If j, k, and n are consecutive integers such that 0 < j < k < n and the units (ones) digit....?

Question

of the product jn is 9, what is the units digit of k ?

Answer 1

You know that j and n are both odd numbers, since their product is odd, and that, since the three are consecutive, that j and n must have a difference of 2.

Only 9,11 or -11,-9 will give you numbers exactly 2 apart whose product ends in 9. that makes k either 10 or -10. In either case the units digit of k is 0.

Answer 2

0 j,k,n are single digit
if jn = 9, what are the values of the 3?
k = j+1
n=j+2
if jn = 9, then j(j+2) = 9
j^2 + 2j - 9 = 0
j = (-2 +- sqrt(4 + 36))/2 = (-2 +- sqrt(40))/2
It seems this problem will not yield whole number for the solution; else, I don't get it correctly.

Answer 3

0<1<3<9

but they're not consecutive...

but if we have
0<3^0<3^1<3^2
the exponents are consecutive

Answer 4

the ones unit of k must be 0:

ones unit of j = 9
ones unit of k = 0
ones unit of n = 1

Best of luck! :)

Answer 5

zero?