Suppose a^2-b^2=91 (a, b integers). If n=a^2+b^2<1000, find the units digit of n.
straight forward...think out of the box....[ a + b ] [ a - b ] = 91 means find the breakdown of 91 .. { 13 * 7 }....thus a + b = 13 and a - b = 7 ---> a = 10 , b = 3 and n = 109
I have answer choices which are 1,3,5,7,9.
then it is be sorry mix up with b and n
no worries. Thank you
, find the units digit of n. that would be 9 also, notice you could have used 91 and 1 (rather than 7 and 13), but this leads to a^2+b^2 being > 1000, so not allowed.
is there a way that you worked it out? Sorry, im having a little trouble understanding
it is a bit tricky. First, do you know you can factor a difference of squares? a^2-b^2= (a-b)(a+b)
yes.
next, notice that (a-b)(a+b) can be thought of as two numbers x= (a-b) and y= (a+b) so x*y= 91
If we factor 91: 1,91 or 7,13 are the only possible choices. (a and b are integer, so x and y are integer) so we could have 1*91= 91 x=1 y=91 or x=7 , y= 13 7*13= 91
follow?
yes so why is the answer 9? What are units digit?
only half way there. x= a-b= 7 (want x to be the smaller of 7 and 13. ) y= a+b=13 (and when we add a+b, we want the bigger number) solve for a and b can you do that?
one way, add the 2 eqs 2a=20, a=10 10-b=7, b=3 now find n n= a^2+b^2= 100+9= 109 find the units digit of n. that would be 9
Thank you, I understand it now. :)
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