Difficult Manhatan Problem

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Difficult Manhatan Problem

by kakz » Mon Oct 24, 2011 9:35 am

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The difference between positive two-digit integer A and the smaller two-digit integer B is twice A's units digit. What is the hundreds digit of the product of A and B?

(1) The tens digit of A is prime.
(2) Ten is not divisible by the tens digit of A.

OA is C

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by neelgandham » Mon Oct 24, 2011 10:15 am
Let A be of the form 10a+b, then B = 10a-b as the difference between positive two-digit integer A and the smaller two-digit integer B is twice A's units digit.

Product of the numbers = (10a+b)(10a-b) = (100*a*a)-(b*b), and b!=0


(1) The tens digit of A is prime.
=> a =2,3,5,7

So, Product of the numbers = (100*a*a)-(b*b) =400-(b^2), if a = 2 and 318<400-(b^2)<399 (varying the vale of b from 1 to 9)
So, Product of the numbers = (100*a*a)-(b*b) =900-(b^2), if a = 3 and 818<900-(b^2)<899 (varying the vale of b from 1 to 9)
So, Product of the numbers = (100*a*a)-(b*b) =2500-(b^2), if a = 5 and 2418<2500-(b^2)<2499 (varying the vale of b from 1 to 9)
So, Product of the numbers = (100*a*a)-(b*b) =4900-(b^2), if a = 7 and 4818<4900-(b^2)<4899 (varying the vale of b from 1 to 9)

So the hundreds digit of the product of A and B can be 3,4,8.Hence,Insufficient !


(2) Ten is not divisible by the tens digit of A.

so a != 1,2,5 and a can be 3,4,6,7,8,9

So, Product of the numbers = (100*a*a)-(b*b) =900-(b^2), if a = 3 and 818<900-(b^2)<899 (varying the vale of b from 1 to 9)
So, Product of the numbers = (100*a*a)-(b*b) =1600-(b^2), if a = 4 and 1518<1600-(b^2)<1599 (varying the vale of b from 1 to 9)
So, Product of the numbers = (100*a*a)-(b*b) =3600-(b^2), if a = 6 and 3518<3600-(b^2)<3599 (varying the vale of b from 1 to 9)
So, Product of the numbers = (100*a*a)-(b*b) =4900-(b^2), if a = 7 and 4818<4900-(b^2)<4899 (varying the vale of b from 1 to 9)
So, Product of the numbers = (100*a*a)-(b*b) =6400-(b^2), if a = 8 and 6318<6400-(b^2)<6399 (varying the vale of b from 1 to 9)
So, Product of the numbers = (100*a*a)-(b*b) =8100-(b^2), if a = 9 and 8018<8100-(b^2)<8099 (varying the vale of b from 1 to 9)

So the hundreds digit of the product of A and B can be 8,5,8,3,0.Hence,Insufficient !

Using both the options, we get the value of a = 3 or 7 (a=2,3,5,7 from 1 and a=3,4,6,7,8,92)

So, Product of the numbers = (100*a*a)-(b*b) =900-(b^2), if a = 3 and 818<900-(b^2)<899 (varying the vale of b from 1 to 9)
So, Product of the numbers = (100*a*a)-(b*b) =4900-(b^2), if a = 7 and 4818<4900-(b^2)<4899 (varying the vale of b from 1 to 9)

The hundreds digit of the product of A and B is 8 in both the cases, hence option C
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by GmatKiss » Mon Oct 24, 2011 11:07 am
Man thats a 1000- Level question for me!

If you answer this in the real CAT, say with less than 3 mins, you will get a bonus :D
BTW, is there any easier way to solve this one?

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by saketk » Mon Oct 24, 2011 11:08 am
kakz wrote:The difference between positive two-digit integer A and the smaller two-digit integer B is twice A's units digit. What is the hundreds digit of the product of A and B?

(1) The tens digit of A is prime.
(2) Ten is not divisible by the tens digit of A.

OA is C

--===== ----

Made a blunder earlier on -- I considered unit digit of A as prime in statement 1. Damn!

from statement 1: we can have the tens digit of A = 2,3,5 & 7

it is given that the difference of A and B is twice that of unit's digit of A i.e. the difference
of (A-B) can be 4,6,10 or 14


for example if A = 22 then B will be 18 and the difference will be 4 (twice of 2)

and the hundred digit of the product will be 3

if A = 53 then B =47 and the difference will be 6 (twice of 3)

and and the hundred digit of the product will be 4.


Hence, Statement 1 is not sufficient


STATEMENT 2: - 10 is not divisible by tens digit of A.

This means that A's tens digit cannot be 1,2 or 5.

Using the numbers again.. we can have A = 42 and B = 38 - difference = 4 (hundred digit of the product = 5)

or A= 33 and B= 27 - difference = 6-- twice of 3 (and the hundred digit of the product will be = 8)

Hence, this statement is also insufficient

Now, let us combine the two statements

A tens can now be --3 or 7 ONLY

for example

A = 33
B = 27

difference = 6-- twice of 3 (and the hundred digit of the product will be = 8)

or A= 74 B = 66

difference = 8 (twice of 4) and the hundred digit of the product will be = 8

Another example --

A= 77
B = 63 ( twice of 7 is 14) and the hundred digit of the product 8


Every time we will get 8

Therefore , the answer should be C

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by neelgandham » Mon Oct 24, 2011 12:44 pm
GmatKiss wrote:Man thats a 1000- Level question for me!

If you answer this in the real CAT, say with less than 3 mins, you will get a bonus :D
BTW, is there any easier way to solve this one?
GmatKiss - The explanation looks lengthy, but I took around 75 seconds to answer this question. I feel it is more to do with the confidence of approach than anything else. I feel so scared of a few basic sentence correction questions that I end up marking randomly.:-) !
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by yashodhan » Wed Oct 26, 2011 12:07 pm
Hey Neel, Did you really follow this method to answer in 75 sec
or did you use numbers? It seems to involve too many calculations to be done in 75 sec.

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by neelgandham » Wed Oct 26, 2011 12:29 pm
yashodhan wrote:Hey Neel, Did you really follow this method to answer in 75 sec
or did you use numbers? It seems to involve too many calculations to be done in 75 sec.
To be precise I took around 83 secs(I record my times :D) on paper, where most of the calculation is more or less eliminated. For example

The step:
if a = 2 and 318<400-(b^2)<399 (varying the vale of b from 1 to 9) implies if a=2, a^2 =4 and the hundreds digit is 4-1, So if a=3 the hundreds digit is 9-1, if a=5 the hundreds digit is 25-1, and so on. On paper I would have 3,8,4 etc and not all the calculations as explained.

Hope it helps !
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by GmatKiss » Thu Oct 27, 2011 1:06 am
Sent a PM to Mitch. Waiting for his solution :)

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by GMATGuruNY » Thu Oct 27, 2011 3:14 am
kakz wrote:The difference between positive two-digit integer A and the smaller two-digit integer B is twice A's units digit. What is the hundreds digit of the product of A and B?

(1) The tens digit of A is prime.
(2) Ten is not divisible by the tens digit of A.

OA is C
Let T = the tens digit of A and U = the units digit of A.
A = 10T+U.
Since the difference between A and B is twice A's units digit:
B = 10T+U - 2U = 10T-U.
Thus:
A*B = (10T+U)(10T-U) = 100T²-U².

Since U≤9, U²<100.
Thus, to determine the hundreds digit of 100T²-U², we need to know the value of T².
To illustrate:
If T²=4, then 100T²-U² = 400-U².
The hundreds digit will be 3.
If T²=9, then 100T²-U² = 900-U².
The hundreds digit will be 8.

Question rephrased: What is T²?

Statement 1: T is prime.
T could be 2,3,5 or 7.
Thus, T² could be 4,9,25,or 49, resulting in different hundreds digits for 100T²-U².
INSUFFICIENT.

Statement 2: 10 is not a multiple of T.
T could be 3,4,6,7,8 or 9.
Thus, T² could be many different values, resulting in different hundreds digits for 100T²-U².
INSUFFICIENT.

Statements 1 and 2 combined:
Either T=3 or T=7.
If T=3, then T² = 9, so that 100T²-U² = 900-U².
If T=7, then T² = 49, so that 100T²-U² = 4900-U².
In each case, the hundreds digit will be 8.
SUFFICIENT.

The correct answer is C.
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by Anurag@Gurome » Thu Oct 27, 2011 3:19 am
kakz wrote:The difference between positive two-digit integer A and the smaller two-digit integer B is twice A's units digit. What is the hundreds digit of the product of A and B?

(1) The tens digit of A is prime.
(2) Ten is not divisible by the tens digit of A.[/spoiler]
There is no need to check for so many possibilities.

Say, A = (10a + b), where a is the tens digit and b is the units digit of a.
Hence, B must be (10a - b)

Thus, AB = (10a + b)(10a - b) = (100a² - b²)

Note that, 1 ≤ b ≤ 9
So, 1 ≤ b² ≤ 81

Statement 1: Possible values of a : 2, 3, 5, and 7
Say, a = 2 and b = 1 --> AB = (100*4 - 1) = 399 --> Hundreds digit is 3
Say, a = 3 and b = 1 --> AB = (100*9 - 1) = 899 --> Hundreds digit is 8

NOT Sufficient

Statement 2: Possible values of a : 1, 3, 4 etc
Say, a = 1 and b = 1 --> AB = (100*1 - 1) = 099 --> Hundreds digit is 0
Say, a = 3 and b = 1 --> AB = (100*9 - 1) = 899 --> Hundreds digit is 8

NOT Sufficient

1 & 2 Together: Possible values of a : 3 and 7
Say, a = 3 and 1 ≤ b² ≤ 81 --> (900 - 81) ≤ AB ≤ (900 - 1) --> 819 ≤ b² ≤ 899 --> Hundreds digit is always 8
Say, a = 7 and 1 ≤ b² ≤ 81 --> (4900 - 81) ≤ AB ≤ (4900 - 1) --> 4819 ≤ b² ≤ 4899 --> Hundreds digit is always 8

SUFFICIENT

The correct answer is C.
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by enjoylife1788 » Thu Oct 27, 2011 10:53 pm
Just amazing how rephrasing helps in such complicated questions. Honestly, I was bogged down when i first saw the question. But, after I saw such simple explanations, I feel like no question on the GMAT is too difficult. :)

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by LalaB » Sun Oct 30, 2011 12:25 am
kakz wrote:The difference between positive two-digit integer A and the smaller two-digit integer B is twice A's units digit. What is the hundreds digit of the product of A and B?

(1) The tens digit of A is prime.
(2) Ten is not divisible by the tens digit of A.

OA is C

A number y x
minus
B number z p
==============================
equals to 2x 2x

so, lets begin -
we know that x-p=2x lets assume that x=1 then p=9 2x=2*1=2 then the result of subtraction is 22

now we r moving to our stmnts-

stmnt 1- y can be 2;3;5;7. 2 and 3 are out,since A number > B number,and we need 22 as a result.if A begins with 2 or 3 , then B cant be a two-digit number.
so, now we have 5 and 7. it means that our number could be either 51 or 71. since we dont know the definite answer , stmnt 1 is insuff

lets go to stmnt 2-

Ten is not divisible by the tens digit of A. of course, it is not suff

but taking stmnt 1 (51 or 71)& stmnt 2 (Ten is not divisible by the tens digit of A)together we have -

oh, that is good! it means ,that 51 cant be our answer choice(since 10 is divisible by 5).
since 51 is out, our answer is 71

so, now we know A is 71, B is 49.so, we can calculate the hundreds digit of the product of A and B

now my question is -what is wrong in my way of thinking? )))

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by GmatMathPro » Sun Oct 30, 2011 8:10 am
LalaB wrote:
A number y x
minus
B number z p
==============================
equals to 2x 2x

so, lets begin -
we know that x-p=2x lets assume that x=1 then p=9 2x=2*1=2 then the result of subtraction is 22
You seem to have misunderstood the problem. If x is the unit's digit of A, then A-B=2x, not 2x 2x. It looks like you're trying to make 2x the ten's and unit's digit of the difference between A and B, but 2x is not any particular digit, it is just the answer is to A-B. For example, it could be that A=29, B=11 so A-B=18, which is 2(9). Or it could be something like A=53 and B=47 and A-B=6, and 6=2(3).

Also, why did you assume that x=1? It's true that the exact value of the unit's digit of A doesn't end up mattering in this problem, but are you really sure of that at this point in the process?
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by LalaB » Sun Oct 30, 2011 10:37 am
GmatMathPro wrote: You seem to have misunderstood the problem. If x is the unit's digit of A, then A-B=2x, not 2x 2x. It looks like you're trying to make 2x the ten's and unit's digit of the difference between A and B, but 2x is not any particular digit, it is just the answer is to A-B. For example, it could be that A=29, B=11 so A-B=18, which is 2(9). Or it could be something like A=53 and B=47 and A-B=6, and 6=2(3).

Also, why did you assume that x=1? It's true that the exact value of the unit's digit of A doesn't end up mattering in this problem, but are you really sure of that at this point in the process?
yes, i assumed that the result of subtraction is a two-digit number.

as for the number 1- it was just an example. it is the 1st number. that is why I began from it.that is all.

so i seem to get a right answer by lucky (( pity ((

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by Brent@GMATPrepNow » Wed Nov 28, 2018 9:21 am
kakz wrote:The difference between positive two-digit integer A and the smaller two-digit integer B is twice A's units digit. What is the hundreds digit of the product of A and B?

(1) The tens digit of A is prime.
(2) Ten is not divisible by the tens digit of A.

OA is C
Target question: What is the hundreds digit of the product AB?

Given: The difference between positive two-digit integer A and the smaller two-digit integer B is twice A's units digit
Let x = the tens digit of A, and let y = the units digit of A
So, the VALUE of A = 10x + y

From the given information, we can write: (10x + y) - B = 2y
Add B to both sides: 10x + y = 2y + B
Subtract 2y from both sides: 10x - y = B

So, A = 10x + y and B = 10x - y
So, the product AB = (10x + y)(10x - y) = 100x² - y²


Statement 1: The tens digit of A is prime.
In other words, x is prime
Let's TEST some values.
Case a: x = 2 (which is prime) and y = 3. In this case, AB = 100(2²) - 3² = 400 - 9 = 391. So, the answer to the target question is the hundreds digit of AB is 3
Case b: x = 3 (which is prime) and y = 1. In this case, AB = 100(3²) - 1² = 900 - 1 = 899. So, the answer to the target question is the hundreds digit of AB is 8
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: Ten is not divisible by the tens digit of A.
10 is not divisible by 3, 4, 6, 7, 8, or 9
In other words, x could equal 3, 4, 6, 7, 8, or 9
Let's TEST some values.
Case a: x = 3 and y = 1. In this case, AB = 100(3²) - 1² = 900 - 1 = 899. So, the answer to the target question is the hundreds digit of AB is 8
Case b: x = 6 and y = 1. In this case, AB = 100(6²) - 1² = 3600 - 1 = 3599. So, the answer to the target question is the hundreds digit of AB is 5
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that x could equal 2, 3, 5 or 7
Statement 2 tells us that x could equal 3, 4, 6, 7, 8, or 9
When we COMBINE the two statements, we see that x must equal EITHER 3 OR 7

IMPORTANT: Many students will incorrectly conclude that, since x can equal EITHER 3 OR 7, then the combined statements are not sufficient.
However, the target question is not asking us for the value of x; the target question is asking for the hundreds digit of AB.
So, let's test the two possible values of x:
Case a: x = 3 and y = any single digit. In this case, AB = 100(3²) - (any single digit)² = 900 - (some number less than 100) = 8??. So, the answer to the target question is the hundreds digit of AB is 8
Case b: x = 7 and y = any single digit. In this case, AB = 100(7²) - (any single digit)² = 4900 - (some number less than 100) = 48??. So, the answer to the target question is the hundreds digit of AB is 8
Aha!!!
In both possible cases, the answer to the target question is the SAME.
So, it MUST be the case that the hundreds digit of AB is 8
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

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