There are 100 players numbered 1 to 100 and 100 baskets numbered 1 to 100. The first player puts one ball in every basket starting from the first basket (i.e. in baskets numbered 1, 2 , 3) . The second player puts 2 balls each in every second basket starting from the second ( i.e. in baskets numbered 2,4,6. . .) The third player puts three balls each in every third basket starting from the third , and so on till the hundredth player.
1. Which basket will have the maximum number of balls?
a.96
b.98
c.100
d.none
e.Not Attempted
2. How many baskets will finally have exactly twice the number of balls as the number on the basket itself?
a.8
b.6
c.4
d.2
e.Not Attempted
3. In how many baskets will exactly two players put the balls?
a.50
b.25
c.15
d.None
e.Not Attempted
How to solve the above question
In this test I scored 32 and it was my first attempt – how good is it
the first question reduces to : which number has the max. number of divisors.
of the three 96 has the max divisors. however None of these could also be an answer.
Rather it should be - which number has the max. sum of all its divisors
yes! missed that part.
1st qs->max sum of divisers which is for 96
3rd qs->no of prime b/w 1&100
2nd qs-> buckets 3rd,6th,12th,24th,48th,96th
6 of them will have twice the no. of balls..
for 1st and 2nd question, ans lie in calculating the sum of factors including that number and 1 also, while for the third one ans will be the total number of primes
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