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Integer Properties - DS
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dayanandkamble
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davidschneider
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rajeshsinghgmat
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It would be helpful if you guys typed up the question correctly. I read the question as 8* 10^(k+j). That really confused me as there could only ever be one answer if both are positive integers. So would be helpful if you guys clearly stated the question on the forum.
Why is everyone complicating this question with rules of divisibility? Why are you even bringing in the point that the you can find out if a number is divisible by 9 if the digit sum adds up to a certain number? The question just asks what is the remainder.
The simple answer is that no matter what the value of k is, it will always be a multiple of 80. Since the remainder is the same whether you do 80/9 or 800/9 or 8000/9, the only difference to the remainder will depend on what the value of J is. Thats all you need to know to answer this question correctly.
Who cares if 800+j is divisible by 9?
There is no need to bring in 'trivia' about what numbers are divisible by 9.
The simple answer is that no matter what the value of k is, it will always be a multiple of 80. Since the remainder is the same whether you do 80/9 or 800/9 or 8000/9, the only difference to the remainder will depend on what the value of J is. Thats all you need to know to answer this question correctly.
Who cares if 800+j is divisible by 9?
There is no need to bring in 'trivia' about what numbers are divisible by 9.
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Brent@GMATPrepNow
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You're right when you say that "the remainder is the same whether you do 80/9 or 800/9 or 8000/9."harpott wrote:Why is everyone complicating this question with rules of divisibility? Why are you even bringing in the point that the you can find out if a number is divisible by 9 if the digit sum adds up to a certain number? The question just asks what is the remainder.
The simple answer is that no matter what the value of k is, it will always be a multiple of 80. Since the remainder is the same whether you do 80/9 or 800/9 or 8000/9, the only difference to the remainder will depend on what the value of J is. Thats all you need to know to answer this question correctly.
Who cares if 800+j is divisible by 9?
There is no need to bring in 'trivia' about what numbers are divisible by 9.
Yes, it's true that 80 divided by 9 leaves us with remainder 8, 800 divided by 9 leaves us with remainder 8, 8000 divided by 9 leaves us with remainder 8, and so on. But how do you know that this is true for ALL possible values? Since the solution depends on this fact, we need some way to verify/prove that your statement is true, otherwise we're just waving our arms.
One way to show that 80, 800, 8000, 80000, etc will leave the same remainder when divided by 9 is to use the fact that numbers divisible by 9 are such that the sum of their digits is divisible by 9. This is why this divisibility rule is mentioned.
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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Mathsbuddy
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In the rules of mathematics, the order of operations require that orders/powers/indices/roots should be calculated BEFORE multiplying, so as there were no brackets (around k+j) the question is sufficient. I often find the 'clarity' in many GMat questions quite subtle, but understanding the rules helps.kiwiana wrote:It would be helpful if you guys typed up the question correctly. I read the question as 8* 10^(k+j). That really confused me as there could only ever be one answer if both are positive integers. So would be helpful if you guys clearly stated the question on the forum.
Here is the "BIDMAS" list that states the order of operations:
Brackets
Indices
Divide
Multiply
Add
Subtract
I hope thisis useful.
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Mathsbuddy
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tsrajkumar
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One way to split (8)* ((10^k)/9)+ j. We know any 10^k divided by 9 will have a remainder of 1. Hence the remainder of the expression is 8*1 =8. This principle can be applied for other values of 8 (1 through 8 but not for 9). So all we need is the value of j to determine the remainder and hence B is sufficient.
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nadiva171987
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Is the rule for adding the sum of the digits and the sum being able to be divided by a specific number evenly only apply to the number 9? Or could this be for all numbers? I'm assuming only 9 because in the case of 11 the sum of the digits is 2 but 11 is not divisible by 2 evenly.
