What is the least number that should be added to 2497 so that the sum is exactly divisible by 5/6 4 and 3?

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Asked by sushilasushila35756 | 01 Aug, 2021, 07:43: PM

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Asked by khushisk25 | 18 Aug, 2020, 07:37: PM

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Answer

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Hint: This problem is based on the concept of L.C.M. of numbers. First find the L.C.M. of the given numbers by which the sum should be exactly divisible. Then divide the number \[2497\] by this L.C.M. the remainder obtained from this division needs to be subtracted from the L.C.M. to get the least number that needs to be added.

Complete step by step solution:

Before proceeding with the problem we first need to know that the L.C.M. of two numbers is the smallest number that is completely divisible by both the numbers.Here we need to find the least number that needs to be added to \[2497\]such that the sum is divisible exactly by \[5,6,4,3\]. Now for the sum to be divisible by all the given numbers it has to be divisible by the L.C.M. of these numbers. Let us first find the L.C.M. of the given numbers.We’ll find the L.C.M. by the division method:\[ 2\left| \!{\underline {\, {5,6,4,3} \,}} \right. \\ 3\left| \!{\underline {\, {5,3,2,3} \,}} \right. \\ \;\;\left| \!{\underline {\, {5,1,2,1} \,}} \right. \\ \]Therefore L.C.M. \[ = \] \[2 \times 3 \times 5 \times 1 \times 2 \times 1\].\[ = 60\]Next divide the number \[2497\] by \[60\]. Doing so we obtain a quotient of \[41\] and a remainder of \[37\]. So \[2497\] is not completely divisible by \[60\], but if the difference: \[\left( {60 - 37} \right)\], i.e. \[23\] is added to \[2497\], then it will be completely divisible by \[60\] and will be exactly divisible by \[5,6,4,3\]. Hence the least number to be added to \[2497\] to make it divisible exactly by \[5,6,4,3\] is \[23\].

Correct option is (c).

Note:

L.C.M. means the lowest common multiple. Alternatively the L.C.M. of the numbers can also be found by the prime factorization method. In the prime factorization method first we need to find the prime factors of each number. Then the L.C.M. will be the product of all prime factors with the highest degree (power).

The least number which should be added to 2497 so that the sum is exactly divisible by 5,6,4 and 3 is:

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