# Divisibility Rules (2,3,5,7,11,13,17,19,...) | Brilliant Math & Science Wiki (2023)

A divisibility rule is a heuristic for determining whether a positive integer can be evenly divided by another (i.e. there is no remainder left over). For example, determining if a number is even is as simple as checking to see if its last digit is 2, 4, 6, 8 or 0. Multiple divisibility rules applied to the same number in this way can help quickly determine its prime factorization without having to guess at its prime factors.

#### Contents

• Basic Divisibility Rules
• Intermediate Divisibility Rules
• Divisibility Rules for Numbers $$\gt 20$$
• Proving Divisibility Rules
• Finding Divisibility Rules

## Basic Divisibility Rules

A positive integer $$N$$ is divisible by

• $$\color{green}{\boxed{\mathbf{2}}}$$ if the last digit of $$N$$ is 2, 4, 6, 8, or 0;
• $$\color{green}{\boxed{\mathbf{3}}}$$ if the sum of digits of $$N$$ is a multiple of 3;
• $$\color{green}{\boxed{\mathbf{4}}}$$ if the last 2 digits of $$N$$ are a multiple of 4;
• $$\color{green}{\boxed{\mathbf{5}}}$$ if the last digit of $$N$$ is either 0 or 5;
• $$\color{green}{\boxed{\mathbf{6}}}$$ if $$N$$ is divisible by both 2 and 3;
• $$\color{green}{\boxed{\mathbf{7}}}$$ if subtracting twice the last digit of $$N$$ from the remaining digits gives a multiple of 7 (e.g. 658 is divisible by 7 because 65 - 2 x 8 = 49, which is a multiple of 7);
• $$\color{green}{\boxed{\mathbf{8}}}$$ if the last 3 digits of $$N$$ are a multiple of 8;
• $$\color{green}{\boxed{\mathbf{9}}}$$ if the sum of digits of $$N$$ is a multiple of 9;
• $$\color{green}{\boxed{\mathbf{10}}}$$ if the last digit of $$N$$ is 0;
• $$\color{green}{\boxed{\mathbf{11}}}$$ if the difference of the alternating sum of digits of $$N$$ is a multiple of 11 (e.g. 2343 is divisible by 11 because 2 - 3 + 4 - 3 = 0, which is a multiple of 11);
• $$\color{green}{\boxed{\mathbf{12}}}$$ if $$N$$ is divisible by both 3 and 4.

Here are some example questions that can be solved using some of the divisibility rules above.

Without performing actual division, show that the number below is an integer:

$\dfrac{1,481,481,468}{12}.$

From the divisibility rules, we know that a number is divisible by 12 if it is divisible by both 3 and 4. Therefore, we just need to check that 1,481,481,468 is divisible by 3 and 4.

Applying the divisibility test for 3, we get that $$1+4+8+1+4+8+1+4+6+8=45,$$ which is divisible by 3. Hence 1,481,481,468 is divisible by 3.

Applying the divisibility test for 4, we get that the last two digits, 68, is divisible by 4. Hence 1,481,481,468 is also divisible by 4.

Now, since we know that 1,481,481,468 is divisible by both 3 and 4, it is divisible by 12. Therefore, $$\frac{1,481,481,468}{12}$$ will be an integer. $$_\square$$

Find all the possible values of $$a$$ such that the number $$\overline{98a6}$$ is a multiple of $$3.$$

From the rules of divisibility, the number $$\overline{98a6}$$ is a multiple of $$3$$ if and only if the sum of its digits $$9 + 8 + a + 6 = 23+a$$ is a multiple of $$3.$$ Since $$0 \leq a \leq 9$$, this implies that $$a = 1, 4, 7$$ are all the possible values. $$_\square$$

Without performing division, explain why the number $$987654321$$ is a multiple of $$9$$.

By the rule of divisibility of $$9,$$ since $$9+8+7+6+5+4+3+2+1 = 45$$ is a multiple of $$9,$$ $$987654321$$ is a multiple of $$9. \ _\square$$

Without performing actual division, show that $$87456399$$ is not divisible by $$11$$.

Applying the divisibility rule of $$11,$$ the difference between the sum of digits at the odd places $$(8+4+6+9 = 27 )$$ and the sum of digits at even places $$(7+5+3+9 = 24)$$ is $$27-24=3,$$ which is not divisible by $$11$$. Hence $$87456399$$ is not divisible by $$11$$. $$_\square$$

For what values of $$a$$ and $$b$$ is $$\overline{12ab}$$ a multiple of $$99?$$

Since $$99 = 9 \times 11$$, the number must be a multiple of $$9$$ and $$11.$$

The divisibility rule of $$9$$ tells us that $$1 + 2 + a + b$$ is a multiple of $$9.$$ Since it is a number from $$3$$ to $$21,$$ it must be either $$9$$ or $$18.$$

(Video) Divisibility rules for 2,3,4,5,6,7,8,9,10,11 and 13

Now, the divisibility rule of $$11$$ tells us that $$1 - 2 + a - b$$ is a multiple of $$11.$$ Since it is a number from $$-10$$ to $$8,$$ it must be $$0.$$

Solving $$\begin{cases} 1 + 2 + a + b = 9 \\ 1 - 2 + a - b = 0, \end{cases}$$ there are no integer solutions.

Solving $$\begin{cases} 1 + 2 + a + b = 18 \\ 1 - 2 + a - b = 0, \end{cases}$$ we get $$a = 8, b = 7$$.

Hence, the only solution is $$1287$$ with $$a = 8$$ and $$b = 7.$$ $$_\square$$

Is $$65973390$$ divisible by $$210?$$

We do not know the divisibility rule of 210. However, we can easily see that $$210=2\times 3\times 5\times7$$, so if 65973390 is divisible by 2, 3, 5, 7, then it is divisible by 210.

• Since the last digit of 65973390 is 0, it is divisible by 2.
• Since $$6+5+9+7+3+3+9+0=42$$, which is divisible by 3, it follows that 65973390 is divisible by 3.
• Since the last digit of 65973390 is 0, hence it is divisible by 5.
• To check divisibility by 7, as the initial step, we calculate $$6597339-2(0)=6597339$$. However, this number is still a little too big for us to tell whether it's divisible by 7. In such cases, we keep applying the divisibility rule again and again until we have a small enough number to work with:\begin{align}659733-2(9)&=659715\\65971-2(5)&=65961\\6596-2(1)&=6594\\659-2(4)&=651\\65-2(1)&=63.\end{align}Now we can see that we are left with $$63,$$ which we can easily identify as a multiple of 7. Hence 65973390 is a multiple of 7 also.

Since 65973390 is divisible by all of 2, 3, 5, 7, it is divisible by $$2\times3\times5\times7=210. \ _\square$$

Try some problems for yourself to see if you understand this topic:

If we know an integer is a multiple of 5, how many possibilities are there for the last two digits of the integer?

Yes No

Is 87985 divisible by 7?

## Intermediate Divisibility Rules

A positive integer $$N$$ is divisible by

• $$\color{orange}{\boxed{\mathbf{13}}}$$ if 4 times the units digit of $$N$$ plus the number obtained by removing the units digit of $$N$$ is a multiple of 13;
• $$\color{orange}{\boxed{\mathbf{17}}}$$ if the units digit subtracted 5 times from the remaining number (excluding the units digit) results in a number that is divisible by 17;
• $$\color{orange}{\boxed{\mathbf{19}}}$$ if doubling the units digit and adding it to the number formed by removing the units digit in the original number is divisible by 19.

You may use these rules repeatedly until you can tell if a number is divisible by another number or not.

It is also useful to note that the notation $$x | y$$ is used to say $$x$$ divides $$y$$, i.e. the fraction $$\frac{y}{x}$$ gives an integer. For example, $$\color{green}{ 3| 6 }$$ is true, but $$\color{red}{ 2| 7}$$ is false.

Is $$45238$$ divisible by $$13?$$

(Video) Divisibility Tricks - Numberphile

From the rules of divisibility, the number $$45238$$ is a multiple of $$13$$ if and only if $$13| (4\times 8+4523) \implies 13| 4555$$. Repeating the divisibility rule, we get $$13| (4\times 5+455) \implies 13| 475$$. Repeating the rule once more, we get $$13| (4\times 5 + 47) \implies 13| 67$$, which is clearly $$\color{red}{\text{false}}$$. Therefore, $$45238$$ is not divisible by $$13$$. $$_\square$$

Is $$2853598728$$ divisible by $$24?$$

$$24$$ is a composite number, so we will have to deal with it in a slightly different way. We can write $$24$$ as $$3 \times 8$$. If a number is divisible by both $$3$$ and $$8$$, then the number is also divisible by $$24$$. We choose $$3$$ and $$8$$ because they are coprime, and also because we know the divisibility rules for $$3$$ and $$8$$.

Let's test if $$2853598728$$ is divisible by $$8$$. The last three digits of the number are $$728$$ which is divisible by $$8$$, so $$2853598728$$ is also divisible by $$8$$. Now let's see if $$2853598728$$ is divisible by $$3$$. The sum of digits of $$2853598728$$ is $$57$$. Since $$57$$ is divisible by $$3$$, $$2853598728$$ is also divisible by $$3$$. Since $$2853598728$$ is divisible by both $$3$$ and $$8$$, we can conclude that $$2853598728$$ is divisible by $$24$$. $$_\square$$

Is $$365226929$$ divisible by $$2976?$$

We do not know the divisibility rule for $$2976$$. However, we know that since $$2976$$ is even, any integral multiple of $$2976$$ will be even. However, $$365226929$$ is not even, so $$365226929$$ is not divisible by $$2976$$. $$_\square$$

Is $$25729875$$ divisible by $$759284521?$$

Since $$0 < 25729875 < 759284521$$, we see that it is between two consecutive multiples of $$759284521$$. Hence $$25729875$$ is not a multiple of $$759284521$$, and $$25729875$$ is not divisible by $$759284521$$. $$_\square$$

Try some problems for yourself to see if you understand this topic:

Yes No Insufficient information This question is flawed

Is 986244 divisible by 13?

## Divisibility Rules for Numbers $$\gt 20$$

$$\color{red}{\boxed{\mathbf{23}}}$$ Add 7 times the last digit to the remaining truncated number. Repeat the step if necessary. If the result is divisible by 23, the original number is also divisible by 23.

• Check for 53935: $$5393+7\times 5 = 5428 \implies 542+7\times 8= 598 \implies 59+ 7\times 8=115,$$ which is 5 times 23. Hence 53935 is divisible by 23.

$$\color{red}{\boxed{\mathbf{29}}}$$ Add 3 times the last digit to the remaining truncated number. Repeat the step if necessary. If the result is divisible by 29, the original number is also divisible by 29.

• Check for 12528: $$1252+3\times 8= 1276 \implies 127+3\times 6= 145\implies 14+ 3\times 5=29,$$ which is divisible by 29. So 12528 is divisible by 29.

$$\color{red}{\boxed{\mathbf{31}}}$$ Subtract 3 times the last digit from the remaining truncated number. Repeat the step if necessary. If the result is divisible by 31, the original number is also divisible by 31.

• Check for 49507: $$4950-3\times 7=4929 \implies 492-3\times 9= 465\implies 46-3\times 5=31.$$ Hence 49507 is divisible by 31.

$$\color{red}{\boxed{\mathbf{37}}}$$ Subtract 11 times the last digit from the remaining truncated number. Repeat the step if necessary. If the result is divisible by 37, the original number is also divisible by 37.

• Check for 11026: We have $$1102 – 11\times 6 =1036.$$ Since $$103 – 11\times 6 =37$$ is divisible by 37, 11026 is divisible by 37.

$$\color{red}{\boxed{\mathbf{41}}}$$ Subtract 4 times the last digit from the remaining truncated number. Repeat the step if necessary. If the result is divisible by 41, the original number is also divisible by 41.

(Video) Divisibility rule of 11.

• Check for 14145: We have $$1414 – 4\times 5 =1394.$$ Since $$139 – 4\times 4 =123$$ is divisible by 41, 14145 is divisible by 41.

$$\color{red}{\boxed{\mathbf{43}}}$$ Add 13 times the last digit to the remaining truncated number. Repeat the step if necessary. If the result is divisible by 43, the original number is also divisible by 43. (This process becomes difficult for most of the people because of multiplication with 13.)

• Check for 11739: We have $$1173+13\times 9= 1290.$$ Since 129 is divisible by 43, the 0 can be ignored. So, 11739 is divisible by 43.

$$\color{red}{\boxed{\mathbf{47}}}$$ Subtract 14 times the last digit from the remaining truncated number. Repeat the step if necessary. If the result is divisible by 47, the original number is also divisible by 47. (This too is difficult to operate for people who are not comfortable with table of 14.)

• Check for 45026: We have $$4502 – 14\times 6 =4418.$$ Since $$441 – 14 \times 8 =329,$$ which is 7 times 47, 45026 is divisible by 47.

## Proving Divisibility Rules

Prove that when a number is divisible by $$7,$$ the result when subtracting twice the last digit from the number formed by the remaining digits is a multiple of $$7.$$

Let the number be $$N= 10a + b$$, where $$b$$ is the last digit of $$N$$ and $$a$$ is the number formed by the remaining digits.

Now we have to prove that if $$a-2b$$ is divisible by $$7,$$ then $$N=10a+b$$ must be also divisible by $$7.$$

Let $$a-2b=7k,$$ where $$k$$ is an integer, and multiply this equation by $$10$$ and add $$b$$ to both sides. Then you'll get

$10a-20b+b=70k+b.$

Now move $$-20b$$ to the right, and you'll get

$10a+b=N=70k+21b,$

which is a multiple of $$7.$$ $$_\square$$

Prove that when a number is divisible by $$8,$$ the last $$3$$ digits are a multiple of $$8.$$

Let the number be $$N= 1000M + 100a + 10b + c$$, where $$a, b, c$$ are digits and $$M$$ is a non-negative integer.

Clearly, $$1000M$$ is a multiple of $$8,$$ since $$8 \mid 2^3 \cdot 5^3$$. Hence, $$N$$ is a multiple of $$8$$ if and only if $$100a + 10b + c$$ is a multiple of $$8.$$ $$_\square$$

Prove that when a number is divisible by $$11,$$ the alternating sum of digits is a multiple of $$11.$$

From factorization , we know that

$10^n - (-1)^n = \big(10 - (-1) \big) \left( 10^{n-1} + 10^{n-2}(-1) + \cdots + (-1)^{n-1} \right)$

is always a multiple of $$11.$$

Hence, if $$N = 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10 a_1 + a_0$$, then

\begin{aligned}N = &\left[\left(10^k - (-1)^k\right) a_k + \left( 10^{k-1} - (-1)^{k-1}\right) a_{k-1} + \cdots + (10+1)a_1 + (1-1)a_0 \right]\\\\& \hspace{4cm}+ (-1)^k a_k + (-1)^{k-1}a_{k-1} + \cdots + (-1)^1 a_1 + (-1)^0 a_0.\end{aligned}

Since the terms in the square brackets consist of multiples of $$11,$$ it follows that $$N$$ is a multiple of $$11$$ if and only if the alternating sum is a multiple of $$11.$$ $$_ \square$$

## Finding Divisibility Rules

Although finding clever divisibility tricks requires a bit of ingenuity, finding general divisibility tests is not hard. In this section, we are only going to focus on finding divisibility tests for primes as once we know test for primes, we can decompose any composite number into primes and test for divisibility.

Find the divisibility test for 13.

Let's first find a test for small 3-digit numbers first and then find a general test. We'll be finding tests which involves separation of last digit. These tests will be similar to the tests of 19, 23, 29, etc.

Let $$\overline { abc }$$ be any number such that $$\overline { abc } =100a+10b+c$$. Now assume that $$\overline { abc }$$ is divisible by 13. Then

\begin{align}\overline { abc } &\equiv 0 \pmod{13}\\ 100a+10b+c &\equiv 0 \pmod{13}\\10\left( 10a+b \right) +c &\equiv 0 \pmod{13}\\ 10\overline { ab } +c &\equiv 0 \pmod{13}.\end{align}

Now that we have separated the last digit from the number, we have to find a way to use it: make the coefficient of $$\overline { ab }$$ 1. In other words, we have to find an integer $$n$$ such that $$10n\equiv 1 \bmod{13}$$. It can be observed that the smallest $$n$$ which satisfies this property is 4. Now we can multiply the original equation by 4 and simplify it:

\begin{align}10\overline { ab } +c &\equiv 0 \pmod{13} \\ 40\overline { ab } +4c &\equiv 0 \pmod{13}\\ \overline { ab } +4c &\equiv 0 \pmod{13}.\end{align}

Aha! We have found out that if $$\overline { abc } \equiv 0 \bmod{13},$$ then $$\overline { ab } +4c\equiv 0 \bmod{13}$$. In other words, to check if a 3-digit number is divisible by 13, we can just remove the last digit, multiply it by 4, and then add it to the rest of the two digits.

Now that we have found out the test for 3-digit numbers, let's find out a general divisibility test. Let $$\overline { { x }_{ n }{ x }_{ n-1 }{ x }_{ n-2 }\dots { x }_{ 2 }{ x }_{ 1 } }$$ be any $$n$$ digit number. Then

\begin{align}\overline { { x }_{ n }{ x }_{ n-1 }{ x }_{ n-2 }\dots { x }_{ 2 }{ x }_{ 1 } } &\equiv 0 \pmod{13}\\ 10\overline { { x }_{ n }{ x }_{ n-1 }{ x }_{ n-2 }\dots { x }_{ 2 } } +{ x }_{ 1 } &\equiv 0 \pmod{13}\\ 40\overline { { x }_{ n }{ x }_{ n-1 }{ x }_{ n-2 }\dots { x }_{ 2 } } +4{ x }_{ 1 } &\equiv 0 \pmod{13}\\ \overline { { x }_{ n }{ x }_{ n-1 }{ x }_{ n-2 }\dots { x }_{ 2 } } +4{ x }_{ 1 } &\equiv 0 \pmod{13}.\end{align}

Hence, a number is a multiple of 13 if we add 4 times the last digit to the rest of the number and the resulting number is still divisible by 13. $$_\square$$

This process can be repeated by any arbitrary prime number to find its divisibility test. Now can you find the divisibility tests for 17, 19, 23 and 29 on your own?

Try some problems for yourself to see if you understand this topic:

One can test if an integer $$n$$ is divisible by prime $$101$$ by subtracting the last two digits (as a number) from $$n$$ with those digits shaved off, and see if the result is divisible by $$101$$. For example, $$162794931$$ is divisible by $$101$$ because $$1627949 - 31 = 1627918$$ and $$101 \, | \, 1627918$$.

What are the smallest factors you should multiply the last two digits of $$n$$ (as a number) with, before subtracting from the shaved-off $$n$$ in the divisibility tests for $$p = 43$$ and $$p = 67?$$

(Video) Divisibility Rules

• Proof of Divisibility Rules
• Prime Numbers
• Prime Factorization
• Remainder
(Video) Divisibility Rule - part 1 ( 2-15 & co-primes concept)

## FAQs

### What is the divisibility rule of 17 and 19? ›

The divisibility rule of 17 states, “If the difference between 5 times the last digit and the rest is either 0 or multiple of 17, then the number is divisible by 17”. For example, for 289, (9×5) – 28 = 17.

What is divisibility rule 11? ›

Divisibility Rules for 11

If the difference of the sum of alternative digits of a number is divisible by 11, then that number is divisible by 11 completely.

What is the divisibility rule of 17? ›

In other words , we can say , a number is divisible by 17 if you multiply the last digit by 5 and subtract that from the rest. If that result is divisible by 17, then your number is divisible by 17.

What is the divisibility rule of 13? ›

Rule: Multiply the last digit, i.e. unit digit by 9 of a number N and subtract it from the rest of the number. If the outcome is divisible by 13 then the number N is divisible by 13. Example: Check whether the number 858 is divisible by 13.

What is the divisibility of 13? ›

If adding four times the last digit to the number formed by the remaining digits is divisible by 13, then the number is divisible by 13. Apart from 13, there are divisibility rules for 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, and so on.

What is the divisibility rule of 17 and 13? ›

Divisibility Test of 13 and 17

To check if a number is divisible by 17, we multiply the units place digit of a number by 5 and find the difference between the product and the rest of the number. If the difference is a multiple of 17 or 0, then the given number is divisible by 17.

What is divisibility of 7? ›

The divisibility rule of 7 states that, if a number is divisible by 7, then “the difference between twice the unit digit of the given number and the remaining part of the given number should be a multiple of 7 or it should be equal to 0”. For example, 798 is divisible by 7. Explanation: The unit digit of 798 is 8.

What is the divisibility rule of 2 to 11? ›

A number is divisible by 6 if it is divisible by both 2 and 3. A number is divisible by 4 if its last two digits are divisible by 4. A number is divisible by 8 if its last three digits are divisible by 8. A number is divisible by 11 if the difference between the sum of alternate pairs of digits is divisible by 11.

What is the divisibility rule of 7 and 13? ›

The original number is divisible by 7 (or 11 or 13) if this alternating sum is divisible by 7 (or 11 or 13 respectively). The alternating sum in our example is 963, which is clearly 9*107, and not divisible by 7, 11, or 13.

What is divisible by 5? ›

Divisibility Rule of 5

If a number ends with 0 or 5, it is divisible by 5. For example, 35, 790, and 55 are all divisible by 5.

### What is the divisibility rule of 5? ›

A number is divisible by 5 only if the units digit is 0 or 5.

Is 17 divisible by 3? ›

Since 17 is not divisible by 3, it means 3194 is not exactly divisible by 3.

What is 18 divisibility? ›

Hint: In the above question we will use the concept of divisibility rule for 18: if a number is divisible by both 2 and 9, then the number is exactly divisible by 18. For divisibility by 2 the last digit of a number ends with 0, 2, 4, 6, 8. For divisibility by 9 the sum of the digits of the number is divisible by 9.

Which of the following is divisible by11? ›

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.

What is the divisibility rule of 7 and 11? ›

If the difference is 0 or a multiple of 7, then the given number is divisible by 7. According to the divisibility rule of 11, a number is divisible by 11 if the difference of the sum of the digits at the odd positions and even positions are either equal to 0 or a multiple of 11.

What is divisible by 7 and 11? ›

∴ 16324 is divisible by both 7 and 11.

How do you test for divisibility for 11? ›

Here an easy way to test for divisibility by 11. Take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11, so is the original number. So, for instance, 2728 has alternating sum of digits 2 – 7 + 2 – 8 = -11.

Is 17 divisible by 2? ›

Since the division does not result in a whole number, this shows us that 17 is not divisible by 2.

What is the remainder when 13 17 17 13? ›

∴ Remainder when 1317+1713 is 3+7=10.

What is a number divisible by 7 11 and 13? ›

7 × 11 × 13 = 1001, when any three digit number multiplied by 1001 then it repeats itself and always completely divisible by 7, 11 and 13. Calculation: Let 2 digit number be prq. Hence that number is 823823.

### What is the divisibility rule of 2? ›

The divisibility rule for 2 states that any number with the last digit of 0, 2, 4, 6, or 8 will be divisible by 2. Simply put, any even number (numbers that end in 0, 2, 4, 6, or 8) is divisible by 2. If the number is not an even number, it is not divisible by two.

What are 7 and 9 divisible by? ›

The LCM of 7 and 9 is 63. To find the least common multiple (LCM) of 7 and 9, we need to find the multiples of 7 and 9 (multiples of 7 = 7, 14, 21, 28 . . . . 63; multiples of 9 = 9, 18, 27, 36 . . . . 63) and choose the smallest multiple that is exactly divisible by 7 and 9, i.e., 63.

What is the divisibility of 15? ›

According to the divisibility rule of 15, a numeral is divisible by 15 if it is divisible by both 3 and 5. Since the unit digit of 1440 is 0, it is divisible by 5. Hence, the sum of digits is 9, it is divisible by 3. Since 1440 is divisible by both 3 and 5, 1440 is divisible by 15.

What is the divisibility of 3? ›

The divisibility rule of 3 states that if the sum of digits of a number is a multiple of 3, the number will be completely divisible by 3.

Who found the divisibility rule of 7? ›

“Multiply the last digit by 5 and add it to the remaining number.” This is the divisibility rule which was said to be “discovered” by a 12-year old Nigerian boy, Chika Ofili for which he was given recognition by his school. Examples: 651: 65 + 1 x 5 = 70.

What is 3 divisible by? ›

The answer to your question is yes. We have calculated all the numbers that 3 is evenly divisible by. The numbers that 3 is divisible by are 1 and 3.

Is 13 is divisible by 3? ›

For example, 13 is a prime number and so it not divisible by 3. The rule for divisibility by 3 works for all numbers no matter how large. For example, here is the number 529, 943.

What is divisible by 5 and 7? ›

35 is divisible by 7 and 5. This can also be done by checking if the number is divisible by 35, since the LCM of 7 and 5 is 35 and any number divisible by 35 is divisible by 7 and 5 and vice versa also.

What is divisible by 6 and 7? ›

Numbers that are divisible by 5 are 5, 10, 15, etc. Numbers that are divisible by 6 are 6, 12, 18, etc. And numbers that are divisible by 7 are 7, 14, 21, etc.

What are 2 and 5 divisible by? ›

If a number ends in 2, 4, 6, 8 or 0, it is divisible by 2. If it ends in 5 or 0, it is divisible by 5. If it ends in 0, it is divisible by 10. If it is divisible by 10, it is also divisible by 2 and 5.

### What is the rule of 5 in math? ›

The divisibility rule of 5 states that if the digit on the units place, that is, the last digit of a given number is 5 or 0, then such a number is divisible by 5. For example, in 39865, the last digit is 5, hence, the number is completely divisible by 5.

What is the rule for 5? ›

The rule of five is a rule of thumb in statistics that estimates the median of a population by choosing a random sample of five from that population. It states that there is a 93.75% chance that the median value of a population is between the smallest and largest values in any random sample of five.

Is 2837393449 divisible by 5? ›

Answer: The correct answer is option D, 11. Step-by-step explanation: The given number is not divisible by 5 because, unit place is neither 5 nor 0.

Is 12 divisible by 3? ›

As 12 is divisible by 3, then 138 is also divisible by 3. Since 138 is divisible by both 3 and 2, 138 it is also divisible by 6.

Is 9 divisible by 3? ›

9 is a multiple of 3, therefore 1098 is divisible by 3.

Is 27 divisible by 9? ›

27 = 9 × 3 and so, 27 is divisible by 9. 27 is divisible by 9 therefore 8595 is also divisible by 9.

What is 16 divisibility rule? ›

If the thousands digit is even, the number formed by the last three digits must be divisible by 16 . For example: 544 , 480 the thousands digit 4 is even and the last three digits form 480 which is divisible by 16 .

Is 3333333 is divisible by 11? ›

A number is divisible by 11, if the difference of the sum of its digits in odd places and the sum of the digits in even places (starting from ones place) is either 0 or a multiple of 11. Consider the number 3333333. Since this number (3) is not divisible by 11, 3333333 is not divisible by 11.

Is 22222 divisible by 11? ›

∴ The difference of the sum of alternative digits of a number is 2, which is not divisible by 11. Hence, 22222 is not divisible by 11.

Is 61809 divisible by 11 solve using divisibility rule? ›

### What are the multiples of 19? ›

Consider the first 10 multiples of 19, i.e. 19, 38, 57, 76, 95, 114, 133, 152, 171 and 190.

What is 19 divided half? ›

Half of 19 is 9.5.

What is 20 divisibility rule? ›

The divisibility rules for 20 will be that the last two digits of the given number has to be divisible by 20 which means numbers having 00, 20, 40, 60, 80 at the end will be divisible by 20.

What is the smallest 5 digit number divisible by 19? ›

Hence, the smallest 5 digit number that is divisible by 19 is 10013.

What are the multiples of 17? ›

The multiples of 17 are 17, 34, 51, 68, 85, 102, 119, 136, 153, 170, etc. We can observe that it is a sequence where the difference between each next number and the preceding number, i.e. two consecutive multiples or products, is equal to 17.

In which table does 57 comes? ›

Table of 57 is a multiplication table that represents the repeated addition when 57 is multiplied by natural number. For example, 57 multiplied by 4 is given by 57 + 57 + 57 + 57 = 228.
...
What is 57 times table?
57 × 1 = 5757
57 × 2 = 2457 + 57 = 114
57 × 3 = 3657 + 57 + 57 = 171
57 × 4 = 4857 + 57 + 57 + 57 = 228
6 more rows

What table comes with 43? ›

Multiplication Table of 43
43×43
43×86
43×129
43×172
43×215
15 more rows

What is Half of 1½? ›

Answer: Half of 1½ is 3/4 as a fraction.

What is 9.5 as a fraction? ›

Solution: 9.5 as a fraction is 19/2.

What is 4 15 as a decimal? ›

Solution: 4/15 as a decimal is 0.27.

### How many 5 digit numbers are there that are divisible by 5? ›

Thus, the total number of five digit numbers divisible by 5 is 120+96=216.

What is the least 5 digit number that is divisible by 91? ›

Detailed Solution

For a number to be divisible by another number, the remainder should be 0. ∴ The lease 5-digit number divisible be 91 is 10010.

How many numbers can be formed which are divisible by 5 and less than 1000? ›

Total number of numbers = 1 + 17 + 136 = 154 ways.

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