Latest Elitmus Aptitude Question SOLUTION: plz correct answer he btana ... find the last two digits of the following expression (1122)^(1122!)

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plz correct answer he btana ...

find the last two digits of the following expression

(1122)^(1122!)

Asked In Elitmus ayush (9 years ago)

Unsolved Read Solution (14)

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Read Solution (14) :

Ram  (9 years ago)

(1122)^(1122!)=(11)^(1122!)*(102)^(1122!)
1122!= will have a even number and zeroes for example:2536200000000
(11)^(1122!)----will have last two digits as 01
because it repeats itself every power of 10 like 11^10,11^20
and so on......last two digits for all these are 01
(102)^(1122!)---will have last two digits as 76
bez it repeats every power of 20 like 102^20,102^40
and so on ...has last two digits for all these are 76
so 01*76=76
so Last two digits of the expression is 76

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ayush  (9 years ago)

exlain to krdo

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Manish  (9 years ago)

1122!=*****00
So last digit is 2^4=*6(As remainder of the value divided by 4 is 0)
And Last 2nd Digit is 2*0(As last digit is 0)
last 2 digit is 06

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Ram  (9 years ago)

answer is 76

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oshank karn  (9 years ago)

1122! is 10*even no from
so (2^10) ^even always give last digit as 76

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ashish patel  (9 years ago)

Ans. iz 56

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ak sharma  (9 years ago)

when result of 28 sep kolkata test ll be declared, any idea guys??????

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aryan  (9 years ago)

answer is 76

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Animesh Gupta  (7 years ago)

1122!= will have a even number and zeroes for example:2536200000000...explain it @Ram or anyone else

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Kamal Kant Mandal  (7 years ago)

(1122)^(1122!)
we can split 1122! in term of 4x+c
there for
1122!=4x+2
1122^2= 125884

so the last 2 digit of the number =84

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Manisha Choudhary  (7 years ago)

correct ans is 84 . go for it :D

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Deepak Bharty  (7 years ago)

1122/100 = 22 ^1122/100

Now R1= 2^1122! /100

R2= 11^1122! / 100

1122! have many zeros in last … as w.k.t after 5! There may have one single zero

So( 2^10)^even no / 100 = 76
e.g 2^10=1024
(1024)^2 mod 100 = 76
(1024)^4 mod 100 = (24)^4 mod 100= 76
(1024)^6 mod 100 =(24)^6 mod 100 =76
so (2^10)^even always gives 76

now 11^1122! /100
see 11^2 = 121
11^3 = __ 31 ( last two digit)
11^4 = … 41
Similarly 11^10 = 01 (last two digit )
Now (11^10) ^even num / 100
= 01 as last two digit
So last two digit = R1*R2 = 76*1 = 76

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Sayantani Maiti  (6 years ago)

We know the last two digits can only result in a number lower than 100,
we know multiplication between two large numbers is basically an accumulation of simple multiplications.
In this case we have (1000+100+20+2)*(1000+100+20+2),
since we can only allow numbers smaller than 100 we can drop 1000 and 100.
We get (20+2)*(20+2) which results in 400+40+40+4,
400 is greater than 100,
so 40+40+4=84.
84 are the last two digits.

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Divya Batra  (6 years ago)

1122*1121*1120!=last digit is 0
and when it is expanded to 1100!,ssecond last digit is also 0.
so answer is 00.

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