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Re: The difference between positive two-digit integer A and the smaller tw [#permalink]
Hi,
No doubt that your solution is correct. But still could you explain me the highlighted portions? Actually could you solve in more simple manner?
Thanks
Celestial

Harley1980 wrote:
Bunuel 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.


Kudos for a correct solution.


Let's \(A = ab\) and \(B =cd\); so \(A = (10a + b)\) and \(B=(10c + d)\)
As we know \(ab-cd = 2b\) so
\((10a + b) - (10c + d) = 2b\)
\(10a - 10c = d + b\)
\(10(a-c)=d+b\)
As we know that \(d\) and \(b\) its one unit digits and can be max \(9\) so their sum can be max \(18\)
So \(a-c\) can't be more than \(1\) and we can infer that:
\(d+b = 10\)
\(a-c = 1\)

And now we should find what will be \(ab*cd\)
\((10a + b)*(10c + d)\)
\(100ac + 10ad + 10bc+bd\) let's substitute \(d\); \(d=10-b\)
\(100ac+100a-10ab+10bc+10b-b^2\) let's substitute \(c\); \(c =a-1\)
\(100a(a-1) + 100a - 10ab+10b(a-1)+10b-b^2\)
\(100a^2-100a+100a-10ab+10ab-10b+10b-b^2\)
\(100a^2-b^2\)

So we should find what will be hundreds digit of the equation: \(100a^2-b^2\)

1 statement) \(a\) = prime and \(b\) can be any number;
\(a = 3\) and \(b = 1\): \(900-1 = 899\); hundreds digit = \(8\);
\(a = 5\) and \(b = 1\): \(2500-1 = 499\); hundreds digit = \(4\);
Insufficient


2 statement) \(10\) not divisible by \(a\); \(a\) can be \(3\), \(4\), \(6\), \(7\), \(8\), \(9\) and \(b\) can be any number;
\(a = 3\) and \(b = 1\): \(900-1 = 899\); hundreds digit = \(8\);
\(a = 4\) and \(b = 1\): \(1600-1 = 1599\); hundreds digit = \(5\);
Insufficient


1+2) \(a\) can be \(3\) and \(7\) and \(b\) can be any number;
\(a = 3\) so \(100*a^2=900\) and \(b^2\) can be number from \(1\) to \(81\) and their difference will be from \(899\) to \(819\); hundreds digit = \(8\);
\(a = 7\) so \(100*a^2=4900\) and \(b^2\) can be number from \(1\) to \(81\) and their difference will be from \(4899\) to \(4819\); hundreds digit = \(8\);
Sufficient

Answer is C
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Re: The difference between positive two-digit integer A and the smaller tw [#permalink]
Given the larger two-digit number be \(a\) and smaller number be \(b\)
hundreds digit of ab ?
\(a\) - \(b\) = \(2\) * units digit of \(a\), so \(a\) - \(b\) could be \(2\), \(4\), ... \(18\)
i.e. if \(a\) = \(21\), then \(b\) = 21 - (2*1) = \(19\)
\(ab\) = \((21\) * \(19)\) = \((20+1)\) * \((20-1)\); expressed in form of \((x+y)(x-y)\)

\(a\) = \(29\), then \(b\) = 29 - (2*9) = \(11\);
\(ab\) = \((29\) * \(11)\) = \((20+9)\) * \((20-9)\); expressed in form of \((x+y)(x-y)\)


Statement (1)
The tens digit of \(a\) is prime

\(a\) = \(21\)... \(29\), \(ab\) = \((20+1)\)\((20-1)\) ... \((20+9)\)\((20-9)\) = \(399\) ... \(319\)

\(a\) = \(31\)... \(39\), \(ab\) = \((30+1)\)\((30-1)\) ... \((30+9)\)\((30-9)\) = \(899\) ... \(819\)

\(a\) = \(51\)... \(59\), \(ab\) = \((50+1)\)\((50-1)\) ... \((50+9)\)\((50-9)\) = \(2499\) ... \(2419\)

\(a\) = \(71\)... \(79\), \(ab\) = \((70+1)\)\((70-1)\) ... \((70+9)\)\((70-9)\) = \(4899\) ... \(4819\)

Not Sufficient

Statement (2)
\(10\) is not divisible by the tens digit of \(a\).
\(10\) is divisible by either \(2\) or \(5\)

\(a\) = \(31\) ... \(39\), ... \(41\) ... \(49\), ... \(91\) ... \(99\)

\(a\) = \(31\)... \(39\), \(ab\) = \((30+1)\)\((30-1)\) ... \((30+9)\)\((30-9)\) = \(899\) ... \(819\)

\(a\) = \(91\)... \(99\), \(ab\) = \((90+1)\)\((90-1)\) ... \((90+9)\)\((90-9)\) = \(8099\) ... \(8019\)

Not Sufficient

From (1) and (2)
\(a\) = \(31\) ... \(39\) and \(71\) ... \(79\)
in both the cases \(ab\)'s hundreds digit is \(8\)
Sufficient

Answer C
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Re: The difference between positive two-digit integer A and the smaller tw [#permalink]
Celestial09 wrote:
Hi,
No doubt that your solution is correct. But still could you explain me the highlighted portions? Actually could you solve in more simple manner?
Thanks
Celestial


Hello Celestial09

I'll be honest. Firstly I decided this tasks by picking numbers, and it took like 2:30 minutes.
And after this I make this provement just because of curiousity and it took quite much time.
And if I meet such task on exam I think I will choose the first approach, because it is more safely and less time consuming for me ;)

You're highlight a lot of text, can you give more information what exactly you don't understand?
For example you've highlight this part:

Quote:
1 statement) \(a\) = prime and \(b\) can be any number;
\(a = 3\) and \(b = 1\): \(900-1 = 899\); hundreds digit = \(8\);
\(a = 5\) and \(b = 1\): \(2500-1 = 499\); hundreds digit = \(4\);
Insufficient


You don't understand why I pick such numbers for \(a\) and \(b\), or why hundreds equal \(8\) and \(4\) or why I make decision that this insufficient?
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Re: The difference between positive two-digit integer A and the smaller tw [#permalink]
Any tricks here chetan2u
Still not able to get this one apart form testing numbers.
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Re: The difference between positive two-digit integer A and the smaller tw [#permalink]
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stonecold wrote:
Any tricks here chetan2u
Still not able to get this one apart form testing numbers.


Hi,
main statement tells you that the numbers can have various possiblities..
1) if units digit of A is 1, the diff will be 2..
so numbers can be 21 &19....... 31 & 29... and so on till 91 & 89...
2) if units digit of A is 2, the diff will be 4..
so numbers can be 22 &18....... 32 & 28... and so on till 92 & 88..

and so on till units digit of A is 9, the diff will be 18..
so numbers can be 29 &11....... 39 & 21... and so on till 99 & 81...

lets see the statements -
I) tens digit of A can be 2, 3, 5, 7, so tens digit of B will be 1,2,4,6 respectively...
various possiblities...
insuff

II) tens digit of A can be 3, 4, 6, 7, 8 or 9 so tens digit of B will be 2,3,5,6,7 or 8 respectively...
various possiblities...
insuff

Combined
tens digit can be 3 or 7...
now the TRICK that can help-
n^2 will be greater than (n-1)(n+1) which will be > (n-2)(n+2)..
ex 30^2=900>31*29=899.....

lets see the two options now..

1) tens digit of A is 3..
numbers are 31&29.....32&28........till 39&21..
so product will be < 30*30, that is 900 and > 40*20, which is 800..
so it will be between 800 and 900...
in all cases the hundreds digit will be 8...

2) tens digit of A is 7..
numbers are 71&69.....72&68........till 79&61..
so product will be < 70*70, that is 4900 and > 80*60, which is 4800..
so it will be between 4800 and 4900...
in all cases the hundreds digit will be 8...

so in both cases, the hundreds digit is 8..
suff
C

hope it helps
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Re: The difference between positive two-digit integer A and the smaller tw [#permalink]
Expert Reply
chetan2u wrote:
stonecold wrote:
Any tricks here chetan2u
Still not able to get this one apart form testing numbers.


Hi,
main statement tells you that the numbers can have various possiblities..
1) if units digit of A is 1, the diff will be 2..
so numbers can be 21 &19....... 31 & 29... and so on till 91 & 89...
2) if units digit of A is 2, the diff will be 4..
so numbers can be 22 &18....... 32 & 28... and so on till 92 & 88..

and so on till units digit of A is 9, the diff will be 18..
so numbers can be 29 &11....... 39 & 21... and so on till 99 & 81...

lets see the statements -
I) tens digit of A can be 2, 3, 5, 7, so tens digit of B will be 1,2,4,6 respectively...
various possiblities...
insuff

II) tens digit of A can be 3, 4, 6, 7, 8 or 9 so tens digit of B will be 2,3,5,6,7 or 8 respectively...
various possiblities...
insuff

Combined
tens digit can be 3 or 7...
now the TRICK that can help-
n^2 will be greater than (n-1)(n+1) which will be > (n-2)(n+2)..
ex 30^2=900>31*29=899.....

lets see the two options now..

1) tens digit of A is 3..
numbers are 31&29.....32&28........till 39&21..
so product will be < 30*30, that is 900 and > 40*20, which is 800..
so it will be between 800 and 900...
in all cases the hundreds digit will be 8...

2) tens digit of A is 7..
numbers are 71&69.....72&68........till 79&61..
so product will be < 70*70, that is 4900 and > 80*60, which is 4800..
so it will be between 4800 and 4900...
in all cases the hundreds digit will be 8...

so in both cases, the hundreds digit is 8..
suff
C

hope it helps


Do we have any easier way to solve this question?

On the actual test, if this question comes, I am not sure how are we gonna handle it. :(
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Re: The difference between positive two-digit integer A and the smaller tw [#permalink]
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Bunuel 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.


A student asked me to respond to this question, so here it goes...

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

Cheers,
Brent
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Re: The difference between positive two-digit integer A and the smaller tw [#permalink]
Hi,

we are given that:

(10a+b)-(10a'+b')=2b
10(a-a')+b-b'=2b

this leaves open two possibilities:

Either, 10(a-a')=b and b-b'=b, so in total, we would get b+b=2b, or 10(a-a')=2b

Case1:

It follows from b-b'=b that b'=0, and from 10(a-a')=b that a=a'=0, because b is a number in (0-9), and if a!=a', we would get something greater equal 10. From these two, it follows b=b' and in total, we have A=B which doesn't work by the task, as A>B per question

Case2:

We know that min(10a-10a') with a!=a' (we have that case already in Case1) is 10. It follows that 2b=10 and b=5, and a=a'+1. So in total we have the pairs (95,85), (85,75),...(15,5) that would work.

(1) We have (75,65),(55,45),... and so on working, so (1) is not enough (take multiplications / one can see that 75*65 doesnt yield same hundred digit as 25*15 directly)
(2) Again, we have (95,85), (85,75),... and so on left. Taking the multiplication we see that it wouldn't work

(1) and (2): We are left with (75,65) and (35,25) only left. Both have 8 as hundred digit, so C
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Re: The difference between positive two-digit integer A and the smaller tw [#permalink]
I think this question is a perfect example of how the GMAT Exam tries to waste your time. It's not about solving this question. It's about training yourself to spot such questions in the actual exam. Just guess something and move on...
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Re: The difference between positive two-digit integer A and the smaller tw [#permalink]
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Re: The difference between positive two-digit integer A and the smaller tw [#permalink]
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