Search results
Results from the WOW.Com Content Network
$\begingroup$ This is the standard way, in the specific meaning of compliance to international standards: ISO 80000-2, clause 2.7-17.
It may be awkward if in an exam the marker doesn't know what it means. The only way in normal notation to write it is 5 ∣ 5(2 + 3m). 5 ∣ 5 (2 + 3 m). I don’t know of a standard symbol for “a is divisible by b.”. This is also new to me.
9. Yes, it is. Formal definition of divisibility is the following: Let a, b ∈ Z. Then a is divisible by b if and only if there exists an integer k such that a = kb. Now, in your case, we have a = b = 0, that is, 0 = k ⋅ 0 and there exists infinitely many such k so yes, 0 is divisible by 0. Share.
so that = (p3 − p) + 3p(p + 1) = (p 3 − p) + 3 p (p + 1) is divided by 6. hence, the answer is correct when n=p+1. When n=p the answer is correct too. We proved that the answer is correct when n=1. so as the mathematical induction, n3 − n n 3 − n is divided by 6 for the all positive integers.
Contrary, 22 cannot be nicely divisible by 3, 22/3 does not have a remainder of zero, 22/3 = "7 R 1", or 7 + 1/3. Go back to the problem and use this reasoning as a template for the solution. Since we know that p(x) is divisible by x (nicely), then we know that the quantity that remains in terms of x must be equal to zero.
2. I presume you mean that n2 − 1 n 2 − 1 is divisible by 8 8 for all odd values of n ∈N n ∈ N. In this case, let n = 2k + 1 n = 2 k + 1 with k ≥ 0 k ≥ 0. Then. n2 − 1 = (2k + 1)2 − 1 = 4k2 + 4k = 4k(k + 1) n 2 − 1 = (2 k + 1) 2 − 1 = 4 k 2 + 4 k = 4 k (k + 1) Now notice that either k k or k + 1 k + 1 is even to deduce ...
I always find myself doing tests with binary numbers (without a calculator, I'm now developing automatas) and I've always asked myself if there was a fast trick to check whether a generic number is divisible by another binary number.
1. Yes. More generally, if V V is a vector space over Q Q and A A is a subgroup (=Z Z -submodule) of V V, then the divisible (essential) hull of A A can naturally be identified to the Q Q -subspace VectQ(A) V e c t Q (A) generated by A A. This is shown by checking that the latter is an (essential) divisible hull of A A.
7. n3 + 3n2 + 2n = (n + 1)(n + 2)n n 3 + 3 n 2 + 2 n = (n + 1) (n + 2) n. One of the factors must be even and one must be a multiple of three. Hence the product is a multiple of both 2 2 and 3 3 and hence is divisible by the least common multiple of 2 2 and 3 3, which turns out to be 6 6. Share. Cite. answered Jul 13, 2015 at 14:27.
m = akak − 1…a1a0. Lemma 1: If m ≥ 18 satisfies. 9 divides ak + ak − 1 + ⋯ + a1 + a0. then 9 divides m. Proof. To get a contradiction assume that 9 does not divides m and take m to be the minimal positive integer where this happens. If ak − 1 = 9 then replacing it with 0, m ′ = m − 9 × Tk − 1 = ak0ak − 2…a1a0.