Answer
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Hint: in order to check if the number is divisible by all of these numbers do not just divide the number by the LCM of these numbers. For finding the largest five numbers divisible by all of these numbers start with the largest 5 digit number and then check for its divisibility, if there is a remainder subtract that from the number.
Complete step-by-step answer:
To find the divisibility by all of the numbers. First let us find out the LCM of all the numbers given.
To find LCM of $7,10,15,21{\text{ and }}28$ . First let us find the prime factorization of each of them.
Prime factorization of $7 = 7$
Prime factorization of $10 = 2 \times 5$
Prime factorization of $15 = 3 \times 5$
Prime factorization of $21 = 3 \times 7$
Prime factorization of $28 = 2 \times 2 \times 7$
LCM of $7,10,15,21{\text{ and }}28 = 2 \times 2 \times 3 \times 5 \times 7 = 420$
Since the LCM of $7,10,15,21{\text{ and }}28 = 2 \times 2 \times 3 \times 5 \times 7 = 420$ , so we will check the divisibility by 420.
To find the largest 5 digit number divisible by $7,10,15,21{\text{ and }}28$.
Let us take the largest 5 digit number and divide it by 420.
Largest 5 digit number $ = 99999$
On dividing 99999 by 420 we get
$99999 = \left( {420 \times 238} \right) + 39$
So we get 39 as a remainder.
Therefore the required number is $99999 - 39 = 99960$
Hence, the largest five digit number that is divisible by $7,10,15,21{\text{ and }}28 = 99960$
So, option C is the correct option.
Note: In order to solve the problem related to finding the largest divisible by a given number of digits, students must start with the largest number and then divide it. Remainder if found must be subtracted. Also in the similar type of problem if we need to find the smaller number then the remainder must be subtracted from the divisor and then should be added to the dividend.
Complete step-by-step answer:
To find the divisibility by all of the numbers. First let us find out the LCM of all the numbers given.
To find LCM of $7,10,15,21{\text{ and }}28$ . First let us find the prime factorization of each of them.
Prime factorization of $7 = 7$
Prime factorization of $10 = 2 \times 5$
Prime factorization of $15 = 3 \times 5$
Prime factorization of $21 = 3 \times 7$
Prime factorization of $28 = 2 \times 2 \times 7$
LCM of $7,10,15,21{\text{ and }}28 = 2 \times 2 \times 3 \times 5 \times 7 = 420$
Since the LCM of $7,10,15,21{\text{ and }}28 = 2 \times 2 \times 3 \times 5 \times 7 = 420$ , so we will check the divisibility by 420.
To find the largest 5 digit number divisible by $7,10,15,21{\text{ and }}28$.
Let us take the largest 5 digit number and divide it by 420.
Largest 5 digit number $ = 99999$
On dividing 99999 by 420 we get
$99999 = \left( {420 \times 238} \right) + 39$
So we get 39 as a remainder.
Therefore the required number is $99999 - 39 = 99960$
Hence, the largest five digit number that is divisible by $7,10,15,21{\text{ and }}28 = 99960$
So, option C is the correct option.
Note: In order to solve the problem related to finding the largest divisible by a given number of digits, students must start with the largest number and then divide it. Remainder if found must be subtracted. Also in the similar type of problem if we need to find the smaller number then the remainder must be subtracted from the divisor and then should be added to the dividend.
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