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Find the least number which when divided by $6,$ $15$ and $18$ leaves the remainder $5$ in each case.

Last updated date: 21st Mar 2023
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Hint: First assume the number to be $x$. Then find the product of each prime factor of the highest power of every number i.e. take LCM of all. After that, we have been given the remainder $5$. Add the remainder $5$ to the LCM. You will get the number i.e. you will get $x$.
Have you ever seen the show Fear Factor? It required contestants to face a variety of fear-inducing stunts to win the grand prize of $\$50000$. At the end of the show, the host would say to the winner, 'Evidently, fear is not a factor for you!' What exactly does that mean? Well, it means that fear doesn't play a part in their actions and decisions. So, then a 'factor' is something that affects an outcome. In mathematics, factors are the numbers that multiply to create another number. The prime factorization of a number, then, is all of the prime numbers that multiply to create the original number. It would be pretty difficult to perform prime factorization if we didn't first refresh our memory on prime numbers. With that being said, a prime number is a number that can only be divided by one and itself. The prime factorization of a number is the product of prime factors that make up that number. So, prime factorization is writing the prime numbers that will multiply together to make a new number as a multiplication problem. Prime factorization is the product of primes that could be multiplied together to make the original number. Two possible ways of getting the list of the primes include a factor tree and upside down division. If a prime number occurs more than once in the factorization, it is usually expressed in exponential form to make it look more compact. We have been given that a number when divided by$6,15$and$18$leaves the remainder$5$. Let the number be$x$. Here, to find$x$we have to take LCM of the given numbers i.e. to take the product of each prime factor of highest power and add$5$. Now taking the product of each prime factor of the highest power of every number. Prime factorization of$6=2\times 315=3\times 518=2\times 3\times 3=2\times {{3}^{2}}$. Now, the product of each prime factor of the highest power$=2\times {{3}^{2}}\times 5=90$. The required number$x=90+5=95$. Here, the required number is$95$. The least number which when divided by$6,15$and$18$leaves the remainder$5$is$95\$.