# Find the least number when divided by 16, 22 and 40 leaves a remainder 7 in each case.

Last updated date: 27th Mar 2023

•

Total views: 306k

•

Views today: 6.83k

Answer

Verified

306k+ views

Hint: In this question first find out the L.C.M of these numbers later on add 7 in the L.C.M to reach the solution of the question.

Complete step-by-step answer:

Let us consider a number p and let us also consider that it is exactly divisible by x, y and z.

So if the number is exactly divisible by given numbers then the number is also divisible by the L.C.M of the numbers.

And the L.C.M is the least number which is exactly divisible by x, y and z.

Now if we want any remainder say (q) after divisible, then add the remainder in the L.C.M of the numbers and the resultant number is the least number when divided by x, y and z leaves the remainder q in each case.

Now the given numbers are 16, 22 and 40.

So first find out the L.C.M of the numbers.

Least common multiple of two numbers is to first list the prime factors of each number. Then multiply each factor the greatest number of times it occurs in either number. If the same factor occurs more than once in both numbers, you multiply the factor the greatest number of times it occurs.

So first factorize the number.

Factors of 16 are

$ \Rightarrow 16 = 2 \times 2 \times 2 \times 2$

Factors of 22 are

$ \Rightarrow 22 = 2 \times 11$

Factors of 40 are

$ \Rightarrow 40 = 2 \times 2 \times 2 \times 5$

So L.C.M of the numbers is

$ \Rightarrow L.C.M = 2 \times 2 \times 2 \times 2 \times 5 \times 11 = 880$

Now add 7 in the L.C.M value we have,

So the required least number when divided by 16, 22 and 40 leaves the remainder 7 in each case is

$ = \left( {880 + 7} \right) = 887$.

So, this is the required answer.

Note: Whenever we face such types of questions the key concept is L.C.M ,so take the L.C.M of the numbers according to the property which is stated above, then use the concept if a number is exactly divisible by the given numbers than it is also divisible by the L.C.M of the numbers and L.C.M of the given numbers is the least number which is exactly divisible by the given numbers and if we want any remainder then simply add the remainder value in the L.C.M value and the resultant value is the required least number when divisible by given numbers leaves the given remainder.

Complete step-by-step answer:

Let us consider a number p and let us also consider that it is exactly divisible by x, y and z.

So if the number is exactly divisible by given numbers then the number is also divisible by the L.C.M of the numbers.

And the L.C.M is the least number which is exactly divisible by x, y and z.

Now if we want any remainder say (q) after divisible, then add the remainder in the L.C.M of the numbers and the resultant number is the least number when divided by x, y and z leaves the remainder q in each case.

Now the given numbers are 16, 22 and 40.

So first find out the L.C.M of the numbers.

Least common multiple of two numbers is to first list the prime factors of each number. Then multiply each factor the greatest number of times it occurs in either number. If the same factor occurs more than once in both numbers, you multiply the factor the greatest number of times it occurs.

So first factorize the number.

Factors of 16 are

$ \Rightarrow 16 = 2 \times 2 \times 2 \times 2$

Factors of 22 are

$ \Rightarrow 22 = 2 \times 11$

Factors of 40 are

$ \Rightarrow 40 = 2 \times 2 \times 2 \times 5$

So L.C.M of the numbers is

$ \Rightarrow L.C.M = 2 \times 2 \times 2 \times 2 \times 5 \times 11 = 880$

Now add 7 in the L.C.M value we have,

So the required least number when divided by 16, 22 and 40 leaves the remainder 7 in each case is

$ = \left( {880 + 7} \right) = 887$.

So, this is the required answer.

Note: Whenever we face such types of questions the key concept is L.C.M ,so take the L.C.M of the numbers according to the property which is stated above, then use the concept if a number is exactly divisible by the given numbers than it is also divisible by the L.C.M of the numbers and L.C.M of the given numbers is the least number which is exactly divisible by the given numbers and if we want any remainder then simply add the remainder value in the L.C.M value and the resultant value is the required least number when divisible by given numbers leaves the given remainder.

Recently Updated Pages

If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

The coordinates of the points A and B are a0 and a0 class 11 maths JEE_Main

Trending doubts

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

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

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

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE

What is per capita income

Change the following sentences into negative and interrogative class 10 english CBSE

A Short Paragraph on our Country India