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What should be the approach to do the below question?

How many positive integers less than 20 can be expressed as the sum of a positive multiple of 2 and a positive multiple of 3? (A) 14 (B) 13 (C) 12 (D) 11 (E) 10

We are looking at the set {1,2,3,4,5,...,19} So all numbers of the form 2+3k (where k>=1) can be considered {5,8,11,14,17} - set 1 Similarly 4+3k (k>=1) gets us {7,10,13,16,19} - set 2 6+3k (k>=1) gets us {9,12,15,18} - set 3 8+3k (k>=1) : already in set 1 10+3k (k>=1) : already in set 2 12+3k (k>=1) : already in set 3 14+3k (k>=1) : already in set 1 16+3k (k>=1) : already in set 2 18+3k (k>=1) : already in set 3

So the full list is {5,7,8,9,10,11,12,13,14,15,16,17,18,19} which is 14 numbers
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What should be the approach to do the below question?

How many positive integers less than 20 can be expressed as the sum of a positive multiple of 2 and a positive multiple of 3? (A) 14 (B) 13 (C) 12 (D) 11 (E) 10

We are looking at the set {1,2,3,4,5,...,19} So all numbers of the form 2+3k (where k>=1) can be considered {5,8,11,14,17} - set 1 Similarly 4+3k (k>=1) gets us {7,10,13,16,19} - set 2 6+3k (k>=1) gets us {9,12,15,18} - set 3 8+3k (k>=1) : already in set 1 10+3k (k>=1) : already in set 2 12+3k (k>=1) : already in set 3 14+3k (k>=1) : already in set 1 16+3k (k>=1) : already in set 2 18+3k (k>=1) : already in set 3

So the full list is {5,7,8,9,10,11,12,13,14,15,16,17,18,19} which is 14 numbers

Thanks for the Questions & Answer.

The mistake I did was that I constructed the equation as Number = 2n+3n [i.e. 5,10,15] so my answer was "3" which was not there in the options. So I realized I m doing st wrong but I could not figure out until I saw the solution above.

The only problem was for me, above solution will take >2 min. Then I realized we can stop at 6+3k , because the # of numbers are already 14 ; the greatest answer option. Is there any other clue to look for?

What should be the approach to do the below question?

How many positive integers less than 20 can be expressed as the sum of a positive multiple of 2 and a positive multiple of 3? (A) 14 (B) 13 (C) 12 (D) 11 (E) 10

We are looking at the set {1,2,3,4,5,...,19} So all numbers of the form 2+3k (where k>=1) can be considered {5,8,11,14,17} - set 1 Similarly 4+3k (k>=1) gets us {7,10,13,16,19} - set 2 6+3k (k>=1) gets us {9,12,15,18} - set 3 8+3k (k>=1) : already in set 1 10+3k (k>=1) : already in set 2 12+3k (k>=1) : already in set 3 14+3k (k>=1) : already in set 1 16+3k (k>=1) : already in set 2 18+3k (k>=1) : already in set 3

So the full list is {5,7,8,9,10,11,12,13,14,15,16,17,18,19} which is 14 numbers

although this solution is very helpful, but still I find the question a bit strange, without the solution it is almost impossible to understand what the question is asking, I tried 2x + 3 and 2+3x as the number of elements, still no luck .

Can anybody make another attempt at this, thank you
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Re: How many positive integers less than 20 can be expressed as [#permalink]

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18 Aug 2012, 08:40

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Mas[m]terGMAT12 wrote:

How many positive integers less than 20 can be expressed as the sum of a positive multiple of 2 and a positive multiple of 3? (A) 14 (B) 13 (C) 12 (D) 11 (E) 10

The numbers must be of the form \(2a+3b,\) where \(a\) and \(b\) are positive integers. The smallest number is \(5 = 2*1 + 3*1.\) Starting with \(5\), we can get all the other numbers by adding either \(2\) or \(3\) to the already existing numbers on our list. Adding either \(2\) or \(3\) to \(2a+3b\) will give another number of the same form. So, after \(5\), we get \(5+2=7, \,5+3=8, \,7+2=9, \,8+2=10,...\) We will get all the numbers up to \(19\) inclusive, except \(1,2,3,4,\)and \(6,\) because once we have \(7\) and \(8,\) by adding \(2\) all the time we can get any odd or even number. We get a total of \(19-5=14\) numbers.

Answer A

Note: In fact, any integer \(n\) greater than 6 has at least one representation of the form \(2a+3b.\) If \(n\) is odd, then \(n-3>2\), so we can take \(b=1\) and \(a=\frac{n-3}{2}.\) If \(n\) is even, being greater than \(6\), \(n-6\) is a positive multiple of \(2\). Now we can take \(b=2\) and \(a=\frac{n-6}{2}.\) If the question would have been the same but for integers less than \(100\), then the answer would be quite easy, \(99 - 5 = 94.\)
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Re: How many positive integers less than 20 can be expressed as [#permalink]

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01 Feb 2014, 21:13

we are looking for all positive numbers less than 20. That means we have 19 numbers. Now, 1,2,3 and 4 can never be expressed as sum of 2 and 3. So we are left with 15 numbers.

By this time i already had spent around 3 min and had to take a shot, so i guessed it to 14.

Btw, i never came across an explanation where people would just guess the answers. I read that guessing is one of the skills that we need to master. Anymore inputs to guessing will be welcomed
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“Confidence comes not from always being right but from not fearing to be wrong.”

Re: How many positive integers less than 20 can be expressed as [#permalink]

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06 Aug 2014, 21:35

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MasterGMAT12 wrote:

How many positive integers less than 20 can be expressed as the sum of a positive multiple of 2 and a positive multiple of 3?

(A) 14 (B) 13 (C) 12 (D) 11 (E) 10

The number = 2a + 3b < 20

When a = 1, b = 1, 2, 3, 4, 5 -> 2a = 2; 3b = 3, 6, 9, 12, 15 -> the number = 5, 8, 11, 14, 17 --> 5 numbers when a =2, b = 1,2,3,4,5 -> ....--> 5 numbers when a =3, b = 1,2,3,4 --> ....--> 4 numbers

Total number is already 14. Look at the answer there is no number greater than 14 --> we dont need to try any more Answer must be A
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Explanation: Positive multiples of 2 are even numbers; the relevant multiples of 3 are 3, 6, 9, 12, 15, and 18. No number smaller than 5 can be expressed as the sum of one and the other, as the smallest options are 2 and 3. Rather than going through every number between 5 and 19, look for patterns. There are 8 odd numbers between 5 and 19, inclusive, and each of them can be expressed as the sum of an even number and 3, so those 8 must be counted. The smallest even number that could be counted is 8 (2 + 6), and by the same reasoning, every even number between 8 and 18,inclusive, must be counted, adding 6 more to our total. That’s 6 + 8 = 14 total numbers, choice (A).

Explanation: Positive multiples of 2 are even numbers; the relevant multiples of 3 are 3, 6, 9, 12, 15, and 18. No number smaller than 5 can be expressed as the sum of one and the other, as the smallest options are 2 and 3. Rather than going through every number between 5 and 19, look for patterns. There are 8 odd numbers between 5 and 19, inclusive, and each of them can be expressed as the sum of an even number and 3, so those 8 must be counted. The smallest even number that could be counted is 8 (2 + 6), and by the same reasoning, every even number between 8 and 18,inclusive, must be counted, adding 6 more to our total. That’s 6 + 8 = 14 total numbers, choice (A).

gmatser1, note that the problem specifies that we are dealing with positive multiples, so we don't need to consider 0. Otherwise, you would have a point. You'll find that little specifications like that (positive, not zero, integer, odd, even, etc.) are very important to take note of!
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Dmitry Farber | Manhattan GMAT Instructor | New York

How many positive integers less than 20 can be expressed as the sum of a positive multiple of 2 and a positive multiple of 3?

(A) 14 (B) 13 (C) 12 (D) 11 (E) 10

Responding to a pm:

I would do this question by enumerating and using pattern recognition.

Note that we need the number to be the sum of a positive multiple of 2 and a positive multiple of 3. The first such number will be 5 (which is 2 + 3). Now, every time we add one or more 2s and/or one or more 3s to 5, we will will one of our desired numbers.

\(5 +2 = 7\)

\(5+3 = 8\)

\(5 + 2*2 = 5 + 4 = 9\)

\(5 + 2 + 3 = 5 + 5 = 10\)

5 + 4 + 2 = 11

5 + 4 + 3 = 12

... Note that you will get all other numbers because the new base number is 5 + 4 = 9 now. You can add 2, 3, 4, 5 and 6. Thereafter, we can consider the new base to be 14 and then again add 2, 3, 4, 5, and 6 and so on... So all numbers including and after 7 can be written in the form 2a + 3b.

In the first 19 positive integers, there are only 5 numbers (1, 2, 3, 4, 6) which you cannot express as 2a + 3b such that a and b are positive integers. SO 14 numbers can be written as a sum of a positive multiple of 2 and a positive multiple of 3.

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