2IIM Online CAT Preparation – Earn your discount Qn 3

This is the day 3 post of the challenge quiz week by 2IIM Online CAT Preparation :

Challenge Quiz Puzzle – Day 3:

Consider two distinct natural numbers x, y both less than 21. What is the probability that either x/y or y/x will be a terminating decimal?


Probability that either x/y or y/x should be a terminating decimal = 1 – probability that neither is a terminating decimal.

A fraction of the form p/q is terminating decimal if and only if the prime factorization of q comprises purely of 2’s and 5’s. The important condition for this rule is the fact that p/q should be in the simplest form, i.e. HCF(p,q) should be 1.

So, first up let us round up the numbers that are made of only 2’s and 5’s. These are 1, 2, 4, 5, 8, 10, 16 and 20. If x or y is one of these terms, then at least one of x/y or y/x would be terminating. So, if we want to find the probability that neither is a terminating decimal x and y should be from the other 12 numbers. So, prima facie the probability that neither is prime looks like it could be . But this ignores the fact that denominators that do not have only 2’s and 5’s can also end up creating terminating decimals. For example, 3/6 = 0.5. We need to somehow create a framework to account for this as well. This is where it gets interesting.

How many pairs do we have like {3, 6}.

Let us write down the remaining 12 numbers first and then think about them – {3, 6, 7, 9, 11, 12, 13, 14, 15, 17, 18, 19}. These can be placed in 3 separate buckets – {3, 6, 9, 12, 15, 18}, {7, 14} and {11, 13, 17, 19}.

Out of the 12C2 pairs that we originally outlined, how many will not work? The 4 prime numbers when combined with any other number from this 12 will always yield a non-terminating decimal. 7 or 14 when combined with anything other 14 or 7 will yield a non-terminating decimal. So, these numbers are easy to account for. Now, within the set {3, 6, 9, 12, 15, 18} , any pair of numbers selected as x,y either x/y or y/x or both will be terminating. Essentially, one 3 will get cancelled and then it will yield a terminating decimal.

So, of the 12C2 that we started off with, 6C2 + 1 will yield terminating decimals.

So, probability of having non-terminating decimals = [12C2 – (6C2 + 1)]/20C2 = (66 – 16 ) /190  = 5/19 .

So, probability that at least one of the two would be a terminating decimal would be 14/19 .

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