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Of the integers between 100 and 799, inclusive, how many do
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15 Jan 2010, 13:11
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Of the integers between 100 and 799, inclusive, how many do not have digit 2 and 5? (A) 310 (B) 320 (C) 410 (D) 420 (E) 520
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Re: how many do not have digit 2 and 5? a bit tricky!!
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15 Jan 2010, 14:13




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Re: how many do not have digit 2 and 5? a bit tricky!!
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16 Jan 2010, 10:57
Bunuel wrote: apoorvasrivastva wrote: Of the integers between 100 and 799, inclusive, how many do not have digit 2 and 5? (A) 310 (B) 320 (C) 410 (D) 420 (E) 520 OA Options for the first digit 72(2 and 5)=5; Options for the second digit 102=8; Options for the third digit 102=8; Total numbers possible 5*8*8=320. Answer: B. Bunuel, I understand your method of getting to 320  but does that seem like it is solving for numbers between 100 and 799 that do not include a 2 or a 5? I think the wording of the question can lead someone to think of the total of numbers between 100799 that do not include both a 2 and 5. Is there a way to solve for that without using brute force? (I came up with 50 different numbers that included atleast one 2 and atleast one 5, but was done using brute force).



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Re: how many do not have digit 2 and 5? a bit tricky!!
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16 Jan 2010, 11:48
xiao85yu wrote: Bunuel wrote: apoorvasrivastva wrote: Of the integers between 100 and 799, inclusive, how many do not have digit 2 and 5? (A) 310 (B) 320 (C) 410 (D) 420 (E) 520 OA Options for the first digit 72(2 and 5)=5; Options for the second digit 102=8; Options for the third digit 102=8; Total numbers possible 5*8*8=320. Answer: B. Bunuel, I understand your method of getting to 320  but does that seem like it is solving for numbers between 100 and 799 that do not include a 2 or a 5? I think the wording of the question can lead someone to think of the total of numbers between 100799 that do not include both a 2 and 5. Is there a way to solve for that without using brute force? (I came up with 50 different numbers that included atleast one 2 and atleast one 5, but was done using brute force). I understand the question so as the number shouldn't include neither 2 nor 5. If it were that the number shouldn't include 2 and 5 together (eg. 352, 425, 525, 572, ...), then as there are 50 such numbers and as there are total 700 numbers between 100 and 799 inclusive > so 70050=650 would be the answer.
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Re: how many do not have digit 2 and 5? a bit tricky!!
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16 Jan 2010, 15:16
Bunuel wrote: xiao85yu wrote: Bunuel, I understand your method of getting to 320  but does that seem like it is solving for numbers between 100 and 799 that do not include a 2 or a 5? I think the wording of the question can lead someone to think of the total of numbers between 100799 that do not include both a 2 and 5. Is there a way to solve for that without using brute force? (I came up with 50 different numbers that included atleast one 2 and atleast one 5, but was done using brute force).
I understand the question so as the number shouldn't include neither 2 nor 5. If it were that the number shouldn't include 2 and 5 together (eg. 352, 425, 525, 572, ...), then as there are 50 such numbers and as there are total 700 numbers between 100 and 799 inclusive > so 70050=650 would be the answer. Understood, but is there a way to figure out 50 numbers have both a 2 and 5 together using combinatoric equations? (I know in this case it's pretty straight forward to just thinking logically about how many there are, but not sure if it can be applied for bigger numbers).



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Re: how many do not have digit 2 and 5? a bit tricky!!
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16 Jan 2010, 18:32
xiao85yu wrote: Understood, but is there a way to figure out 50 numbers have both a 2 and 5 together using combinatoric equations? (I know in this case it's pretty straight forward to just thinking logically about how many there are, but not sure if it can be applied for bigger numbers).
For every hundred there are 2 combinations (X25 and X52) except the 200s and 500s which have more. So, we have 2x5 = 10 (the 1,3,4,6,7 hundreds). For 200, the combination is (1)(1)(10C1) + (1)(10C1)(1) = 10 + 10 = 20. This is the case of 25X and 2X5 (only 1 choice for choosing 2, 1 choice of choosing 5 and 10 choices for choosing X). Do the same for the 500s, and we have 40. Add the 10 for the other 000s and we get 50.



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Re: how many do not have digit 2 and 5? a bit tricky!!
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16 Jan 2010, 22:51
mrblack wrote: xiao85yu wrote: Understood, but is there a way to figure out 50 numbers have both a 2 and 5 together using combinatoric equations? (I know in this case it's pretty straight forward to just thinking logically about how many there are, but not sure if it can be applied for bigger numbers).
For every hundred there are 2 combinations (X25 and X52) except the 200s and 500s which have more. So, we have 2x5 = 10 (the 1,3,4,6,7 hundreds). For 200, the combination is (1)(1)(10C1) + (1)(10C1)(1) = 10 + 10 = 20. This is the case of 25X and 2X5 (only 1 choice for choosing 2, 1 choice of choosing 5 and 10 choices for choosing X). Do the same for the 500s, and we have 40. Add the 10 for the other 000s and we get 50. Thanks! Yeah that was how I got 50, but I guess I just logically thought it through without writing down the equation (1)(10C1)(1)+(1)(1)(10C1) but it makes sense now that you put it into words.



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Re: how many do not have digit 2 and 5? a bit tricky!!
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16 Jan 2010, 23:18
Bunuel wrote: Options for the first digit 72(2 and 5)=5; Options for the second digit 102=8; Options for the third digit 102=8;
Total numbers possible 5*8*8=320.
I understand that for each digit place  ones, tens and hundreds  there are two numbers that we can exclude: 2 and 5. Therefore, there exists 8 possible choices (102) to place in each of those digit places. This explains your options for the second and third digit. However, I do not understand how you arrived at 72 in 'options for the first digit.' Would you please explain where the number 7 comes from? Thanks.



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Re: how many do not have digit 2 and 5? a bit tricky!!
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17 Jan 2010, 01:07
handsomebrute wrote: Bunuel wrote: Options for the first digit 72(2 and 5)=5; Options for the second digit 102=8; Options for the third digit 102=8;
Total numbers possible 5*8*8=320.
I understand that for each digit place  ones, tens and hundreds  there are two numbers that we can exclude: 2 and 5. Therefore, there exists 8 possible choices (102) to place in each of those digit places. This explains your options for the second and third digit. However, I do not understand how you arrived at 72 in 'options for the first digit.' Would you please explain where the number 7 comes from? Thanks. We have the numbers between 100 and 799 inclusive. First digit for these numbers can have 7 values: 1, 2, 3 ,4 ,5 ,6 , and 7. But as we need to avoid the numbers with 2 and 5, we should exclude them from the list, so only 5 values will be left: 1, 3, 4, 6, and 7.
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Re: how many do not have digit 2 and 5? a bit tricky!!
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17 Jan 2010, 18:46
Whoops Thanks for the explanation!



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Re: how many do not have digit 2 and 5? a bit tricky!!
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22 Feb 2010, 05:08
Bunuel wrote: I understand the question so as the number shouldn't include neither 2 nor 5. If it were that the number shouldn't include 2 and 5 together (eg. 352, 425, 525, 572, ...), then as there are 50 such numbers and as there are total 700 numbers between 100 and 799 inclusive > so 70050=650 would be the answer. @Bunuel How can we have 50 such numbers where 2 & 5 can neither come together? According to me it should be 48. From 100799 (excluding the series of 200 & 500) we have 10 such numbers. In the series of 200 we have 10 such numbers that have 5 in the units place & 10 such numbers that have 5 in the tens place. But number 255 got counted twice because we have 5 both in units place as well as tens place. Therefore, there will be only 19 such numbers for 200 series & 19 for 500 series, totaling to 48. Please explain!



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Re: how many do not have digit 2 and 5? a bit tricky!!
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22 Feb 2010, 09:32
honeyrai wrote: Bunuel wrote: I understand the question so as the number shouldn't include neither 2 nor 5. If it were that the number shouldn't include 2 and 5 together (eg. 352, 425, 525, 572, ...), then as there are 50 such numbers and as there are total 700 numbers between 100 and 799 inclusive > so 70050=650 would be the answer. @Bunuel How can we have 50 such numbers where 2 & 5 can neither come together? According to me it should be 48. From 100799 (excluding the series of 200 & 500) we have 10 such numbers. In the series of 200 we have 10 such numbers that have 5 in the units place & 10 such numbers that have 5 in the tens place. But number 255 got counted twice because we have 5 both in units place as well as tens place. Therefore, there will be only 19 such numbers for 200 series & 19 for 500 series, totaling to 48. Please explain! I didn't count this myself (note that it's not what the question is asking), just used the number provided by other member, but yes you are right.
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Re: Of the integers between 100 and 799, inclusive, how many do
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30 Aug 2018, 17:55
apoorvasrivastva wrote: Of the integers between 100 and 799, inclusive, how many do not have digit 2 and 5?
(A) 310 (B) 320 (C) 410 (D) 420 (E) 520 Let’s start with hundreds digit; we have 5 choices (1, 3, 4, 6, 7). Each of the tens and units digits has 8 choices (any of the 10 digits except 2 and 5). Thus, there are 5 x 8 x 8 = 320 numbers that do not contain 2 and 5. Answer: B
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Re: Of the integers between 100 and 799, inclusive, how many do &nbs
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