Number Theory problem

Question

The last digit in LCM of !!(3^2002 - 1) and (3^2002 + 1)!!.

!!a)" "8!!

!!b)" "5!!

!!c)" "4!!

!!d)" "0!!

e) None

Solution

In the given numbers !!(3^2002 - 1) and (3^2002 + 1)!! we can see that after simplification we will get two consecutive even numbers.

Since HCF of two consecutive even numbers is !!2!!.

Also product of LCM and HCF of two numbers are equal to the product of those two numbers.

Hence, !!LCMxxHCF = (3^2002 - 1)xx(3^2002 + 1)!!

!!=> LCM = ((3^2002 - 1)xx(3^2002 + 1))/(HCF)!!

!!=> LCM = ((3^2002)^2 - 1)/2!!

!!=> LCM = (3^4004 - 1)/2!!

!!=> LCM = ((3^4)^1001 - 1^1001)/2!!............ (i)

We know that !!(a^n - b^n) = (a-b)xxfurther" "terms!!.

Using the same concept in equation (i):

!!(3^4 - 1)!! will be the factor !!(3^4)^1001 - 1^1001!!

Therefore, !!80!! will be the factor !!(3^4)^1001 - 1^1001!!

Similarly !!40!! will be the factor !!((3^4)^1001 - 1^1001)/2!!

Hence the last digit of LCM will be !!0!!

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