What # when you multiply it by 5 and add 6 then multiply the result by 4 and add 9 gives you a # that when...?

Question

you multiply it by 5 and subract 165 gives you a # that when you knock off the last two digits brings you back to the original #?

Answer 1

Let's translate the English into mathematical notation.

Let's let the original number be x.
Then 4(5x + 6) + 9 = y, a new number.
5y - 165 = z, yet another number. However, when we "knock off the last two digits" of z, we get x again. In other words,

int (z/100) = x
which may be also written
100x ≤ z < 101x, assuming that x is an integer.

Let us now combine our equations and eliminate y to get
5(4(5x + 6) + 9) - 165 = z

Before proceeding, this looks like a good time to simplify the left side of the equation, giving us
100x = z

If you make this substitution into the inequality, the inequality will be true no matter what value x takes. This will work for negative and zero values for x as well as positive ones, so you can start with any integer and massage it according to the directions and get back to the original number. Only the condition that you "knock off the last two digits" to get back to the original number restricts the solution set to integers. If this had been replaced by a simple divide by 100, any real number would have worked.

Answer 2

n = #

4(5n + 6) +9 = x

5x-165 = y

n = 1

4(5 + 6) +9 = 53 = x

5(53) - 165 = 100

knock off the last 2 digits and you get 1...


n = 1

Answer 3

snickers should have explained what mathematical operation is 'knock off'. This is absurd to interpret knock off as 'divide' or just remove or subtract or what.
[(5x + 6)*4 + 9]*5 -- 165 = 100x + 165 -- 165 = 100x now what is knock off?

Answer 4

any # will work
when you do all of this you get the orginal number. it does not mater what the # is