How many positive integers less than 20 can be expressed as

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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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

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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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?

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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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

I don't understand this one. According to Bunuel here https://gmatclub.com/forum/is-0-zero-to-be-considered-as-a-multiple-of-every-number-104179.html, zero is a multiple of all numbers. Doesn't that mean the answer to this should be 19 (all are multiples except for one).

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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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.

Answer (A)

Positive multiple of 2 = 2,4,6,8,10,12,14,16,18,20
positive multiple of 3 = 3,6,9,12,15,18

So, various sums = 5, 8, 11, 14, 17, 7, 10, 13, 16, 19, 9, 12, 15, 18

This is a total of 14.

I think it can be done very simply with the help of this equation ..
2a+3b .....the highest possible value of a we can plug ( assuming the lowest value of b which is b=1) is '8' likewise, the highest value of 'b' can be 6 ( assuming..... a=1)
8+6 =14

Posted from my mobile device

I think it can be done very simply with the help of this equation ..
2a+3b .....the highest possible value of a we can plug ( assuming the lowest value of b which is b=1) is '8' likewise, the highest value of 'b' can be 6
(assuming..... a=1)
8+6 =14

Posted from my mobile device

MIHI95

Please explain why 6 is the highest value for b since 2(1) + 3(6) = 20 but the question says the number is less than 20.

febinlaw Actually, the 2a + 3b approach doesn't work as presented. First, as you say, b can't be 6. Second, there isn't any clear meaning to adding the two potential values. Just knowing the number of values for each variable doesn't allow us to add them and get a total range of values for the expression. For each value of a, only certain values of b are allowed. For instance, if a=1, b can be anything from 1 to 5, but if a=3, b can only range from 1 to 4. Even then, we can't just count up all the values or we'll end up with more than 19 numbers! We have to take out the overlap. We're much better off working with the actual numbers from 1 to 19.
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Here's a fairly quick solution. Start by listing out the relevant multiples of 3:

3, 6, 9, 12, 15, 18

If we start with 3, we see that we can keep adding multiples of 2 to generate all the odd values above 3: 5, 7, 9, 11, . . .
If we start with 6, we can proceed in the same way to get all the evens above 6: 8, 10, 12 . . .

Since we've covered both the odds and the evens, the only numbers left out are the odds below 5 (1,3) and the evens below 8 (2,4,6). The other 14 numbers all work.
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We need to find 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

Let's try to express each number from 5 to 19 as a positive multiple of 2 and 3
We are considering only between 5 and 19 because
- 5 is the least number which can be formed with positive multiples of 2 and 3
- 19 because we need to find numbers less than 20

Using following format: Number = multiple of 2 + multiple of 3

5 = 2 + 3
6 = NOT POSSIBLE to express
7 = 4 + 3
8 = 2 + 6
9 = 6 + 3
10 = 4 + 6
11 = 8 + 3
12 = 6 + 6
13 = 10 + 3
14 = 8 + 6
15 = 12 + 3
16 = 10 + 6
17 = 14 + 3
18 = 12 + 6
19 = 16 + 3

Total numbers = (19-5 + 1 ) - 1 = 14

So, Answer will be A
Hope it helps!

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