
State whether True or False:
If a number is divisible by \[9\], it must be divisible by \[3\]
A. True
B. False
Answer
579.6k+ views
Hint: Here we use the concept that any number \[a\] when divided by \[b\] can be written as \[a = bq + r\] where \[q\] is the quotient and \[r\] is the remainder. A number is completely divisible by any number if its remainder is zero i.e. it can be written in the form \[a = bq\]. So we write number divisible by \[9\] then further factoring nine we write if the number is divisible by factors of \[9\]
Complete step-by-step answer:
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.
So, the correct answer is “Option A”.
Note: Students many times make mistake when they assume looking at the word divisible by a number and they write the number in fraction form which is wrong, keep in mind that whenever a number is divisible by another number then the first number can be written as a multiple of second number.
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.
Complete step-by-step answer:
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.
So, the correct answer is “Option A”.
Note: Students many times make mistake when they assume looking at the word divisible by a number and they write the number in fraction form which is wrong, keep in mind that whenever a number is divisible by another number then the first number can be written as a multiple of second number.
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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