If you do a quick list of first integers which have 9 at the end of a square is 3, 7...and so on. 7 fits so a is 7
so 7^2 = 49
8^2 = 64.
9^2 = 81.
So units digit is 1.
unit digit
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tohellandback
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unit digit of a must be 3 or 7 because unit digit of a^2 is 9.
unit digit of (a+1) must be 2 or 8 because (a+2)^2 has the unit digit 4
from this, unit digit of a must be 7
7+2=9
9^2 has the unit digit 1
so A
unit digit of (a+1) must be 2 or 8 because (a+2)^2 has the unit digit 4
from this, unit digit of a must be 7
7+2=9
9^2 has the unit digit 1
so A
The powers of two are bloody impolite!!
This problem was printed in OG Quant Review p80 #142. My question is, am I the only person confused by the wording of this problem? At first I really couldn't figure out what the heck they were asking.
They should have said something like the (unit digit of a)^2. the [(unit digit of a)+1]^2, etc. Basically I'm saying that they messed up the parenthesis and really threw me off.
They should have said something like the (unit digit of a)^2. the [(unit digit of a)+1]^2, etc. Basically I'm saying that they messed up the parenthesis and really threw me off.