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The integers a, b, c, and d can each be equal to 0, 1, 2, or 3, indepe
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24 Apr 2015, 02:07
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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
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Re: The integers a, b, c, and d can each be equal to 0, 1, 2, or 3, indepe
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24 Apr 2015, 07:46
Bunuel wrote: 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 Picking numbers approach: 1) let's start from biggest numbers: max \(d\) can be \(2\) and we have \(2*64 = 128\); max \(c\) can be \(2\) so \(2*16=32\); \(128+32 = 160\) and we need \(5\) so \(b=1\) and \(a=1\). \(1*1+4*1+16*2+64*2=165\) We've found first variant and should seek for another possible scenarios: \(d = 1\) so \(1*64 = 64\) and for \(c\) we pick \(3\) so \(3 * 16 = 48\); \(64+48 =112\) and we need \(53\) and we can't make such number from \(4b\) and \(a\), because maximum can be \(4*3+1*3 = 15\) Sufficient 2) Max \(a\) can be \(1\) so \(1*64\); max \(b\) can \(1\) so \(1*16= 16\) and we have \(80\) and need \(10\) more so \(c =2\) and \(d = 2\). \(64*1+16*1+4*2+1*2=90\) And if we try another variants we can see that we don't have other variants. Sufficient Answer is D.
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Re: The integers a, b, c, and d can each be equal to 0, 1, 2, or 3, indepe
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25 Apr 2015, 00:58
Bunuel wrote: 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 (1) a + 4b + 16c + 64d = 165 a=1,b=1, c=2,d=2 (2) 64a + 16b + 4c + d = 90 a=1, b=1,c=2,d=2 Answer D.
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Re: The integers a, b, c, and d can each be equal to 0, 1, 2, or 3, indepe
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25 Apr 2015, 19:44
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
From Statement 1 , we know that the sum of integers is equal to an ODD number (165) Hence, one of the 4 integers has to be an odd number ==> Since 4b, 16c and 64d always result in even, a must take an odd value i.e 1 or 3 only
a = {1,3}
a + 4b + 16c + 64d = 165 > Lets start with 'd' . The possible values for d are 0,1,2 and 3.
d can take only 0,1 & 2 because d=3 gives a value >165. And since there is no way we can achieve 165 by substituting values 0 & 1 , d must be 2. ==> d = 2
a + 4b + 16c = 37 > Lets go with 'c' now. The possible values for c are 0,1,2 and 3.
c can take only 0,1 & 2 because c=3 gives a value >37. And since there is no way we can achieve 37 by substituting values 0 & 1 , c must be 2. ==> c = 2
a + 4b = 5  > We know a must be odd i.e 1 or 3 only and possible values for b are 0,1,2 and 3.
Substituting a =3 , we get 4b = 1 (not possible because b is an integer)
Therefore, a = 1 and b = 1
Statement 1 is sufficient
(2) 64a + 16b + 4c + d = 90
From Statement 2 , we know that the sum of integers is equal to an EVEN number (90) Hence, all 4 integers have to be an even ==> Since 64a, 16b and 4c always result in even, d must take an even value i.e 0 or 2 only
a = {0,2}
64a + 16b + 4c + d = 90 > Lets start with 'a' . The possible values for a are 0,1,2 and 3.
a can take only 0 & 1 because a= 2 or 3 gives a value >90. And since there is no way we can achieve 90 by substituting value 0 , a must be 1. ==> a = 1
16b + 4c + d = 26 > Lets go with 'b' now. The possible values for b are 0,1,2 and 3.
b can take only 0 & 1 because b = 2 or 3 gives a value >26. And since there is no way we can achieve 26 by substituting value 0 , b must be 1. ==> c = 1
4c + d = 10  > We know d must be even i.e 0 or 2 only and possible values for c are 0,1,2 and 3.
Substituting d =0 , we get 4c = 10 (not possible because c is an integer)
Therefore, d = 2 and c = 2
Statement 2 is sufficient
Answer is (D)



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Re: The integers a, b, c, and d can each be equal to 0, 1, 2, or 3, indepe
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25 Apr 2015, 22:51
The integers a, b, c, and d can each be equal to 0, 1, 2, or 3, independently (a + 1)(b + 1)(c + 1)(d + 1) = ?
Statement (1): a + 4b + 16c + 64d = 165 a + 4(b + 4c + 16d) = 165 odd + even = odd a = 1 or 3 if a = 1, 164 is divisible by 4, if a = 3, 162 is not divisible by 4 > a = 1
b + 4c + 16d = 41 b + 4(c + 4d) = 41 odd + even = odd b = 1 or 3 if b = 1, 40 is divisible by 4, if b = 3, 38 is not divisible by 4, > b = 1
c + 4d = 10 even + even = even c = 0 or 2 if c = 0, 10 is not divisible by 4, if c = 2, 8 is divisible by 4, > c = 2
4d = 8 > d = 2 Sufficient
Statement (2) 64a + 16b + 4c + d = 90 4(16a + 4b + c) + d = 90 even + even = 90 d = 0 or 2 if d = 0, 90 is not divisible by 4, if d = 2, 88 is divisible by 4, > d = 2
16a + 4b + c = 22 4(4a + b) + c = 22 even + even = even c = 0 or 2 if c = 0, 22 is not divisible by 4 if c = 2, 20 is divisible by 4 > c = 2
4a + b = 5 even + odd = odd b = 1 or 3 if b = 1, 4 is divisible by 4 b = 3, 2 is not divisible by 4 > b = 1
4a = 4 > a = 1 Sufficient
Answer D



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Re: The integers a, b, c, and d can each be equal to 0, 1, 2, or 3, indepe
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27 Apr 2015, 00:56
Bunuel wrote: 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 MANHATTAN GMAT OFFICIAL SOLUTION: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 base10 numbers that you can express any positive integer uniquely as a sum of 09 units, 09 tens, 09 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.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



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Re: The integers a, b, c, and d can each be equal to 0, 1, 2, or 3, indepe
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