One Counterexample Destroys a Rule That Looks Obvious

Because a rule claims to hold for every number, so a single case where it fails destroys it completely. "Every number divisible by 4 is divisible by 8" collapses the moment you notice 12.

That single example is called a counterexample, and finding one is a complete proof. This page covers everything in the CBSE Class 8 Mathematics chapter's first part: divisibility of multiples and factors, sums and differences, combining divisors, and counterexamples.

Two related properties, and both are easy to see once stated.

Every multiple of a divisible number is divisible too. If $a$ is divisible by $b$, then every multiple of $a$ is also divisible by $b$.

Since $12$ is divisible by $3$, so are $24$, $36$, $48$ and $120$ — every multiple of 12.

The reason is that a multiple of 12 is a multiple of 12 groups of 3, so it splits into 3s exactly.

A number is divisible by every factor of its divisor. If $a$ is divisible by $b$, then $a$ is divisible by every factor of $b$.

Since $36$ is divisible by $12$, and the factors of 12 are $1,2,3,4,6,12$, the number 36 is divisible by all of them:

$$\frac{36}{2}=18,\frac{36}{3}=12,\frac{36}{4}=9,\frac{36}{6}=6$$

All whole numbers, as predicted.

Worked use. A number is known to be divisible by $24$. Without knowing the number, you can say at once that it is divisible by $1,2,3,4,6,8,12$ and $24$, since those are the factors of 24.

Packing is the everyday version: a quantity that fills boxes of 12 exactly will also fill boxes of 6, of 4, of 3 and of 2 exactly.

The direction is what must not slip. Divisibility passes upward to multiples and downward to factors — but a number divisible by 4 need not be divisible by 8, because 8 is a multiple of 4, not a factor of it.

Because if both numbers are built from groups of $a$, then so is their total and so is what is left when one is taken from the other.

The property: if $a$ divides $m$ and $a$ divides $n$, then $a$ divides $m+n$ and $a$ divides $m-n$.

Worked example. $3$ divides $12$ and $3$ divides $18$. Therefore:

$$3\mid (12+18)=30,3\mid (18-12)=6$$

Both check out, since $30÷3=10$ and $6÷3=2$.

Worked example. Is $4$ a divisor of $24+36$? Since $4$ divides $24$ and $4$ divides $36$:

$$4\mid 60,\text{and indeed }\frac{60}{4}=15$$

Worked example on a compound expression. Is $7\times 15+7\times 23$ divisible by $7$? Both terms are multiples of 7, so their sum is:

$$7(15+23)=7\times 38=266,\frac{266}{7}=38$$

Worked example with a mixed case. Is $25+14$ divisible by $5$? Here $5$ divides $25$ but not $14$, so the property does not apply — and $39$ is indeed not divisible by 5.

The condition is that $a$ must divide both numbers, and that is where the property is misused. If $a$ divides only one of them, nothing follows about the sum — $3$ divides $9$ but not $10$, and their sum $19$ is not divisible by 3 either. So check both before applying it.

When the two divisors are co-prime — sharing no common factor except 1. Then a number divisible by both is divisible by their product, which is also their LCM.

Worked example. A number is divisible by $3$ and by $4$. Since $3$ and $4$ are co-prime, it is divisible by

$$3\times 4=12$$

Testing on $84$: the digit sum $8+4=12$ is divisible by 3, and $84=4\times 21$ is divisible by 4. So 84 should be divisible by 12, and indeed

$$\frac{84}{12}=7$$

Worked example. A number divisible by $2$ and $9$, which are co-prime, is divisible by $18$. Testing $126$: it is even, and $1+2+6=9$ is divisible by 9, so $126÷18=7$.

Worked example. A number divisible by $5$ and $6$ is divisible by $30$, since 5 and 6 are co-prime.

Now the failure case, which is the heart of this section. A number divisible by $2$ and $4$ is not necessarily divisible by $8$, because $2$ and $4$ are not co-prime — both contain a factor of 2.

Counterexample: $12$ is divisible by 2 and by 4, but

$$\frac{12}{8}=1.5$$

So 12 is not divisible by 8. The correct combination for non-co-prime divisors is their LCM, and the LCM of 2 and 4 is $4$, not 8 — which is all you may conclude.

Find one number that satisfies everything the claim assumes but fails what it concludes. That is enough, because the claim asserted it held for all numbers.

Claim: Every number divisible by 4 is divisible by 8.

Take $12$. It is divisible by 4, since $12÷4=3$. But $12÷8=1.5$, not a whole number. So the claim is false, and $12$ is the counterexample. So are $20$, $28$ and $36$.

Claim: If a number is divisible by 2 and by 6, it is divisible by 12.

Take $18$. It is divisible by 2 and by 6, yet $18÷12=1.5$. False — and it fails because 2 and 6 are not co-prime.

Claim: If a number is divisible by 3 and by 9, it is divisible by 27.

Take $9$ itself. It is divisible by 3 and by 9, but not by 27. False.

Claim: The sum of two numbers divisible by 5 is divisible by 10.

Take $5$ and $10$. Both are divisible by 5, and their sum is $15$. But

$$\frac{15}{10}=1.5$$

so the sum is not divisible by 10. The claim is false, and $5+10$ is the counterexample. What is actually true is the sum-and-difference property from earlier: the sum is divisible by 5.

Notice how easily this one could have been missed. The pairs $5+15=20$, $10+20=30$ and $15+25=40$ all give multiples of 10, so three confirming examples would have suggested the claim was safe — and it is not.

The rule about evidence is the point of the whole section. One counterexample disproves; any number of confirming examples never proves. Checking 3, 12 and 24 for the first claim would have found no problem at all — the disproof needed exactly the right number.