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Question

Answers

If a number is divisible by \[9\], it must be divisible by \[3\]

A. True

B. False

Answer
Verified

General equation of a number \[a\] completely divisible by \[b\] is \[a = bq\].

Since, we are given a number is divisible by \[9\], so put the value \[b = 9\] in the above equation.

We can write \[a = 9q\]

Since, we can write nine in simpler form i.e. \[9 = 3 \times 3\]

Therefore, we can write \[a = 9q = (3 \times 3)q\]

Group together all the factors other than \[3\]

\[a = 3 \times (3q)\]

Assuming the factor \[3q = p\]

We can write \[a = 3p\]

which is of the form \[a = bq\], where \[b = 3\]

Therefore, number \[a\] is divisible by \[3\]

Since, the number on the LHS of the equation is the same, we can say the number that is divisible by \[9\] is also divisible by \[3\] .

So, the statement in the question is true.

Alternate method:

We can also show this solution by taking an example

Say a number is divisible by \[9\], let us take that number to be \[36\]

We can write \[36 = 9 \times 4\]

Since we know \[9 = 3 \times 3\]

Therefore, we can write

\[36 = 3 \times 3 \times 4\]

Grouping together all factors other than three

\[

36 = 3 \times (3 \times 4) \\

36 = 3 \times 12 \\

\]

Therefore, the number is divisible by \[3\] as it is written in the form of multiple of three.

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