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In how many ways can the number 18900 can be split in two factors which are relative prime (or co-prime).

Answer
VerifiedVerified
508.5k+ views
Hint:Find the prime-factorization of 18900. Thus find the total number of prime numbers formed and take it as ‘n’. The ways to split into 2 factors which are relative prime are given by \[{{2}^{n-1}}\].

Complete step-by-step answer:
We have been given the number 18900.
Relative prime or co-prime can be explained with the integers for example a and b. They are said to be co-prime if the only positive integer that divides both of them is 1. For example 14 and 15 are co-prime, being commonly divisible only by 1. While 14 and 21 are not co-prime, because they are both divisible by 7.
Now the number 18900 can be split with the help of prime factorization which is finding the prime numbers multiplied together to make the original number.
\[\therefore 18900=2\times 2\times 3\times 3\times 3\times 5\times 5\times 7=\left( 2\times 2 \right)\times \left( 3\times 3\times 3 \right)\times \left( 5\times 5 \right)\times 7={{2}^{2}}\times {{3}^{3}}\times {{5}^{2}}\times 7\].
Thus we got the prime factorization of 18900 as,
\[\therefore 18900={{2}^{2}}\times {{3}^{3}}\times {{5}^{2}}\times 7\].
From this we can say that there are 4 prime numbers here, 2, 3, 5 and 7. From this we can say that there are a total of 4 prime numbers. Thus here n = 4, where n is the total number of co-prime numbers.
Thus the ways in which we can split the number 18900 into two factors which are relative prime are \[{{2}^{n-1}}\].
\[\therefore \]We know that n = 4.
\[\therefore {{2}^{n-1}}={{2}^{4-1}}={{2}^{3}}=2\times 2\times 2=8.\]
Thus the total number of ways in which number 18900 can be split into 2 factors which are relative prime = 8.

Note:For 1, all the factors of each prime have to stay together, otherwise the two factors will not be co-prime. We can consider it to be 18900 = \[4\times 27\times 25\times 7\]. One factor will be the subset of \[\left[ 4,27,25,7 \right]\] and the other factor will be at rest. Each factorization gets counted twice, once when each factor is the first one selected.

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