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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
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02 Jan 2011, 02:38
gmatpapa wrote: Hi Bunuel,
I'm a little confused. The general formula above, does it give us the number of integers inclusive of the upper and lower limits or non inclusive of the limits?
For example: what is the number of integers between 1 and 5, inclusive? The answer is 5, Which is also substantiated by the formula above. You should read the solution carefully: it doesn't matter whether the the range is inclusive or exclusive, the formula gives you the correct answer in any case. How many multiples of 5 are there between 5 and 35, not inclusive? Last multiple of 5 IN the range is 30 (not 35 but 30 as 35 is not IN the range); First multiple of 5 IN the range is 10 (not 5 but 10 as 5 is not IN the range); \(\frac{3010}{5}+1=5\): 10, 15, 20, 25 and 30. How many multiples of 5 are there between 5 and 35, inclusive? Last multiple of 5 IN the range is 35; First multiple of 5 IN the range is 5; \(\frac{355}{5}+1=7\): 5, 10, 15, 20, 25, 30 and 35.
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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
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02 Jan 2011, 11:17
A quick demonstration of the formula and why it is intuitive: How many numbers are there from 3 to 7 (inclusive)? 3, 4, 5, 6, 7 i.e. 5 numbers When we do (7  3 = 4), we remove 3 from the list. I I I I I I I : 7 sticks [strike] I I I[/strike] I I I I : (7  3 = 4) sticks The third stick has been removed but we need it to be included too. So we add 1 to the result 4 + 1 = 5 sticks These 5 sticks are: I I I I I Now, how many numbers are there from 3 to 7 (exclusive)? 4, 5, 6 i.e. 3 numbers When we do (7  3 = 4), we still have 7 in the list. I I I I I I I : 7 sticks [strike] I I I[/strike] I I I I : (7  3 = 4) sticks The seventh stick is still there which has to be removed. So we subtract 1 from the result 4  1 = 3 sticks These 3 sticks are: I I I or you can think of the exclusive case as (inclusive case  2) You say in inclusive case: (Last term  First term + 1): Both end terms are included here. Then you subtract 2 to remove both the end terms and you get: (Last term  First term + 1)  2 = Last term  First term  1 I have also explained this in the post: http://gmatquant.blogspot.com/2010/11/fourprongs.htmlCheck out the subtraction part.
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Re: M01 Question 18 [#permalink]
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04 Feb 2011, 11:27
Bunuel wrote: jayasimhaperformer wrote: a 7*15=105 b 106/7=15.1 c 127/7=18.1 d 128/7=18.12 aprox e 142/7=20. aprox so a is right & a straight ans as its div by 7*15 Hi, and welcome to Gmat Club. It seems that you misinterpreted the question. The question is "How many of the threedigit numbers are divisible by 7?" It's not necessary the answer itself to be divisible by 7 GENERALLY: \(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \  \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\). For example: how many multiples of 5 are there between 7 and 35, not inclusive? Last multiple of 5 IN the range is 30; First multiple of 5 IN the range is 5; \(\frac{30(5)}{5}+1=8\). OR:How many multiples of 7 are there between 28 and 1, not inclusive? Last multiple of 7 IN the range is 7; First multiple of 7 IN the range is 21; \(\frac{7(21)}{7}+1=3\). Back to the original question:Last 3digit multiple of 7 is 994; First 3digit multiple of 7 is 105; So # of 3digt multiples of 7 is \(\frac{994105}{7}+1=128\). Answer: D. Hope it helps. Hello Bunuel, nice explanation as usual is there any shortcut to find last and first multiples? for example how to find 994 ad 105 of 7? or we have to use the long method of division? thx 4 response



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Re: M01 Question 18 [#permalink]
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04 Feb 2011, 11:46



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Re: M01 Question 18 [#permalink]
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04 Feb 2011, 12:03
Bunuel wrote: tinki wrote: Hello Bunuel, nice explanation as usual is there any shortcut to find last and first multiples? for example how to find 994 ad 105 of 7? or we have to use the long method of division? thx 4 response Well, it depends multiple of which number you want to find and the range you are looking in. Common sense, and divisibility rules should help you in this but sometimes 'trial and error' is also a good method. Check for example divisibility rules here: mathnumbertheory88376.htmli did not know about 7 and 11 . nowhere seen the rule before. GREAT HELP . THANKS A LOT A LOT A LOT FOR SWIFT RESPONSES +kudo from me



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Re: M01 Question 18 [#permalink]
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10 May 2011, 11:11
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gmatprep09 wrote: How many of the threedigit numbers are divisible by 7? (A) 105 (B) 106 (C) 127 (D) 128 (E) 142 Source: GMAT Club Tests  hardest GMAT questions Description of the technique: Q: How many 3 digits numbers are divisible by 7 OR How many multiples of 7 are there between 100 and 999, inclusive? Choose the first number greater than equal to 100 that is divisible by 7. Remainder of 100/7 is 2. 1002+7=105 "105" is the first number greater than equal to 100 that is divisible by 7. Choose the first number less than equal to 999 that is divisible by 7. Remainder of 999/7 is 5. 9995=994 "994" is the first number less than equal to 999 that is divisible by 7. \(n=\frac{994105}{7}+1=127+1=128\) Ans: "D" ************************ This rules hold good for all numbers. Find the numbers divisible by 3 between 1 and 18, inclusive. First Number greater than equal to 1 that is divisible by 3 = 3 First Number less than equal to 18 that is divisible by 3 = 18 \(Number of multiples = \frac{183}{3}+1=5+1=6\) Validate: Count of {3,6,9,12,15,18} = 6 Try it with any number and you should get the count.
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Re: M01 Question 18 [#permalink]
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22 Sep 2011, 06:27
hockeyplaya182 wrote: 1) Take all numbers between 1 and 999 divisible by 7: 999/7=142 2) Subtract the number that are not three digits: 99/7=14
14214=128 Great explanation. Thanks.



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Re: M01 Question 18 [#permalink]
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22 Sep 2011, 10:07
Here is a simple Solution :
The count of numbers<=999 divisible by seven = rounded[999/7] = 142
The number of 2 digit numbers divesible by 7 = rounded[99/7] = 14
The number of 3 digit numbers divisible by 7 = 142  14 = 128



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Re: 3 Digit Numbers Divisible by 7 (and others) [#permalink]
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22 Sep 2011, 10:30
james12345 wrote: I'm kept my lame questions to a minimum. I see why plus 1 work for inclusive. I don't see for exclusive why minus 1 would work. If both of the end number "excluded" are actually multiples of the number in question it changes it. but that should be the case for inclusive as well. in other words, how many multiples of 7 between 8 and 22 exclusive means 9 to 21 which means the answer is 2. 219 is 12 which divided by 7 is 1 so plus 1 = 2. however 7 to 28 exclusive means 827 = 19 which is 2 7s which is actually correct.
It makes me very nervous to apply formulas without being crystal clear as to why they're right. Maybe I've misunderstood part of the premise of the question. The key point is that in a simple subtraction, we are acting inclusive in one side and exclusive in the other side. For example, when we say 106=4, we have not considered 6 itself (exclusive in one side) but we have considered 10 (inclusive in the other side). This is the rule. So, if you figure that out, you see that when GMAT says exclusive, we have to just exclude the side which was not originally excluded. Since one side is not excluded in a normal subtraction, we just subtract 1, not 2. On the other hand, when GMAT says inclusive, we have already included one side, so we just add 1, not 2, to include the other side. Hope it helps
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Re: M01 Question 18 [#permalink]
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24 Sep 2012, 05:18
Ans is 128. 14*7 = 98 next will be a 3 digit factor.. and 994 /7 = 142 next will be a 4 digit.. so 142 14 = 128
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Re: M01 Question 18 [#permalink]
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24 Sep 2012, 05:22
Total 3 digit nos 100 to 999 among them total digits div.by 7 ={(last digit div.by 7 Minus 1st digit div by 7)/7 }+1 {(994105)/7}+1 = 128
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Re: M01 Question 18 [#permalink]
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24 Sep 2012, 08:47
T How do we find out the largest 3 digit number that is divisible by 7 ? Well one way is trial and error, but that is not too efficient. Rather , let's divide 999 by 7 to find the remainder. Subtract the remainder from 999 and voila ! For eg : When 999 is divided by , the remainder is 5. Subtract from 999, the answer is 994. If m/n leaves "r" as the remainder, then mr is divisible by n. What is the logic behind this ? Well when you divided a number by another divisor, the remainder represents the piece that is not divisible by the divisor. Remove the offending remainder from the quotient and the new quotient will be divisible by the divisor
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Re: M01 Question 18 [#permalink]
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27 Feb 2013, 03:54
bb wrote: Is something not clear in the Official Explanation?
Approach One: Divide all of the threedigit numbers \(999  100 + 1 = 900\) (Don't forget to add 1 to get the number of all the 3digit numbers) by 7, which is 128.57, and then round it off to 128.
Approach Two: Find the first and the last multiples of 7 WITHIN the range (which is between 105 and 994). Find the positive difference, divide by 7, and add 1: \(\frac{994105}{7}+1=128\) Hi, I don't understand how and why approach 1 works. Please help explain why it works and provide some examples.
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Re: M01 Question 18 [#permalink]
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23 Sep 2013, 08:34
bb wrote: Is something not clear in the Official Explanation?
Approach One: Divide all of the threedigit numbers \(999  100 + 1 = 900\) (Don't forget to add 1 to get the number of all the 3digit numbers) by 7, which is 128.57, and then round it off to 128.
Approach Two: Find the first and the last multiples of 7 WITHIN the range (which is between 105 and 994). Find the positive difference, divide by 7, and add 1: \(\frac{994105}{7}+1=128\) Be careful using approach one because it can lead to wrong answers. It is best used as an approximation because it will be within 1 of the correct answer. How many twodigit numbers that start with 3 are divisible by 7: 39  30 + 1 = 10. Divide by 7 to get 1.43. Round down to 1 (Correct because we have 35 only.) How many twodigit numbers that start with 4 are divisible by 7: 49  40 + 1 = 10. Divide by 7 to get 1.43. Round down to 1 (Incorrect because we have 42 and 49.)



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Re: M01 Question 18 [#permalink]
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23 Sep 2013, 20:05
How many of the threedigit numbers are divisible by 7?
(A) 105 (B) 106 (C) 127 (D) 128 (E) 142
Firstly,find the first 3 digit number that is divisible by 7 ie 105 and the last 3 digit number that is divisible by 7 is 994.
so, series become like this:
105,112,.....................994.
l = a + (n1) d
994 = 105+(n1)7 994105+7=7n n=896/7=128
so, right answer is
128.




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