Only the Last Three Digits Decide Divisibility by Eight

Because **$1000$ is itself divisible by $8$**, so everything above the hundreds place divides cleanly no matter what it is.

Take $4128$. Split it as $4000+128$. The $4000$ is $4\times 1000$, and since $8$ divides $1000$ it divides $4000$ — for any thousands digit. So the whole number is divisible by $8$ exactly when the leftover $128$ is, and $128÷8=16$ settles it.

Every divisibility test in this chapter works the same way: find what the place values give you for free, and the test is whatever is left over. This page covers the second part of the ICSE Class 8 Mathematics chapter on playing with numbers.

Nine tests, and each looks at a different feature of the number.

- **By $2$ — the units digit** is $0$, $2$, $4$, $6$ or $8$.
- **By $3$ — the sum of the digits** is divisible by $3$.
- **By $4$ — the number formed by the last two digits** is divisible by $4$.
- **By $5$ — the units digit** is $0$ or $5$.
- **By $6$ — the number is divisible by both $2$ and $3$.
- By $8$ — the number formed by the last three digits** is divisible by $8$.
- **By $9$ — the sum of the digits** is divisible by $9$.
- **By $10$ — the units digit** is $0$.
- **By $11$ — the difference between the sum of the digits in the odd places and the sum in the even places** is $0$ or a multiple of $11$.

Worked example 1 — testing one number against several divisors. Test $61254$.

- Digit sum: $6+1+2+5+4=18$. Since $18$ is divisible by both $3$ and $9$, the number is divisible by **$3$ and by $9$.
- Units digit** is $4$, so it is divisible by **$2$**.
- Divisible by $2$ and $3$, so it is divisible by **$6$.
- Last two digits** are $54$, and $54÷4=13.5$, so it is not divisible by $4$.

Checking the claims: $61254÷9=6806$ and $61254÷6=10209$, both exact.

Worked example 2 — by eight. Test $4128$. The last three digits are $128$, and $128÷8=16$, so $4128$ is divisible by $8$. Checking: $4128÷8=516$.

Worked example 3 — by eleven. Test $91828$.

Counting places from the right, the digits are $8$, $2$, $8$, $1$, $9$.

$$\text{odd places }(1\text{st},3\text{rd},5\text{th}):8+8+9=25$$

$$\text{even places }(2\text{nd},4\text{th}):2+1=3$$

$$\text{difference}=25-3=22$$

Since $22$ is a multiple of $11$, the number is divisible by $11$. Checking: $91828÷11=8348$.

**Why the test for $6$ needs both parts.** A number divisible by $2$ need not be divisible by $6$, and neither need one divisible by $3$ — $14$ and $9$ show each case. Only both together work, and the reason is that $6=2\times 3$ with $2$ and $3$ having no common factor.

A warning about combining tests this way. The same trick does not work for every composite number. A number divisible by $2$ and by $4$ is not necessarily divisible by $8$ — $12$ is divisible by both and not by $8$ — because $2$ and $4$ share a factor. So $6=2\times 3$ can be split and $8=2\times 4$ cannot.

Apply the relevant test with the missing digit as a letter, then solve for the digit.

**Worked example 1 — divisible by $6$.** Find the least value of $x$ so that $\overline{21x8}$ is divisible by $6$.

Divisibility by $6$ needs both $2$ and $3$.

- **By $2$**: the units digit is $8$, which is even, so this holds whatever $x$ is.
- **By $3$**: the digit sum is $2+1+x+8=11+x$, which must be a multiple of $3$. So $11+x\in \{12,15,18\}$, giving $x=1$, $4$ or $7$.

The least value is $x=1$, giving $2118$. Checking: $2118÷6=353$. Correct.

Notice the question has three valid answers, and it asked for the least. Listing all of them and then picking is safer than stopping at the first one found.

**Worked example 2 — divisible by $9$.** Find $x$ so that $\overline{31x5}$ is divisible by $9$.

The digit sum is $3+1+x+5=9+x$, which must be a multiple of $9$. So $9+x\in \{9,18\}$, giving $x=0$ or $x=9$.

Taking the least, $x=0$ gives $3105$, and $3105÷9=345$. Correct.

**Worked example 3 — divisible by $11$.** Find $x$ so that $\overline{1234x}$ is divisible by $11$.

Counting from the right, the digits are $x$, $4$, $3$, $2$, $1$.

$$\text{odd places}:x+3+1=x+4$$

$$\text{even places}:4+2=6$$

$$\text{difference}=(x+4)-6=x-2$$

This must be $0$ or a multiple of $11$. Since $x$ is a single digit, $x-2=11$ is impossible, so

$$x-2=0\Rightarrow x=2$$

Checking: $12342÷11=1122$. Correct — and here the answer is unique.

**Worked example 4 — divisible by $4$.** Find $x$ so that $\overline{24x8}$ is divisible by $4$.

Only the last two digits matter, so $\overline{x8}$ must be divisible by $4$. Testing: $08$, $28$, $48$, $68$ and $88$ are all divisible by $4$, while $18$, $38$, $58$, $78$ and $98$ are not.

So $x$ must be even: $x\in \{0,2,4,6,8\}$. The least is $x=0$, giving $2408$, and $2408÷4=602$. Correct.

The pattern across these four. The number of answers depends on the test. A digit-sum test gives several answers spaced by $3$ or $9$; the alternating-sum test for $11$ usually gives exactly one, because $11$ is larger than any single digit can compensate for. So expect one answer for $11$ and several for $3$, $4$ and $9$ — and read whether the question wants the least, the greatest, or all of them.

Write the number in generalised form and split it into a part that is obviously divisible and a remainder. The test is always about the remainder.

**Divisibility by $9$ and by $3$.** For a three-digit number:

$$100a+10b+c=(99a+9b)+(a+b+c)$$

$$=9(11a+b)+(a+b+c)$$

The first part is a multiple of $9$ whatever the digits are. So the whole number is divisible by **$9$ exactly when the digit sum** $a+b+c$ is — and since $9$ is a multiple of $3$, the same split proves the test for **$3$.

This also explains why the rule works for a number of any** length: $999$, $9999$ and so on are all multiples of $9$, so every place value contributes a multiple of $9$ plus its own digit.

**Divisibility by $11$.** Split differently:

$$100a+10b+c=(99a+11b)+(a-b+c)$$

$$=11(9a+b)+(a-b+c)$$

The first part is a multiple of $11$. So the number is divisible by **$11$** exactly when $a-b+c$ — the alternating sum of the digits — is. And that alternating sum is precisely odd places minus even places.

The reason the signs alternate is that $10$ leaves a remainder of $-1$ when compared with $11$: $10=11-1$, $100=99+1$, $1000=1001-1$, and so the place values contribute $+1,-1,+1,-1$ in turn.

**Divisibility by $4$.** A four-digit number can be written as

$$\overline{abcd}=100\times \overline{ab}+\overline{cd}$$

Since $100$ is divisible by $4$, the first term is a multiple of $4$ whatever $\overline{ab}$ is. So only the last two digits matter.

**Divisibility by $8$.** The same argument with $1000$:

$$\overline{abcd}=1000a+\overline{bcd}$$

and $1000$ is divisible by $8$, so only the last three digits matter.

**Divisibility by $2$, $5$ and $10$.** Every place value except the units is a multiple of $10$, and $10$ is divisible by $2$, by $5$ and by $10$. So the whole number minus its units digit is always divisible by each of them, and only the units digit decides.

The single idea behind all seven proofs. Each test asks what do the place values give me for free? For $9$ it is $9,99,999$; for $11$ it is $11,99,1001$; for $4$ it is every multiple of $100$; for $8$ every multiple of $1000$. Whatever is left after removing the free part is the test — which is why the tests look so different from one another despite being proved the same way.