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Post subject: GMAT Number Theory Posted: Mon Mar 07, 2011 5:13 pm 

Joined: Sun May 30, 2010 3:15 am Posts: 424

If a, b, c, d, e, f, g, and h represent any of the digits from 1 to 9, inclusive and the 5digit number abcd9 is divisible by the 3digit number ef7, then the quotient could be
A. gh3 B. gh4 C. gh5 D. gh6 E. gh7
(E) (To understand units and tens, in 75 7 is tens and 5 is units.) This question is testing your knowledge of the fact that when you multiply multidigit integers, the units digit of the result is the same as the units digit on the product of the units digits of the original integers. For example, the units digit of the result of 1,237 × 653 is 1, because the units digit on the result of 7 × 3 is 1.
So if abcd9 is divisible by ef7, we know that: (abcd9) / (ef7) = some integer.
Rearranging this equation, we have: ef7 × (some integer) = abcd9.
Let’s think about possibilities for the units digit of this unknown integer, given the rule above: 7 × 1 = 7 7 × 2 = 14 7 × 3 = 21 … 7 × 7 = 49.
We can see that if the unknown integer has a units digit of 7, then abcd9 may be divisible by ef7. gh7 is the only answer with a units digit of 7.
The correct answer is choice (E).  According to the solution: "Because the 5digit number's units digit is 9, and the divisor's units digit is 7, the quotient must have a units digit of 7." I did not understand this. Could you make it more clear to understand?


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Gennadiy

Post subject: Re: math (t. 4, qt. 8): number theory Posted: Mon Mar 07, 2011 5:45 pm 

Joined: Sun May 30, 2010 2:23 am Posts: 498

The're 3 key moments I'd like to explain to you.
1. Any multidigit integer, e.g. 3digit integer, can be presented as: abc = a00 + b0 + c, which is the same as abc = a × 100 + b × 10 + c.
2. If we multiply two multidigit integers, then the units digit of the product is solemnly defined by units digits of the integers.  Why is it so? All letters represent digits. a..bc × d..ef = (a..b0 + c)(d..e0 + f) = a..b0 × d..e0 + a..b0 × f + c × d..e0 + c × f = (a..b0 × d..e + a..b × f + c × d..e) × 10 + c × f. The first term (...) × 10 doesn't affect the units digit. The second term, c × f, does. So the units digit of the product is the same as the units digit of c × f. 
3. Now, let's review the given question.
If we denote the units digit of the quotient by x, then We have: abcd9 = ef7 × ghx As we already know from the second statement, the units digit of 7 × x must be 9.  Why is it so, again? ghx × ef7 = (gh × 10 + x)(ef × 10 + 7) = (gh × ef × 10 + gh × 7 + x × ef) × 10 + x × 7  Multiplication table (as also the possible answer choices) gives us only one possible value of x, 7: 7 × 7 = 49.
(7 × 3 = 21, 7 × 4 = 28, 7 × 5 = 35, 7 × 6 = 42)


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