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Maths Puzzle
HCF of 3240,3600 and a third number is 36 and their LCM is (2^4)x(3^5)x(5^2)x(7^2).
The third number is?
a> 2^2 x 3^5 x 7^2
b> 2^2 x 5^3 x 7^2
c> 2^5 x 5^2 x 7^2
d> 2^3 x 3^5 x 7^2
Read Solution (Total 2)
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- as we know that... prodct of nmbrs always equal to the prodct of hcf and lcm
so,
3240*3600*unknown nmbr= 36*LCM,
unknown nmbr= 36*lcm/(3240*3600),
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and i hope u can calculate next step to make ur answer, - 12 years agoHelpfull: Yes(1) No(12)
- HCF = 36 = (2^2) x (3^2)
LCM = (2^4) x (3^5) x (5^2) x (7^2)
3240 = (2^3) x (3^4) x (5^1)
3600 = (2^4) x (5^2) x (3^2)
x = (2^2) x (3^5) x (7^2) ( a )
Explanation:
1.Finding power of 2:
HCF is the product of least powers of common prime factors. The given product contains (2 ^ 2),
which(2 ^ 2) is not contained by any of the given two numbers.Therefore it should come from x.
2.Finding powers of 3:
LCM is the product of highest powers of all the factors. The given product contains (3 ^ 5),
which(3 ^ 5) is not contained by any of the given two numbers.Therefore it should come from x.
3.Finding powers of 5:
x should not contain powers of 5 as HCF does not contain it.
4.Finding powers of 7:
Since LCM contains 7 ^ 2 , which is not contained by any of the given two numbers.
So, x should contain it. - 6 years agoHelpfull: Yes(0) No(0)
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