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What is the smallest number that, when divided by $35,56$ and $91$ leaves remainder of $7$ in each case?

Last updated date: 19th Sep 2024
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Hint: To find the smallest number which when divided by $35,56,91$ leaves remainder $7$ in each case, calculate the LCM of $35,56,91$ and add $7$ to the LCM of $35,56,91$.

We have to find the smallest number which when divided by $35,56,91$ leaves remainder $7$ in each case. To do so, we will firstly evaluate the smallest number which is a multiple of each of $35,56,91$.
Thus, we will evaluate the LCM of $35,56,91$.

To find the LCM of three numbers, we will use the prime factorization method. We will write the prime factorization of each of the numbers in exponential form. Then, we will align the common prime factor base whenever possible. For the numbers with a common prime factor base, select the prime number that has the highest power. The prime factor with highest power implies that it occurs the most in the list. If a distinct prime factor has no matching prime factor base in the list, include this factor with its exponent in the collection of numbers. Multiply all the numbers which were collected earlier to get the LCM of three numbers.

Factorizing each of the numbers, we have $35=5\times 7,56={{2}^{3}}\times 7,91=7\times 13$ .

Thus, the LCM of $35,56,91$ is $7\times 5\times {{2}^{3}}\times 13=3640$.
The LCM of $35,56,91$ is $3640$.

We have to find the smallest number which leaves a remainder $7$ when divided by $35,56,91$, add $7$ to the LCM of numbers $35,56,91$.
Thus, we have $3640+7=3647$.

Hence, the smallest number which leaves a remainder $7$ when divided by $35,56,91$ is $3647$.

Note: Least Common Multiple (LCM) of any two numbers is the smallest positive number that is divisible by both the numbers. We can also find LCM of two numbers using list method. We should carefully calculate the LCM of three numbers; otherwise we will get an incorrect answer.