Related concepts and properties of 1. integer divisibility
Definition of (1) divisibility: For two integers A and d(d≠0), if there is an integer P, it is said that D is divisible by A, or A is divisible by D, and it is recorded as D | A. ..
If d is not divisible by a, it is recorded as d a, such as 2|6, 4 6.
(2) Nature
1) If b|a, then b|(-a) and bm|am are arbitrary nonzero integers m.
2) If a|b, b|a, then | a | = | b |
3) if b|a and c|b, then c | a.
4) If b|ac and (a, b)= 1((a, b)= 1 indicate that A and B are coprime, then B | c；;
5) If b|ac and b is a prime number, then b|a or b | c；;
6) If c|a and c|b, then c|(ma+nb), where m and n are arbitrary integers (this property can also be extended to the sum of more terms).
Example 1 (1987 Beijing Junior Two Mathematics Competition) X, Y and Z are integers. If 1 1 | (7x+2y-5z), verify: 1 1 | (3x-7y)
Prove that ∵ 4 (3x-7y+12z)+3 (7x+2y-5z) =11(3x-2y+3z).
And11|11(3x-2y+3z),
And 1 1 | (7x+2y-5z),
∴ 1 1 | 4(3x-7y+ 12z)
( 1 1,4)= 1.
∴ 1 1 |(3x-7y+ 12z)。
2. Proof method of divisibility problem
Divisibility of (1) utilization number (see lecture 2)
The value of Example 2( 1980 Canadian Competition Question) is 72 |.
Solution 72=8×9, and (8,9) =1,so we only need to discuss the divisible values of 8 and 9.
If 8 |, you can get 8 | by division, and b = 2.
If 9 |, then 9 | (a+6+7+9+2) gives a=3.
(2) Using the property of continuous integer product
The product of any two consecutive integers must be the product of odd and even numbers, so it must be divisible by 2.
(2) Any three consecutive integers have at least one even number and at least one multiple of 3, so their product can be divisible by 2 or 3, so it can also be divisible by 2×3=6.
This property can be extended to the continuous product of any integer.
Example 3( 1956 Beijing Competition Question) proves that any integer n is an integer, and the remainder 2 is divided by 3.
certificate
∵ is the product of two consecutive integers and must be divisible by 2.
∴ n is an integer of any integer,
∵ is an integer, that is, the original formula is an integer.
It's also VIII
,
2n, 2n+ 1 and 2n+2 are three consecutive integers, and their products must be multiples of 3, and 2 and 3 are coprime.
∴ is an integer divisible by 3.
Therefore, when it is divided by 3, it is greater than 2.
Example 4 If an integer A is not divisible by 2 and 3, a2+23 is divisible by 24.
To prove that ∵a2+23=(a2- 1)+24, just prove that a2- 1 is divisible by 24.
∵2 .∴a is an odd number. Let a=2k+ 1(k is an integer),
Then a2-1= (2k+1) 2-1= 4k2+4k = 4k (k+1).
∵k and k+ 1 are two consecutive integers, so k(k+ 1) must be divisible by 2.
∴8|4k(k+ 1), which means 8|(a2- 1).
∫(A- 1), a, (a+ 1) are three consecutive integers, and their products must be divisible by 3, that is, 3 | a (a-1) (a+1) = a (a2-).
∵3 a, ∴3|(a2- 1).3 and 8 are coprime, ∴24|(a2- 1), that is, a2+23 can be divisible by 24.
(3) Using the parity of integers
Next, we apply the knowledge of integer parity introduced in the third lecture to solve several integer problems.
Example 5 proves that there are no such integers a, b, c and d:
a b c d-a= ①
a b c d-b= ②
a b c d-c= ③
a b c d-d= ④
Proved by ①, a(bcd- 1)=.
The right end is odd, the left end A is odd, and bcd- 1 is odd.
Similarly, from ②, ③ and ④, we can see that B, C and D must be odd, then bcd is odd, bcd- 1 must be even, and then a(bcd- 1) must be even, which contradicts the right end of Formula ①. So the proposition was proved.
Example 6 (1985 Hefei Junior High School Mathematics Competition) has n real numbers, x 1, x2, …, xn, and each real number is either+1 or-1.
and
Try to prove that n is a multiple of 4.
Proof hypothesis (i= 1, 2, …, n- 1),
Yi is not+1 or-1, but y 1+y2+...+yn = 0, so the number of+1 is the same as that of-1, so n=2k. And y 1y2y3...yn.
∴n is a multiple of 4.
Other methods:
Integer A can be divisible by integer B, that is, B contains factor A, so it is a very natural idea to prove that A can be divisible by B by various formulas and deformation means.
Example 7 (The 4th Mathematics Invitational Tournament in the United States) What is the maximum value of a positive integer n that enables n3+ 100 to be divisible by n+ 10?
Solution n3+100 = (n+10) (N2-10n+100)-900.
If n+ 100 is divisible by n+ 10, then 900 can also be divisible by n+ 10. Further, when the value of n+ 10 is the maximum, the value of n is correspondingly the maximum. Because the maximum factor of 900 is 900, n+60.
Example 8 (Shanghai 1989 Senior Two Mathematics Competition) Let A, B and C be integers satisfying the inequality 1 < A < B < C, (AB-1) (BC-1) (CA-1).
Solution: (ab-1) (BC-1) (ca-1)
= a2b2c 2-ABC(a+b+c)+a b+ AC+BC- 1，①
∫ABC |(a b- 1)(BC- 1)(ca- 1)。
There is a positive integer k, so
ab+ac+bc- 1=kabc，②
k= < < <