Determine the smallest positive integer n, which has exactly 144 distinct divisors and there are 10 consecutive integers among these divisors.

We know that after the prime factorization of that number when the powers of those prime factors +1 when multiplied with each other would result in 144.
144 = (a+1)(b+1). . .
We also know that there are 10 consecutive numbers among these divisors so the number must be divisible by 2,3,5,7.

I don’t know what to do now , what to do ?

What you say is true, but you should apply it. Factor $$144$$. You know that there are at least $$4$$ distinct prime factors. There are not many patterns of exponents of primes to choose from, then not many ways to choose the primes so they include $$2,3,5,7$$. Make a list of the patterns of exponents, then the smallest numbers that satisfy each pattern. Sort them and list the divisors of each, starting from the smallest. I suspect you will not need to search far. What is the factorization of $$LCM(1,2,3ldots 10)?$$ What is the minimum you need to multiply it by to get $$144$$ distinct factors? That gives an upper bound to the answer.

Correct answer by Ross Millikan on December 25, 2020

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