What is the smallest number that, when divided by \[35,56\] and \[91\] leaves remainder of \[7\] in each case?
Last updated date: 19th Mar 2023
•
Total views: 306k
•
Views today: 6.84k
Answer
306k+ views
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 \[35,56,91\].
Complete step-by-step answer:
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.
Complete step-by-step answer:
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.
Recently Updated Pages
If abc are pthqth and rth terms of a GP then left fraccb class 11 maths JEE_Main

If the pthqth and rth term of a GP are abc respectively class 11 maths JEE_Main

If abcdare any four consecutive coefficients of any class 11 maths JEE_Main

If A1A2 are the two AMs between two numbers a and b class 11 maths JEE_Main

If pthqthrth and sth terms of an AP be in GP then p class 11 maths JEE_Main

One root of the equation cos x x + frac12 0 lies in class 11 maths JEE_Main

Trending doubts
What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?
