If a=3^{129}\times5^{128}\times7^{22}, b=3^{128}\times5^{129}\times7^{21}, c=3^{127}\times5^{129}\times7^{23} and \mathrm{d}=3^{127}\times5^{128}\times7^{24} then H.C.F. of a, b, c and d is :

Correct Answer is : 3127×5128×721

3^{128}\times5^{128}\times7^{21}

3^{128}\times5^{129}\times7^{21}

3^{127}\times5^{128}\times7^{21}

3^{127}\times5^{128}\times7^{22}

CTET 2 Math and Science Sep 2015 Paper

Question : 54 of 120

Marks: +1, -0

a=3^{129}×5^{128}×7^{22}

b=3^{128}×5^{129}×7^{21}

c=3^{127}×5^{129}×7^{23}

and

\mathrm{d}=3^{127}×5^{128}×7^{24}

then H.C.F. of a, b, c and d is :

Solution:

Given that

a=3^{129}×5^{128}×7^{22}

b=3^{128}×5^{129}×7^{21}

c=3^{127}×5^{129}×7^{23}

and

d=3^{127}×5^{128}×7^{24}

Now, HCF of a, b, c and d
= Product of the minimum

=3^{127}×5^{128}×7^{21}

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