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23 Apr 2015, 03:57
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C
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Difficulty:
95%
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Question Stats:
32% (02:50) correct
68%
(02:58)
wrong
based on 376
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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.
(1) The tens digit of A is prime.
(2) Ten is not divisible by the tens digit of A.
27 Apr 2015, 02:27
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Bunuel
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.
(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.
23 Apr 2015, 11:13
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Bunuel
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.
(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
