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(#M40038862) CAT Numerical Ability question number system Keep an EYE

find the last two digit of 136^132!

Asked In CAT Abir Bandyopadhyay (10 years ago)

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jiten choudhary  (10 years ago)

last two digit

6^1=06
6^2=36
6^3=16
6^4=96
6^5=76
6^6=56
6^7=36
6^8=16
6^9=96
6^10=76
6^11=56

repeatation of last two digit is after every 5 so 132!/5 is perfect divisible by 5 so there is no remainder so last two digit is 6^5=76

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Abhinav khare  (10 years ago)

36
the unit place is always going to be 6 for tens place just after going through several iterations the pattern is foun to be 3..53197..531975...so except the first no. rest is repeating with a cycle of 5 so iviing 131 by 5 remainer is 1 so tens place is 3 and unit place is 6

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Abhishek Mohanty  (8 years ago)

136^132!

we can get the last 2 digits by dividing it by 100

i.e------>136^132!%100--------->36^132!%100

now by binomial theorem----36^132!=(6+30)^132!=6^132! + 132!*(6^(132!-1))*30^1+.......

now if we see,from 2nd term itself every every term is divisible by 100

so 6^132!%100------:

6^1=06
6^2=36
6^3=16
6^4=96
6^5=76
6^6=56
6^7=36
6^8=16
6^9=96
6^10=76
6^11=56

repeatation of last two digits is after every 5 cycles so 132!/5 is perfect divisible by 5 so last two digits are 6^5=76

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