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E
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Difficulty:
85%
(hard)
Question Stats:
38% (02:30) correct
62%
(01:59)
wrong
based on 39
sessions
History
Date
Time
Result
Not Attempted Yet
18 Jun 2020, 10:59
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OUT OF SCOPE OF GMAT
21 = 21 mod 100
21^2 = 41 mod 100 [ (2+2)1......as we won't get carryover, since unit digit is 1]
21^3 = 61 mod 100 [[ (4+2)1......as we won't get carryover, since unit digit is 1]
21^4 = 81 mod 100
21^5 = 01 mod 100
After this point, cycle will repeat.
So basically we have to find the remainder of 3^50 when divided by 5.
\(3^{50} = 9^{25} = (-1) mod 5 = 4 mod 5\)
\(3^{50} = 5k +4\)
\(21^{5k+4} = 21^{5k} * 21^4\)
\(21^{5k} * 21^4 mod 100 = 01*81 mod 100 = 81 mod 100\)
Last second digit of \((21^3)^{50}\) is 8.
21 = 21 mod 100
21^2 = 41 mod 100 [ (2+2)1......as we won't get carryover, since unit digit is 1]
21^3 = 61 mod 100 [[ (4+2)1......as we won't get carryover, since unit digit is 1]
21^4 = 81 mod 100
21^5 = 01 mod 100
After this point, cycle will repeat.
So basically we have to find the remainder of 3^50 when divided by 5.
\(3^{50} = 9^{25} = (-1) mod 5 = 4 mod 5\)
\(3^{50} = 5k +4\)
\(21^{5k+4} = 21^{5k} * 21^4\)
\(21^{5k} * 21^4 mod 100 = 01*81 mod 100 = 81 mod 100\)
Last second digit of \((21^3)^{50}\) is 8.
mrr0821
yashikaaggarwal
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18 Jun 2020, 11:19
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The tens digit of any number ending with 1 is the unit digit of the product of the unit digit of the power& tens digit of the base; with the unit digit as 1
For tens digit of number ending with 3, repeatedly square no.unit we get 1 as unit digit, here 3^4 = 81
So, (3^4)^12*3^2
=> 81^12*3^2
=> 61*9 => 49
Tens digit of 3^50 is 49
Tens digit of 21^49 = 81
Where 8 is tens Digit and 1 is units
Answer is E
Posted from my mobile device
For tens digit of number ending with 3, repeatedly square no.unit we get 1 as unit digit, here 3^4 = 81
So, (3^4)^12*3^2
=> 81^12*3^2
=> 61*9 => 49
Tens digit of 3^50 is 49
Tens digit of 21^49 = 81
Where 8 is tens Digit and 1 is units
Answer is E
Posted from my mobile device
