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How many positive integers less than 2 x 10^4 are there in which each [#permalink]
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31 Jul 2017, 21:51
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Re: How many positive integers less than 2 x 10^4 are there in which each [#permalink]
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31 Jul 2017, 22:16
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The positive integers less than 20000, can be 1 digit, 2 digit, 3 digit, 4 digit and 5 digit.
Now each digit in the above integers must be a prime number.
1 Digit : Possible numbers  2, 3, 5, 7 : Total = 4
2 Digits: Both the hundreds and Units digit can take any value from 2, 3, 5, 7: So Total : 4 * 4 = 16
3 Digits: The Thousands, hundreds and Units digit can take any value from 2, 3, 5, 7: So Total : 4 * 4 * 4 = 64
4 Digits: The Ten Thousands, Thousands, hundreds and Units digit can take any value from 2, 3, 5, 7: So Total : 4 * 4 * 4 * 4 = 256
5 Digits: As the least possible 5 digit number has to to start with 2, which will make the 5 digit number greater than 20000, hence there is 0, 5 digit numbers less than 20000, with each digit as a prime number. Total = 0
Therefore adding all : 4 + 16 + 64 + 256 = 340
Answer is C



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Re: How many positive integers less than 2 x 10^4 are there in which each [#permalink]
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31 Jul 2017, 22:42
Bunuel wrote: How many positive integers less than 2*10^4 are there in which each digit is a prime number?
(A) 256 (B) 326 (C) 340 (D) 625 (E) 775 20000 (5 digits) No number can be formed of 5 digits as 1 is not a prime number and at First place only Prime number 2 can come single digit prime numbers are 2,3,5,7 4 digit numbers with prime numbers will be So 4^4 = 256 3 digit numbers with prime numbers will be 4^3 = 64 2 digit numbers with prime numbers will be 4^2 = 16 1 digit numbers with prime numbers will be 4 Total = 256 + 64 + 16 + 4 = 340 C
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How many positive integers less than 2 x 10^4 are there in which each [#permalink]
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01 Aug 2017, 20:39
Numbers less than 20,000 will have 5 digits or less. This is a generic representation of this number: _ _ _ _ _ Clearly, we can't have a prime number for the 1st place since that would take the value more than 20k. Therefore, we start with 2nd place and so on. We can have only 2,3,5,7 as single digit prime numbers filling in the places. So for a 4 digits number: we have 4 options for each digit. This gives us \(4 . 4 . 4 . 4 = 256\) possible numbers. And for a 3 digits number: we have 4 options for each digit. This gives us \(4 . 4 . 4 = 64\) possible numbers. And for a 2 digits number: we have 4 options for each digit. This gives us \(4 . 4 = 16\) possible numbers. And for a 1 digit number: we have 4 options for the digit. This gives us \(4 . 1 = 4\) possible numbers. Adding all possible numbers, we get 340. Hence C.
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How many positive integers less than 2*10^4 are there in which each di [#permalink]
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29 Oct 2017, 01:15
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How many positive integers less than 2*10^4 are there in which each di [#permalink]
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29 Oct 2017, 01:32
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Integers less than 20000 in which each digit is a prime number : 2,3,5,7,22,23,25,27,32,33,35,37......
possible options Single digit > 4 2 digit > 4 *4 = 16 3 digit > 4*4*4 = 64 4 digit > 4*4*4*4 =256 5 digit : 10000 till 20000 > But here as 1st digit is always 1 which is not prime.. we cannot consider numbers between 10000 and 20000
Total : 4+16+64+256 = 340
Answer: C
Last edited by Nikkb on 29 Oct 2017, 02:09, edited 2 times in total.



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How many positive integers less than 2*10^4 are there in which each di [#permalink]
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29 Oct 2017, 02:04
is this the correct method 2* 10^4=20,000 single digit 2,3,5,7 two digit = 4*4 ie 16 three digit numbers 4*4*4= 64 four digit =4*4*4*4=256 we have reached 10,000 as tenth thousnadths digit is 1 all numbers beyond will have this number(1x,xxx) as non prime which makes them null and void also the range 20,000 forbids us to consider 22,222 so we have to consider 256+64+16+4=340 Am i correct



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Re: How many positive integers less than 2*10^4 are there in which each di [#permalink]
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29 Oct 2017, 05:09
_  4 ways _ _  4 x 4 = 16 ways _ _ _  4x4x4 = 64 ways _ _ _ _  4x4x4x4= 256ways total = 340 Sent from my SMG615F using GMAT Club Forum mobile app




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