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The sum of prime numbers that are greater than 60 but less

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The sum of prime numbers that are greater than 60 but less [#permalink]

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New post 23 Jul 2012, 03:39
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Re: The sum of prime numbers that are greater than 60 but less [#permalink]

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Re: The sum of prime numbers that are greater than 60 but less [#permalink]

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New post 23 Jul 2012, 09:55
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B.) 128

I did this two ways.

The first way used a combination of pure processing power and logic. You know the even numbers - 62, 64, 66, and 68 will never be prime. 65 will always be divisible by 5. That leaves 61, 63, 67, and 69. 63 and 69 are divisible by 9 and 3 respectively (just plug and chug). That leaves you with 2 primes - 61 and 67 - the sum of which is 128.

The second way - applying divisibility rules - seems more elegant and more efficient in the long run:

For 2: All even numbers. This eliminates 62, 64, 66, and 68.
For 3: Add up all the digits of the number. If the sum is divisible by 3, the number is divisible by 3. This eliminates both 63 and 69.
For 4: If the last 2 digits of the number are divisible by 4, the entire number is divisible by 4. This does not apply to any of our remaining numbers - 61, 65, and 67.
For 5: All numbers ending in 5 or 0. This eliminates 65, leaving only 61 and 67 in the running for primes.
For 6: If the number is divisible by BOTH 2 & 3 (see first two rules), then the number is also divisible by 6. This does not apply to 61 and 67 since neither is divisible by 2 or 3.
For 7: No concise rule that I know of. I recommend long-division or mental math. Does anyone know a quick way to check for divisibility by7? Regardless, you can run through the multiples of 7 and realize none fall on 61 or 67 so this divisor doesn't check out.
For 8: If the last 3 digits are divisible by 8, then the whole number is divisible by 8. I didn't stick to this rule when using the divisibility rules approach. You know 2 is a prime factor of 8 meaning any number that has 8 as a multiple must be even. Since neither 61 nor 67 is even, 8 cannot be a divisor.
For 9: Add up all the digits and see if the resulting sum is divisible by 9. This does not work when applied to 61 or 67 and is redundant for 63 (we knocked 63 out in the "For 3" rule).

Therefore the only 2 primes in the set are 61 and 67.

Does anyone know a good way to check for divisibility by 7? Thanks.
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Re: The sum of prime numbers that are greater than 60 but less [#permalink]

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New post 27 Jul 2012, 06:06
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The sum of prime numbers that are greater than 60 but less than 70 is

(A) 67
(B) 128
(C) 191
(D) 197
(E) 260

The only prime numbers between 60 and 70 are 61 and 67 --> 61+67=128.

Answer: B.

Club909 wrote:
Does anyone know a good way to check for divisibility by 7? Thanks.


Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.

For example, let's check whether 1,519 is divisible by 7: 151-2*9=133. Since 133 is divisible by 7 (133=7*19), then so is 1,519.

For more check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.
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Re: The sum of prime numbers that are greater than 60 but less [#permalink]

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Re: The sum of prime numbers that are greater than 60 but less [#permalink]

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New post 09 Sep 2014, 10:02
Bunuel wrote:
The sum of prime numbers that are greater than 60 but less than 70 is

(A) 67
(B) 128
(C) 191
(D) 197
(E) 260

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2 is the only even prime number. so 61,63,67,69 -- 63 & 69 both are divisible by 3 hence 61 & 67 =128..
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Re: The sum of prime numbers that are greater than 60 but less [#permalink]

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New post 09 Sep 2014, 18:30
Just 2 prime numbers 61 & 67 in the given range

61+67 = 128

Answer = B

OA has listed 67; easy to find other :)
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Re: The sum of prime numbers that are greater than 60 but less [#permalink]

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Re: The sum of prime numbers that are greater than 60 but less [#permalink]

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New post 19 May 2016, 09:23
Bunuel wrote:
The sum of prime numbers that are greater than 60 but less than 70 is

(A) 67
(B) 128
(C) 191
(D) 197
(E) 260


A prime number is a number that has only two factors: 1 and itself. Therefore, a prime number is divisible by two numbers only.

Let's list the numbers from 61 to 69.

61, 62, 63, 64, 65, 66, 67, 68, 69

Immediately we can eliminate the EVEN NUMBERS because they are divisible by 2 and thus are not prime.

We are now left with: 61, 63, 65, 67, 69

We can next eliminate 65 because 65 is a multiple of 5.

We are now left with 61, 63, 67, 69.

To eliminate any remaining values, we would look at those that are multiples of 3. If you don’t know an easy way to do this, just start with a number that is an obvious multiple of 3, such as 60, and then keep adding 3.

We see that 60, 63, 66, 69 are all multiples of 3 and therefore are not prime.

Thus, we can eliminate 63 and 69 from the list because they are not prime.

Finally, we are left with 61 and 67, and we must determine whether they are divisible by 7. They are not, and therefore they must be both prime. Thus, the sum of 61 and 67 is 128.

Answer B.

Here is a useful rule: If a two-digit number is a prime, it can’t be divisible by any of the single-digit primes. That is, it can’t be divisible by 2, 3, 5 and 7. In other words, if you have a two-digit number that is not divisible by 2, 3, 5 and 7, it must be a prime. If you have trouble seeing that 61 and 67 are prime, I would suggest that you review your multiplication tables. Doing so will allow you to quickly see that 61 and 67 are not multiples of a given single-digit number, such as 7.
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Re: The sum of prime numbers that are greater than 60 but less [#permalink]

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New post 06 Jul 2016, 02:03
[quote="Bunuel"]The sum of prime numbers that are greater than 60 but less than 70 is

(A) 67
(B) 128
(C) 191
(D) 197
(E) 260



you may notice 61 and 67 are prime numbers here and the sum of them is 128
Re: The sum of prime numbers that are greater than 60 but less   [#permalink] 06 Jul 2016, 02:03
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