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# Question Based on Sum of factors and Number of Prime Numbers

This post contains Questions Based on Sum of factors and Number of Prime Numbers. Answer the QUestions given from the information available.

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Q.) There are 100 players numbered 1 to 100 and 100 baskets numbered 1 to 100. The first player puts one ball each in every basket starting from the first basket(i.e. in the baskets numbered 1,2,3,……), the second player puts two balls each in every second basket starting from the second basket (i.e. in the baskets numbered 2,4,6,….), the third player puts three balls each in every third basket starting from the third basket (i.e. in the baskets numbered 3,6,9 ……..) , and so on till the hundredth player.

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(Q.1.)Which basket will finally have the max. no. of balls?
i)96
ii)98
iii)100
iv)90

Solution : The no of balls in each basket is actually the sum of all the factors of the number on the basket. Hence, the basket numbered n, with the sum of factors of n being the max. will be the answer.

        sum of factors of (n = ${a}^{p} * {b}^{q} * {c}^{r}$)

= $\frac{({a}^{(p+1)} – 1)}{a-1}$ $\frac{({b}^{(q+1)} – 1)}{b-1}$ $\frac{({c}^{(r+1)} – 1)}{c-1}$


Hence, the sum of factors of 96 is the maximum.

Hence , OPTION(i)

Q.2.) How many baskets will finally have exactly twice the no. of balls as the no. on the basket itself?
(i)8
(ii)6
(iii)4
(iv)2

Solution : N = 6,28,496……. So, two baskets numbered 6 & 28.
Hence , option(iv)

Q.3.) In how many baskets will exactly two players put the balls?
(i)50
(ii)25
(iii)15

Solution : Only two players will put the balls in a basket numbered ‘n’, when ‘n’ is prime.
1 – 50 -> 15 prime no.
50 – 100 -> 10 prime no.
total = 15 + 10 =25
Hence, OPTION(ii)