How do you know 111 is divisible by 3?

We have 1 + 1 + 1 = 3. Since the sum of the digits is divisible by the 3, therefore the number 111 is also divisible by 3.

How do you find numbers divisible by 3?

According to the divisibility rule of 3, a number is said to be divisible by 3 if the sum of all digits of that number is divisible by 3. For example, the number 495 is exactly divisible by 3. The sum of all digits are 4 + 9 + 5 = 18 and 18 is exactly divided by 3.

Why is a number divisible by 3 if its digits sum is divisible by 3?

As a result since both the first and the second sum are divisible by three, the integer number itself is divisible by 3. 6*99+7*9+(6+7+2) = (6*11+7)*9 + 15 = [(6*11+7)*3]*3 + 5*3 = 224*3, hence 672 is divisible by three as 6+7+2=15=5*3 is divisible by 3.

Which number is exactly divisible by 11?

Consider the following numbers which are divisible by 11, using the test of divisibility by 11: (i) 154, (ii) 814, (iii) 957, (iv) 1023, (v) 1122, (vi) 1749, (vii) 53856, (viii) 592845, (ix) 5048593, (x) 98521258. -1 is divisible by 11.

How to prove that $10^N -1$ is divisible by 11?

Prove by induction that $10^n -1$ is divisible by 11 for every even natural number n. $0 otin N$ Base Case: n = 2, since it is the first even natural number. $10^2 -1 = 99$ which is divisible by 11. Assume $n =k $ is true for some $k \\in N$. Now prove $n=k+1$ is true.

How do you show that an integer is divisible by 3?

Example (2.3.1)Show that an integer is divisible by 3 if and only if the sum of its digits is a multiple of 3. Let $n=a_0a_1\\ldots a_k$be the decimal representation of an integer $n$.

Is the sum of $3$ divisible by 3?

The divisibility rule for $3$ is well-known: if you add up the digits of $n$ and the sum is divisible by $3$, then $n$ is divisible by three. This is quite helpful for determining if really large numbers are multiples of three, because we can recursively apply this rule:

How do you prove $11|10^{2k}-1$ by induction?

Write $n=2k$. Then, you need to prove by induction that $11|10^{2k}-1$. Do induction by $k$. Otherwise, if you want to do induction by $n$, note that if $n$ is even, the next even number is $n+2$.