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Which of the following are/is prime?

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Which of the following are/is prime? [#permalink]

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New post 21 Feb 2013, 13:48
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Which of the following are/is prime?

I. 143
II. 147
III. 149

(A) II only
(B) III only
(C) I & II
(D) I & III
(E) I, II, & III


In order to handle questions like this without a calculator, it is very helpful to know
(a) divisibility tricks = http://magoosh.com/gmat/2012/gmat-divis ... shortcuts/
(b) factoring tricks = http://magoosh.com/gmat/2012/advanced-n ... -the-gmat/

Mike :-)
[Reveal] Spoiler: OA

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Re: Which of the following are/is prime? [#permalink]

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New post 21 Feb 2013, 15:19
mikemcgarry wrote:
Which of the following are/is prime?

I. 143
II. 147
III. 149

(A) II only
(B) III only
(C) I & II
(D) I & III
(E) I, II, & III


In order to handle questions like this without a calculator, it is very helpful to know
(a) divisibility tricks = http://magoosh.com/gmat/2012/gmat-divis ... shortcuts/
(b) factoring tricks = http://magoosh.com/gmat/2012/advanced-n ... -the-gmat/

Mike :-)


Step 1, look over the three numbers quickly, realize 147 is divisible by 3. We know II is not prime.
Step 2, look at the answer choices. Realize the answer is (B) or (D). Therefore III must be prime.
Step 3, Verify if I is prime. Build a division table with other prime numbers and get your answer!
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Re: Which of the following are/is prime? [#permalink]

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New post 16 Mar 2013, 11:18
An interesting way to solve these kind of problems is to eliminate the obvious answers
1) 143 = 13*11 so we know it is not prime quickly eliminate all options having 1 as a choice
which leaves us with 2 ) and 3) as possible answers
From here on we find 147 = 3* 49 so we eliminate 147 which leaves us with 149 as the only possible solution

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Re: Which of the following are/is prime? [#permalink]

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New post 16 Mar 2013, 15:13
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Which of the following are/is prime?

I. 143:
- sum of digits multiple of 3? NO
- divisible by 7? NO
- divisible by 11? YES, 13x11=143 --> Not prime
II. 147
- sum of digits multiple of 3? YES, 1+4+7=12 --> Not prime

III. 149
- sum of digits multiple of 3? NO
- divisible by 7? NO
- divisible by 11? NO
- divisible by 13? NO

B is the only possible solution

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Re: Which of the following are/is prime? [#permalink]

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Re: Which of the following are/is prime? [#permalink]

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New post 22 Mar 2017, 07:01
mikemcgarry wrote:
Which of the following are/is prime?

I. 143
II. 147
III. 149

(A) II only
(B) III only
(C) I & II
(D) I & III
(E) I, II, & III


I. 143 = 13*11
II. 147 = 21*7

III. 149

Among the given options 149 is a prime number...

Thus, answer must be (III) 149
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Which of the following are/is prime? [#permalink]

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New post 25 Mar 2017, 11:27
Abhishek009 wrote:

I. 143 = 13*11
II. 147 = 21*7




How did you find this out?

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Re: Which of the following are/is prime? [#permalink]

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New post 26 Mar 2017, 00:22
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Which of the following are/is prime?

I. 143
II. 147
III. 149

(A) II only
(B) III only
(C) I & II
(D) I & III
(E) I, II, & III

143= 11*13
147=3*49
149= prime

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Re: Which of the following are/is prime? [#permalink]

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New post 26 Mar 2017, 10:04
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joondez wrote:
Abhishek009 wrote:
I. 143 = 13*11
II. 147 = 21*7


How did you find this out?

Dear joondez,

I'm happy to help. :-)

My friend, one of the most important skills on the GMAT Quant is developing number sense.

For example, when I look at 147, I notice that 147 = 140 + 7. Well, 140 is divisible by 7--it's equal to 20*7. Of course 7 is divisible by 7. Therefore:
147 = 140 + 7 = 20*7 + 7 = (20 + 1)*7 = 21*7
Part of number sense is being able to "chop up" numbers to see what is divisible by what. Here's a video that explains more.
Multiples

For 143, we use another trick. You may be familiar with the algebraic formula know as "Difference of Two Squares."
\(P^2 - Q^2 = (P + Q)(P - Q)\)
Well, we can use this for algebra, but we also can use this for numbers. See:
Advanced (Non-Calculator!) Factoring on the GMAT

We know that \(12^2 = 144\), and \(143 = 144 - 1\). Therefore
\(143 = 144 - 1 = 12^2 - 1^2 = (12 + 1)(12 - 1) = 13*11\)

BTW, another trick we can use: suppose we have a three-digit number abc, where those are the three digits. If it's true that b = a + c, then it has to be true that the number is divisible by 11. For example, consider the number 473. Any number of this kind we can write in the form (10*N) + N for some positive integer N. For this number,
473 = 430 + 43 = 43*10 + 43 = 43(10 + 1) = 43*11

Don't just memorize these patterns: really make sure you understand them.

Does all this make sense?
Mike :-)
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Re: Which of the following are/is prime? [#permalink]

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New post 27 Mar 2017, 08:43
mikemcgarry wrote:
Which of the following are/is prime?

I. 143
II. 147
III. 149

(A) II only
(B) III only
(C) I & II
(D) I & III
(E) I, II, & III


In order to handle questions like this without a calculator, it is very helpful to know
(a) divisibility tricks = http://magoosh.com/gmat/2012/gmat-divis ... shortcuts/
(b) factoring tricks = http://magoosh.com/gmat/2012/advanced-n ... -the-gmat/

Mike :-)



B.

III only.

Another way is to remember all higher prime numbers will by 6X-1 or 6X+1. Since 6 goes into 144 and 150 the potential primes around that area are 143,145,149 and 151.

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Re: Which of the following are/is prime? [#permalink]

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New post 28 Mar 2017, 16:22
mikemcgarry wrote:
joondez wrote:
Abhishek009 wrote:
I. 143 = 13*11
II. 147 = 21*7


How did you find this out?

Dear joondez,

I'm happy to help. :-)

My friend, one of the most important skills on the GMAT Quant is developing number sense.

For example, when I look at 147, I notice that 147 = 140 + 7. Well, 140 is divisible by 7--it's equal to 20*7. Of course 7 is divisible by 7. Therefore:
147 = 140 + 7 = 20*7 + 7 = (20 + 1)*7 = 21*7
Part of number sense is being able to "chop up" numbers to see what is divisible by what. Here's a video that explains more.
Multiples

For 143, we use another trick. You may be familiar with the algebraic formula know as "Difference of Two Squares."
\(P^2 - Q^2 = (P + Q)(P - Q)\)
Well, we can use this for algebra, but we also can use this for numbers. See:
Advanced (Non-Calculator!) Factoring on the GMAT

We know that \(12^2 = 144\), and \(143 = 144 - 1\). Therefore
\(143 = 144 - 1 = 12^2 - 1^2 = (12 + 1)(12 - 1) = 13*11\)

BTW, another trick we can use: suppose we have a three-digit number abc, where those are the three digits. If it's true that b = a + c, then it has to be true that the number is divisible by 11. For example, consider the number 473. Any number of this kind we can write in the form (10*N) + N for some positive integer N. For this number,
473 = 430 + 43 = 43*10 + 43 = 43(10 + 1) = 43*11

Don't just memorize these patterns: really make sure you understand them.

Does all this make sense?
Mike :-)


Your explanation was extremely clear and helped me a lot. Thank you! I have a lot to improve in my number sense

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Re: Which of the following are/is prime? [#permalink]

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New post 02 Apr 2017, 23:30
(1) 143
Not divisible by 3 as the sum of the digits (1+4+3=8) which is not divisible by 3.
It ends in 3, thus not divisible by 2. (not even)
It ends in 3, thus not divisible by 5. (the number should end in 0 or 5)
Not divisible by 7 either as leaves a remainder of 3.
Divisible by 11: 1+3-4=0 which is divisible by 11.
Thus not prime.

(2) 147
Sum of digits is 12 which is divisible by 3.
Thus not prime.

(3) 149
Its a prime number as it is not divisible by any of the prime numbers less than sq. root of 149.

How to know by what all prime numbers we should check its divisibility?

e.g. 149 lies between square of 144 (12^2) and 169 (13^2)
Square root of 149: 12<sq.root 149<13
Therefore we have to check its divisibility by all the prime numbers less than the lower limit i.e. 12.

Prime numbers less than 12 are 2,3,5,7, and 11
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Re: Which of the following are/is prime?   [#permalink] 02 Apr 2017, 23:30
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