$18$ is divisible by both $2$ and $3$ .It is also divisible by $2 \times 3 = 6.$ Similarly, a number is divisible by both $4$ and $6.$ Can we say that the number must also be divisible by $4 \times 6 = 24$ ?If not, give an example to justify your answer.

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Solution

Numbers divisible by $4$ are $4, 8, 12, 16, 20, 24, 28, 32, 36, . . .$

Numbers divisible by $6$ are $6, 12, 18, 24, 30, 36, 42, 48, 54, . . .$
Numbers divisible by $4$ and $6$ are $12, 24, 36, . . .$

We see that $12, 24, 36, . . .$ are not divisible by $24$

So, a number divisible by $4$ and $6$ is not divisible by $24$ since $4$ and $6$ are not co-prime.

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18 is divisible by both 2 and 3. It is also divisible by $2 \times 3 = 6.$ Similarly, a number is divisible by 4 and 6. Can we say that the number must be divisible by $4 \times 6 = 24 ?$ If not, give an example to justify your answer.

True or False.

4) If a number is divisible by both 9 and 10. It must be divisible by 90.

5) A number is divisible by 18 if it is divisible by both 3 and 6.

Q. Give an example of a number
(i) which is divisible by 2 but not by 4.
(ii) which is divisible by 4 but not by 8.
(iii) which is divisible by both 2 and 8 but not by 16.
(iii) which is divisible by both 3 and 6 but not by 18.

(i) All numbers divisible by 3 must also be divisible by 6

(ii) All numbers divisible by 6 must also be divisible by 3

Which of the following statements are true?

(a) If a number is divisible by 3, it must be divisible by 9.

(b) If a number is divisible by 9, it must be divisible by 3.

(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.

(e) If two numbers are co-primes, at least one of them must be prime.

(f) All numbers which are divisible by 4 must also be divisible by 8.

(g) All numbers which are divisible by 8 must also be divisible by 4.

(h) If a number exactly divides two numbers separately, it must exactly divide their sum.

(i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.

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18 is divisible by both 2 and 3.It is also divisible by 2×3=6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4×6=24?If not, give an example to justify your answer.

Numbers divisible by 4 are 4,8,12,16,20,24,28,32,36,...Numbers divisible by 6 are 6,12,18,24,30,36,42,48,54,...Numbers divisible by 4 and 6 are 12,24,36,...We see that 12,24,36,... are not divisible by 24So, a number divisible by 4 and 6 is not divisible by 24 since 4 and 6 are not co-prime.

$18$ is divisible by both $2$ and $3$ .It is also divisible by $2 \times 3 = 6.$ Similarly, a number is divisible by both $4$ and $6.$ Can we say that the number must also be divisible by $4 \times 6 = 24$ ?If not, give an example to justify your answer.