- View Source1. 7^45 = 7(7^44) = 7(2401)^11 = 7(01)^11 = 7(01) = 07.

2. 3^90 + 5^90 = 9^45 + 25^45 is always divisible by (9 + 25) = 34. So remainder is zero.

3. 4 values - 1, 5, 7, 11.**From:**Rajesh Balasubramanian <rajesh@...>**To:**ascent4cat@yahoogroups.com**Sent:**Mon, November 8, 2010 10:57:25 PM**Subject:**[2IIM CAT Prep] CAT Number Theory - Questions on RemaindersHi all,Here are a few more questions on Number Theory based on remainders. They have also been posted at ou blog - http://iimcat.blogspot.com/2010/11/questions-on-number-theory-remainder.html.1. What are the last two digits of the number 7^{45 }?2. What is the remainder when we divide 3^{90}+ 5^{90}by 34?3. N^{2 }leaves a remainder of 1 when divided by 24. What are the possible remainders we can get if we divide N by 12?Cheers all,

Rajesh

99626 48484**New CAT2011 batches starting @ Chennai**

Nov 13th @ Velachery, Nov 20th @ Mylapore