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arith issue with 3 digit

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by selango » Wed Jun 16, 2010 8:16 am
Let the 3 digit number be x

x=7q+5

Plugging value of Q, we can find least and high value of x

q=14,x=103

q=142,x=999

So we need to find number of values bw 14 and 142,Its 129

Hence 129

what is OA
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by kvcpk » Wed Jun 16, 2010 8:17 am
francoisph wrote:How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?

how to find quickly? that the smallest div by 7 and remainder 5 is 103
largest is 999
thks
Just multiply 7 by some number and add 5 to it.. you know rougly what to multiply it with if you need to get a 3 digit number. I dont think thr is shortcut for this.
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by albatross86 » Wed Jun 16, 2010 8:19 am
These will be 2 less than the multiples of 7. eg. 103 = 105 -2. 999 = 1001 - 2

Thus it is an arithmetic progression with first term 103, last term 999 and difference of 7 between consecutive terms.

an = a1 + (n-1)d
999 = 103 + (n-1)*7
=> n = 129.... Ans
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by amising6 » Wed Jun 16, 2010 8:25 am
now c we are looking for the number of type 7k+5 where k can take integer value

now simply divide 100 (smallest 3 digit number) by 7 you get 14 forget remainder
now apply this k=14 in 7k+5 and u will have 103

similiarly largest 3 digit number 999/7=142 now put k=142 in 7k+5 you will get 999
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by Stuart@KaplanGMAT » Wed Jun 16, 2010 12:25 pm
francoisph wrote:How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?

how to find quickly? that the smallest div by 7 and remainder 5 is 103
largest is 999



thks
The 3 digit integers range from 100 to 999. There are 900 of them.

900/7 = 128rem4. So, the answer must be either 128 or 129, depending on where 100 and 999 fall in the cycle of multiples of 7.

Here's where answer choices would be very useful (please always post the choices and the source of your question): if only one of those two appeared among the choices, we'd have our sure point and be finished.

Assuming that they both appear (which normally happens on the GMAT), we'd look at the biggest number under 100 divisible by 7: 98. So, the smallest three-digit number that will give a remainder of 5 is 103.

Since 103 is the 4th number in our sequence (starting at 100), and since we have a remainder of 4, we now know that "remainder 4" occurs 1 extra time, leading to our final answer:

128 + 1 = 129
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by sumanr84 » Wed Jun 16, 2010 9:10 pm
Stuart Kovinsky wrote:
The 3 digit integers range from 100 to 999. There are 900 of them.

900/7 = 128rem4. So, the answer must be either 128 or 129, depending on where 100 and 999 fall in the cycle of multiples of 7.
Wow !! I like this approach..
I am on a break !!
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