nick1816
If \(7^{37}\) is written as 4A + B, where A and B are natural numbers and B < 4, then which of the following statements is/are necessarily true?
I) A is even.
II) A is odd.
III) B is divisible by 3.
A. I only
B. II only
C. III only
D. I and III
E. II and III
So B can be 1, 2 or 3.
But 4A+b will be odd, only when B is odd... So, B can be only 1 or 3.Now, let us check what are the last two digits of power of 7 and if there are any pattern in them..
\(7^1=07=4*1+3......7^2=49=4*12+1.....7^3=343=4*85+1........7^4=343*7=43*7=01......7^5=01*7=07....\)
So \(7^{even}=4*even+1\) and \(7^{odd}=4*odd+3\)
(I) there is a straight pattern \(7^{odd}=4*odd+3=4A+B\)
Thus A is odd and B is 3... I and III are true
E
(II) Slightly longish but concept may help in some other question
There is a pattern...07, 49, 43, 01, 07...
The last two digits of \(7^{37}=7^{36+1}=7^{4*9+1}\) will be same as that of \(7^1\) or 07..
So we have possible values of B as 1 and 3..
(I) 1
07=4A+1......06=4A....
Not possible as the integers ending with 06 will not be multiple of 4(II) 3
07=4A+3......04=4A....
Possible as the integers ending with 04 will be multiple of 4 and it also tells us that A is odd...
(III) We can use binomial theorem, but it is not tested on GMAT, so we should use that if we are thorough with the application.