timothyhenman1 wrote:
If w, x, y, and z are the digits of the four-digit number N, a positive integer, what is the remainder when N is divided by 9?
(1) w + x + y + z = 13
(2) N + 5 is divisible by 9
Nice question. Answer is (D) - Each alone is sufficient. Here's why:
(1) SUFFICIENT.The rule for divisibility by 9 is that you add up all the digits of the given number. If this addition is divisible by nine, the number too is divisible by nine. In case the addition works out to be a number > 9, add the digits again. The same works for remainders - for 9, the remainder you get when you add digits and divide the result by 9 is the same remainder you get after dividing the original number by 9.
In our case, w+x+y+z = 13
Add digits again - 1+3 = 4
So, remainder after dividing by 9 is exactly 4 => SUFFICIENT.
Now for the analytical part - WHY?
A four digit number can be represented as 1000w + 100x + 10y + z.
Let's take the Remainder operation on "number" and call it Rem(number/9). Note that we can add remainders as long as we make the final adjustment to make them < original number (which is 9 here).
So, overall remainder is = Rem(1000w/9) + Rem (100x/9) + Rem(10y/9) +z
For any multiple of 10, you'll notice that 10^n - 1 is always divisible by 9 (e.g. 9, 99, 999 and so on). So, Rem(1000w/9) becomes Rem(1000/9)*Rem(w/9) = 1*Rem(w/9)
As w is a single digit, Rem(w/9) = w (even if w is 9, remainder can be technically 9. we will make the adjustment in the final phase)
Similarly, Rem(100x/9) = x, Rem (10y/9) = y and Rem(z/9) = z
So, overall remainder = Rem{(w+x+y+z)/9}
That's what we did above. We took the remainder after adding all the digits. This is complicated to remember, so just focus on the part before the incurably curious mind asks "WHY". It's a rule worth remembering.
(2) SUFFICIENTIf N+5 is divisible by 9, it means N+5 = 9k (where k is any integer). You can just stop here and say that this is sufficient, as if you know this then intuitively you know that N+4 would leave a remainder of 8, N+3 would leave 7 and so on.. with N/9 leaving a remainder of 4. Hence, SUFFICIENT.
Here's a more analytical treatment (and unnecessary at that - you MUST save time on DS if you can. If you are sure with the above approach, don't even do this) -
So, N = 9k - 5.
i.e. N/9 = k -5/9
compare this with Dividend / Divisor = Quotient + Remainder/Divisor
We get Remainder = -5.
For negative remainders, we add back the divisor to get the proper positive remainder. So, remainder = -5 +9 = 4. Therefore, SUFFICIENT.