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Q1. Let x1,x2,x3,...,x100 be hundred integers such that the sum of any five integers is equal to 20. Then

  1. the largest xi equals 5;
  2. the smallest xi equals 3;
  3. x17=x83; 4.none of the foregoing statement is true.

Q2. The smallest positive integer n with 24 divisors (where 1 and n are also considered as divisors of n) is

  1. 420;
  2. 240;
  3. 360;
  4. 480.

I could not figure out the first question.But for the second one, the answer says it is 360. So I tried to solve it in this way that is:

360 = 3*3*2*2*2*5

So, the total no of divisors = enter image description here

                         = 64

So I took the factors of the 360 and each and every factor is a divisor of 360. And I used the combinations to create new divisors for the 360. So I want to know what is wrong with this method.

Sid
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  • Question $1$ : No idea for a proof yet, but I think all summands have to be $4$. This would imply answer $3)$ – Peter Nov 09 '16 at 16:50
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    Hint for question 2 : If the prime factorization of $n$ is $p_1^{a_1} \cdots p_n^{a_n}$, then the number of divisors of $n$ is $(a_1+1)\cdots (a_n+1)$ – Peter Nov 09 '16 at 16:52

1 Answers1

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For the first one, you should prove that all the numbers are equal. Find two sets of five that only have one element different, then those two elements must be equal. Keep going. For the second, look up the divisor function. The number of factors of a number depends on the exponents of the primes in its factorization.

Ross Millikan
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