Question

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

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.

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.