How do you prove something is a multiple of 5?

How do you prove something is a multiple of 5?

The smallest multiple of every number is the number itself, that is, 5 × 1 = 5. If a number is a multiple of 5, it will have its last digit as 0 or 5. A multiple is termed to be a common multiple if it is common for two or more numbers. Example: 5 × 2 = 10, 2 × 5 = 10; 10 is a common multiple of 5 and 2.

How do you prove divisibility by 3?

is divisible by 3 or 9 if the sum of its digits is divisible by 3 or 9, respectively. Note that this does not work for higher powers of 3. For instance, the sum of the digits of 1899 is divisible by 27, but 1899 is not itself divisible by 27.

How do you show that something is divisible by 2?

Divisibility Rules for some Selected Integers

1. Divisibility by 1: Every number is divisible by 1 1 1.
2. Divisibility by 2: The number should have 0 , 2 , 4 , 6 , 0, \ 2, \ 4, \ 6, 0, 2, 4, 6, or 8 8 8 as the units digit.
3. Divisibility by 3: The sum of digits of the number must be divisible by 3 3 3.

Why do numbers divisible by 3 add up to 3?

When you add the digits of any number that is divisible by three, that sum of those digits also appears to be divisible by three (with no remainder). That sum 33 is divisible by three and so is the original number 289752.

Why do all multiples of 3 add up to 3?

Because the digits multiply by ( a multiple of three, plus one), for each digit. The first part of this is obviously divisible by 3. So for the whole thing to be divisible, the sum of the digits has to be as well.

How do you calculate multiples?

To find multiples of a number, multiply the number by any whole number. For example, 5 × 3 = 15 and so, 15 is the third multiple of 5. For example, the first 5 multiples of 4 are 4, 8, 12, 16 and 20. 1 × 4 = 4, therefore the 1st multiple of 4 is 4.

How do you prove $n^3+2n$ is divisible by $3?

Induction: Assume that for an arbitrary natural number $n$, $n^3+ 2n$ is divisible by $3.$. Induction Hypothesis: To prove this for $n+1,$ first try to express $( n + 1 )^3 + 2( n + 1 )$ in terms of $n^3 + 2n$ and use the induction hypothesis.

How to prove that a(n) holds for all positive integers n?

Let A(n) be an assertion concerning the integer n. If we want to show that A(n) holds for all positive integer n, we can proceed as follows: Induction basis: Show that the assertion A(1) holds. Induction step: For all positive integers n, show that A(n) implies A(n+1). 3 Standard Example

How do you prove that f(n) is constant?

This reduces the induction to a trivial lemma: f ( n) is constant if f ( n + 1) = f ( n) for all n ∈ N, a trivial telescopic induction – see here and here for more on this telescopy viewpoint. Given the n th case, you want to consider the ( n + 1) th case, which involves the number ( n + 1) 3 + 2 ( n + 1).

How do you prove that b(n+1) holds?

Expanding the right hand side yields n3/3 + 3n2/2 + 13n/6 + 1 One easily verifies that this is equal to (n+1)(n+2)(2(n+1)+1)/6 Thus, B(n+1) holds. Therefore, the proof follows by induction on n. 8 Tip How can you verify whether your algebra is correct?