a and b are positive integers. Which unit digit(s) of a result(s) in the maximum possible absolute difference in the unit values of ab and a(b+1)? Choose all that apply.
1
2
3
6
7
9
This question is from the Practice Quant Extreme section, I think the explanation is wrong. In Greg’s explanation - I understand that the max unit difference of 9 must be there.
You put, b = 9 (where you check for the unit of a =1) and b = 10 (where you check for the unit of a =9). But how are you imagining b values?
According to me, it should go like this -
∣unit digit of ab−(unit digit of ab+unit digit of a)∣=∣−unit digit of a∣=unit digit of a.
The goal is to maximize the unit digit of a. Ans should be only 9 here.
That isn’t what the question is asking though?
By imagining b values i mean-
when you are checking for unit of a=1 option, you assume unit of b=9.
What if b is not equal 9, let’s say it is unit of b = 3. Then the max absolute difference = 1.(3+1) - 1.3 = 1 which is not equal to 9 and is not optimized.
It’s the same argument for the unit of a = 9.
If the question was -
Which unit digit of “a” and unit digit of “b” result in the maximum possible absolute difference in the unit values of “ab” and “a(b+1)”?
then the answers are correct which is = (unit of a, unit of b) = (1,9) and (9,0).
Please correct me if I am wrong!
This is not necessary. It’s a “could” case, implied by the use of the word “possible”. This means that the (1, 9) case is equally valid as (9, 0). If the wording implied a “must” case, then things may change.
ahh I see, makes sense, the word “possible” resolves this, didn’t notice the word, thanks a lot !!