What is the smallest number which when divided by 12 18 27 gives a remainder of 5 each time?

Answer

Verified

Hint: Here, we will first find the least common multiple, which is the smallest positive number that is a multiple of two or more numbers. Then we will add the remainder left by dividing the given divisor from the obtained L.C.M. to find the required value.

Complete step-by-step solution:

Here, we have to find the least number which when divided by 18, 24, 30, and 12 leaves 4 as the remainder in each case.Before solving this question, we must know what the division theorem is. Division theorem states that “If ‘n’ is an integer and ‘d’ is a positive integer, there exist unique integers ‘q’ and ‘r’ such that,$n = dq + r$ where $0 \le r < d$Here, ‘n’ is the number or the dividend, ‘d’ is the divisor, ‘q’ is the quotient and ‘r’ is the remainder.We will now find the least common multiple of the given numbers, 18, 24, 30, and 12.$\begin{array}{l}2\left| \!{\underline {\, {18,24,30,12} \,}} \right. \\2\left| \!{\underline {\, {9,12,15,6} \,}} \right. \\2\left| \!{\underline {\, {9,6,15,3} \,}} \right. \\3\left| \!{\underline {\, {9,3,15,3} \,}} \right. \\3\left| \!{\underline {\, {3,1,5,1} \,}} \right. \\5\left| \!{\underline {\, {1,1,5,1} \,}} \right. \\\,\,\,1,1,1,1\end{array}$We will find the product of the above multiplies to find the least common multiple.$ \Rightarrow $ L.C.M. $ = 2 \times 2\, \times 2 \times 3 \times 3 \times 5$On multiplying the terms, we get$ \Rightarrow $ L.C.M. $ = 360$Since we get 4 as the remainder from all the numbers, we will now find the required number.Adding the number 4 from the obtained least common multiple, we get$ \Rightarrow 360 + 4 = 364$

Hence, the smallest number which when divided by 18, 24, 30, and 12 gives a remainder of 4 every time is 364.

Note: Here, the student should first understand what is asked in the question before solving the question. After that, we should figure out that if a number n leaves remainder r when divided by a number q then $n + q - r$ will be a multiple of q and we should apply this concept correctly. Moreover, we should find L.C.M. correctly and calculate the required number correctly.

Question 6 Exercise 8(C)

Answer:

Solution 6:

The least number which is exactly divisible by each given number is their LCM

Required number is LCM of 12, 15, 18, 24 and 36.

LCM = least required number = $2\times2\times3\times3\times5\times2=360$

Hence, required number = 360

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