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The difference between positive two-digit integer A and the smaller tw

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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.


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[Reveal] Spoiler: OA

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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.


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]

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New post 24 Apr 2015, 07:05
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]

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New post 24 Apr 2015, 07:20
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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The difference between positive two-digit integer A and the smaller tw [#permalink]

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New post 24 Apr 2015, 08:13
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]

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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.


Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Write A as 10x + y, where x is the tens digit and y is the units digit. We know that B is a smaller two-digit integer, and that B is smaller by a particular amount: twice A’s units digit, or 2y.

So B is 10x + y – 2y = 10x – y.

We can think about a few constraints. Since B is “smaller,” we know that y cannot be 0 (otherwise, B wouldn’t be smaller than A). Likewise, we know that x is not 0 (otherwise, A would not be a two-digit integer). Finally, we know that x is not 1. Otherwise, B would not be a two-digit integer either, since B = 10x – y. If x were equal to 1, then B would equal 10 – a positive digit, which would be a single-digit number.

We want to know the hundreds digit of the product of A and B. Write this product in terms of x and y:

AB = (10x + y)(10x – y) = 100x^2 – y^2.

Notice that we get the difference of squares. Also, since y is a single positive digit, the greatest y^2 can be is 81, while the smallest is 1. Meanwhile, x^2 is multiplied by 100. For instance, if x = 4, then x^2 = 16, and the first term above is 1,600. Then we subtract a number between 1 and 81 (inclusive), so we get 1,599 down to 1,519. In either case, we have a specific hundreds digit (5) that doesn’t depend on y, the units digit, at all.

In fact, AB’s hundreds digit is completely dictated by the units digit of x^2.

So the question can be rephrased: what is the units digit of x^2, where x is the tens digit of A?

Statement (1): NOT SUFFICIENT. We know that x could be 2, 3, 5, or 7. These are the only prime digits. Squaring those digits, we get 4, 9, 25, and 49. The units digits are different, so this statement is not sufficient.

Statement (2): NOT SUFFICIENT. We know that x could be 3, 4, 6, 7, 8, or 9 (since 10 is divisible by 2 and 5). Squaring a few of these digits, we get 9, 16 – stop. The units digits are again different, so the statement is not sufficient.

Statements (1) and (2) together: SUFFICIENT. Putting the statements together, we know that x could be 3 or 7. The squares are 9 and 49. Since the units digits are the same, we have sufficiency.

The correct answer is C.
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The difference between positive two-digit integer A and the smaller tw [#permalink]

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New post 21 Aug 2016, 11:24
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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New post 27 Aug 2016, 09:02
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]

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New post 27 Aug 2016, 10:35
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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Re: The difference between positive two-digit integer A and the smaller tw   [#permalink] 16 Sep 2017, 05:01
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