other multiples of 3 are composite numbers. Therefore, if a prime number greater than 3 is divided by 6, the remainder can only be 1 or 5

If you divide 6 by other prime numbers, the remainder must be 1 or 5, right?

No

If an integer has the following properties: (1) the difference between this number and 1 is a prime number; (2) the quotient of this number divided by 2 is also a prime number; (3) the remainder of this number divided by 9 is 5. We call this integer a lucky number?

The multiples of 9 within 100 are: 9, 18, 27, 36, 45, 54, 63, 72, 80, 81, 90, 99. The remainder of (3) divided by 9 is 5:14, 23, 32, 41, 50, 59, 68, 77, 86, 95. The difference between (1) and 1 is prime. The quotient of (2) divided by 2 is prime

If a number is divided by 6, the quotient is 13 times of the remainder, and the remainder is the smallest prime number?

The smallest prime number is 2
This number
=2×13×6+2
=156+2
=158

The quotient of a number divided by 9 is the same as the remainder. The remainder is 15. The prime number is ()

The remainder cannot be greater than the divisor
In this problem, the divisor is 9 and the remainder is 15
Moreover, the problem of finding a prime number has nothing to do with the subject condition
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