Question

Remainder when $ {2^{30}}{3^{30}} $ is divisible by 7
 $ \left( a \right) $ 1
 $ \left( b \right) $ 2
 $ \left( c \right) $ 4
 $ \left( d \right) $ 6

Answer 1

Hint: In this particular question first simplify the equation so that the power is of only a single digit number now divide this digit, square of digit, cube of digit by given divisor and observe the pattern so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given number = $ {2^{30}}{3^{30}} $
Now we have to find the remainder when this is divided by 7.
Now as we know that $ {a^x}.{b^x} = {\left( {a.b} \right)^x} $ so use this property in the above equation we have,
 $ \Rightarrow {2^{30}}{3^{30}} = {\left( {2.3} \right)^{30}} = {\left( 6 \right)^{30}} $
Now as we know that when 6 divide by 7 the remainder is 6.
Now when $ {6^2} $ is divide by 7 the remainder is
 $ \Rightarrow \dfrac{{{6^2}}}{7} = \dfrac{{36}}{7} = 5\dfrac{1}{7} $ , so the remainder is 1.
Now when $ {6^3} $ is divide by 7 the remainder is
 $ \Rightarrow \dfrac{{{6^3}}}{7} = \dfrac{{216}}{7} = 30\dfrac{6}{7} $ , so the remainder is 6.
Now when $ {6^4} $ is divide by 7 the remainder is
 $ \Rightarrow \dfrac{{{6^4}}}{7} = \dfrac{{1296}}{7} = 185\dfrac{1}{7} $ , so the remainder is 1.
So as we see that remainder is either 1 or 6 when divisible by 7.
Now as we see that when the power of 6 is odd the remainder is 6 and when the power of 6 is even the remainder is 1.
Now the power of given equation is 30 which is even, so when we divide $ \left[ {{2^{30}}{3^{30}} = {{\left( {2.3} \right)}^{30}} = {{\left( 6 \right)}^{30}}} \right] $ by 7 the remainder is 1.
So this is the required answer.
Hence option (A) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall that if a number is written in the form of $ a\dfrac{b}{c} $ , then a is quotient, b is remainder and c is divisor, so divide the number by 7 one by one as above and observe carefully we will get the required remainder.