
Find the smallest 4-digit number which is divisible by 18, 24 and 32.
Answer
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Hint: First find the smallest number which is divisible by 18, 24 and 32. Then increase that number to its multiple until you have the four digit number. This four digit number is the smallest four digit number divisible by 18, 24 and 32.
Complete step-by-step answer:
To find out the number which is the multiple of 18, 24 and 32 at the same time we have to search for the prime factors which are contained in those numbers. We will do prime factorization of all three numbers. Prime factorization means writing a number as a product of positive powers of distinct primes. For eg- $12={{2}^{2}}\times 3$. This is the unique way of factorization. For any number, there exists unique prime factorization. So we get,
$\begin{align}
& 18=2\times {{3}^{2}} \\
& 24={{2}^{3}}\times 3 \\
& 32={{2}^{5}} \\
\end{align}$
For any number to be divisible by 18 it must have 2 at least one time and 3 at least two times in its prime factorization. Similarly for a number to be divisible by 24, it must have 2 at least three times and 3 at least one times in its prime factorization. Similarly for a number to be divisible by 32, it must have 2 at least 5 times.
For getting the common multiple of 18, 24 and 32 numbers must satisfy all the conditions. For this number should have a maximum of what is needed by each condition.
So the least number which is divisible by all the three numbers is
$N={{2}^{5}}\times {{3}^{2}}=32\times 9=288$
Now we have to get the four digit number, this can be done by searching for the multiple of 288 until we get a four digit number.
We have
$\begin{align}
& 2N=288\times 2=576 \\
& 3N=288\times 3=864 \\
& 4N=288\times 4=1152 \\
\end{align}$
So we get a four digit number. This is the smallest four digit number divisible by 18, 24 and 32.
Hence 1152 is the smallest four digit number divisible by 18, 24 and 32.
Note: Students generally make mistakes in factorization of a number which is the crucial step in this question. Factorization always starts from 2 and moves on to higher primes. Take only prime numbers in the factorization process.
Complete step-by-step answer:
To find out the number which is the multiple of 18, 24 and 32 at the same time we have to search for the prime factors which are contained in those numbers. We will do prime factorization of all three numbers. Prime factorization means writing a number as a product of positive powers of distinct primes. For eg- $12={{2}^{2}}\times 3$. This is the unique way of factorization. For any number, there exists unique prime factorization. So we get,
$\begin{align}
& 18=2\times {{3}^{2}} \\
& 24={{2}^{3}}\times 3 \\
& 32={{2}^{5}} \\
\end{align}$
For any number to be divisible by 18 it must have 2 at least one time and 3 at least two times in its prime factorization. Similarly for a number to be divisible by 24, it must have 2 at least three times and 3 at least one times in its prime factorization. Similarly for a number to be divisible by 32, it must have 2 at least 5 times.
For getting the common multiple of 18, 24 and 32 numbers must satisfy all the conditions. For this number should have a maximum of what is needed by each condition.
So the least number which is divisible by all the three numbers is
$N={{2}^{5}}\times {{3}^{2}}=32\times 9=288$
Now we have to get the four digit number, this can be done by searching for the multiple of 288 until we get a four digit number.
We have
$\begin{align}
& 2N=288\times 2=576 \\
& 3N=288\times 3=864 \\
& 4N=288\times 4=1152 \\
\end{align}$
So we get a four digit number. This is the smallest four digit number divisible by 18, 24 and 32.
Hence 1152 is the smallest four digit number divisible by 18, 24 and 32.
Note: Students generally make mistakes in factorization of a number which is the crucial step in this question. Factorization always starts from 2 and moves on to higher primes. Take only prime numbers in the factorization process.
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