Manhattan GMAT Challenge Problem of the Week – 12 March 2012
Here is a new Challenge Problem! If you want to win prizes, try entering our Challenge Problem Showdown. The more people that enter our challenge, the better the prizes!
Question
The integers a, b, c, and d can each be equal to 0, 1, 2, or 3, independently. What is the value of (a + 1)(b + 1)(c + 1)(d + 1)?
(1) a + 4b + 16c + 64d = 165
(2) 64a + 16b + 4c + d = 90
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed.
Answer
After you note the constraints on a, b, c, and d, focus on the question. Because multiplying the four variables (or, rather, 1 plus each of those variables) together can produce a ton of possible outcomes, there’s not much use in listing all those outcomes out and looking for patterns. Instead, just rephrase to a simple “What are the values of a, b, c, and d?” (Technically, that’s an oversimplification, but it will do for now—and when you rephrase, don’t ever forget completely about the original phrasing of the question.)
Statement (1): SUFFICIENT. Surprisingly, this equation provides enough information to find the values of all four variables. A good way to see why is to consider the following variation on the problem. Imagine that you have three new variables (x, y, and z), and that these variables can take on any integer value from 0 to 9, inclusive. Can you solve for all three variables from the following equation?
100x + 10y + z = 243
Yes, you could! x = 2, y = 4, and z = 3. We’re all used to the idea with typical base-10 numbers that you can express any positive integer uniquely as a sum of 0-9 units, 0-9 tens, 0-9 hundreds, etc. In other words, you can express any integer uniquely as a sum of powers of ten, if you ensure that you never take more than 9 copies of any power of ten. (In other words, you always take a digit number of copies—for instance, in 243, you take 2 hundreds, 4 tens, and 3 units.)
You can do the same thing with any base, as long as you adjust the possible number of copies down. For instance, to express a positive integer as a unique sum of powers of 4, you must limit yourself to taking no more than 3 copies of any power. (In the same way, you can’t take 10 hundreds when you write out a number with digits: the way to do so is to take 1 thousand.)
If you knew all this in advance, great! What if you didn’t? You could get there by fiddling. You want to get to 165. Build up from the biggest power. Make d equal to 2, so 64d = 128.
a + 4b + 16c + 128= 165
Okay, now you have to get to 165 – 128 = 37. Make c = 2, so 16c = 32.
a + 4b + 32 = 37
Finally, make a = 1 and b = 1, and you’re there.
There’s no other way to get to 165, as you can confirm by trying other possible values of the four variables. Thus, you can find a unique value of the expression in the question.
Statement (2): SUFFICIENT. For the same reasons, you can solve uniquely for the values of the four variables. Again, they are a = 1, b = 1, c = 2, and d = 2.
The correct answer is D.
Special Announcement: If you want to win prizes for answering our Challenge Problems, try entering our Challenge Problem Showdown. Each week, we draw a winner from all the correct answers. The winner receives a number of our our Strategy Guides. The more people enter, the better the prize. Provided the winner gives consent, we will post his or her name on our Facebook page.


3 comments
chaitu0610 on March 18th, 2012 at 9:17 pm
This problem can be solved in an alternate method..
1)Sufficient
a + 4b + 16c + 64d = 165
a+4(b+4c+16d)=165
hence a is odd. a is either 1 0r 3
if a=3 then 4(b+4c+16d)= 162 which is not possible as 162/4 is not an integer.
so a=1 and 4(b+4c+16d)=164.
b+4c+16d=41
b+4(c+4d)=41
applying same logic as above we get b=1.
c+4d=10
since 4d is even c is even but not 0 since 10/4 is not an integer. so we get c=2 and d=2
2)Sufficient
64a + 16b + 4c + d = 90
4(16a+4b+c)+d=90
d is even and not 0 as 90/4 is not an integer so d is 2
16a+4b+c=22
4(4a+b)+c=22
we get c=2
4a+b=5
b is odd but not 3 as 2/4 is not an integer
b=1
so a=1
buy123 on April 1st, 2012 at 10:58 am
Can somebody please explain the first alternative to solve this problem. It is still unclear to me. Thank you very much.
chaitu0610 on April 1st, 2012 at 10:51 pm
hi given that a,b,c,d are integers in the range {0,3}
a + 4b + 16c + 64d = 165
a+4(b+4c+16d)=165
4(b+4c+16d)is even so a is odd. of the range we have a can be either 1 or 3.
consider a=3,we have 4(b+4c+16d)=162. (b+4c+16d)=162/4
(b+4c+16d) should be an integer as b,c,d are all integers but that will not be the case when a=3.
now substitute a=1. the condition is satisfied.
Use the same line of reasoning for the subsequent equations u get.
Hope this was helpful.