Suppose a^2-b^2=91 (a, b integers). If n=a^2+b^2

Question

Suppose a^2-b^2=91 (a, b integers). If n=a^2+b^2<1000, find the units digit of n.

anonymous · Answer

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

anonymous · Answer

I have answer choices which are 1,3,5,7,9.

anonymous · Answer

then it is be sorry mix up with b and n

anonymous · Answer

no worries. Thank you

phi · Answer

, 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.

anonymous · Answer

is there a way that you worked it out? Sorry, im having a little trouble understanding

phi · Answer

it is a bit tricky.
First, do you know you can factor a difference of squares?
a^2-b^2= (a-b)(a+b)

anonymous · Answer

yes.

phi · Answer

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

phi · Answer

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

phi · Answer

follow?

anonymous · Answer

yes so why is the answer 9? What are units digit?

phi · Answer

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?

phi · Answer

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

anonymous · Answer

Thank you, I understand it now. :)