A number N when factorised can be written as N=a^4*b^3*c^7 . What is the number of perfect squares which are factors of N (The three prime numbers a,b, c > 2)?
Answers
Step-by-step explanation:
a⁴b³c^7 can be written as
a⁴b²c^6 * bc
now perfect square numbers for a4 are 2
which are a2 and a4
for b2 it is 1. only b2
for c6 it is 3. they are c2,c4,c6
so total required number is=
1(for 1)+2+1+3+2*1*3
=7+6=13
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Answer:
24
Step-by-step explanation:
firstly we will define these terms
1) factors - is simply a divisor which can divide a particular number
2) prime factorisation - reducing a given number to its smallest possible prime number and multiplying them .
example - prime factorisation of 12 = 2×2×3 = 2^2 × 3
3) perfect square - for a number to be called a perfect square it needs to have two pairs of its own
example we have the following nos 11×11×2×3×5×2×3×5×7 = 11^2 × 2^2 × 3^2 × 5^2 as 4 perfect squares × 7 as it does not have another number 7 to pair with.
4) 1 is considered a factor of every number and also a perfect square as 1×1=1 , any number with the power 0 equals to 1; 5^0 =1, 9999^0 = 1
explanation to question
N = a^4 × b^3 × c^7
taking one by one
a^4 has four sets of itself (a×a×a×a)
hence we can make 3 factors out of it - a^0, a^2, a^4 . taking an example say 3^4 = 3^0, 3^2, 3^4
similarly for b^3 = b^0 × b^2 (one b gets left out as we have to form even pairs ) thus having 2 factors
for c^7 = c^0, c^2, c^4, c^6 = 4 factors
lastly we have to multiply all the factors to get the final result which is 3× 2 × 4 = 24