If (x # y) represents the remainder that results when the positive integer x is divided by the positive integer y, what is the sum of all the possible values of y such that (16 # y) = 1?

23

Step-by-step explanation:

We can check all the possibilities.

It is not necessary to consider y>16, because in this case, 16#y=16 as 16 is too small to be split in  y  parts.

Now, 1,2,4, 8 and 16 are factors of 16. When you divide 16 by any of the previous integers, the remainder is zero so we discard these.

When y=3, 16=5(3)+1, 16#3=1 so we add y=3. From this, 16=3(5)+1 thus 16#5=1 and we add y=5.

We discard y=6 as 16#6=4 (using that 16=6(2)+4). We also discard y=7 because 16=2(7)+2 then 16#7=2.

For y=9,10,11,12,13,14, when dividing the quotient is one so 16#y=16-y>1 and these values are discarded. However, we add y=15 because 16=15(1)+1 and  16#15=1.

Adding the y values, the sum is 3+5+15=23.

idk man good luck man

2/3x + 7 = 29

step-by-step explanation:

two thirds = 2/3

plus a number = x

at least 29 = 29

there for you should get 2/3x + 7 = 29