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Of the three-digit integers greater than 750, how many have at least t

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Of the three-digit integers greater than 750, how many have at least t  [#permalink]

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New post 22 Jul 2018, 01:43
7
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A
B
C
D
E

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Question Stats:

32% (03:01) correct 68% (02:27) wrong based on 44 sessions

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Of the three-digit integers greater than 750, how many have at least two digits that are equal to each other?

(A) 56
(B) 70
(C) 72
(D) 74
(E) 78
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Of the three-digit integers greater than 750, how many have at least t  [#permalink]

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New post 22 Jul 2018, 04:42
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Bismarck wrote:
Of the three-digit integers greater than 750, how many have at least two digits that are equal to each other?

(A) 56
(B) 70
(C) 72
(D) 74
(E) 78


The total possibilities of numbers greater than 750 where all digits are different

From 750 to 799 - 1*4*8 = 32 (as 7 can't be part of the ten's or one's digit)
From 800 to 899 - 1*9*8 = 72 (the second digit can take any other values except 8)
From 900 to 999 - 1*9*8 = 72 (the second digit can take any other values except 9)

The 3-digit numbers greater than 750 with all digits different are 32+72+72 = 176
The 3-digit numbers greater than 750 with at least 2 digits that are equal are 250 - 176 or 74.

Therefore, the 3-digit numbers greater than 750 where at least 2 digits are equal to each other are 74(Option D)
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Re: Of the three-digit integers greater than 750, how many have at least t  [#permalink]

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New post 22 Jul 2018, 03:52
1
Bismarck wrote:
Of the three-digit integers greater than 750, how many have at least two digits that are equal to each other?

(A) 56
(B) 70
(C) 72
(D) 74
(E) 78



Let us find between any hundreds..
1) hundreds and tens same xx(10), so 10
2) Hundreds and ones are same x(10)x, so 10
Subtract 1 from total above as xxx is common, so 10+10-1=19
3) tens and ones are same hundreds digit is X and remaining two can be any of the other 9 digits so x(9)(1), so total 9

Total = 19+9=28

800-899 .. 28
900-999... 28


750-800..
Numbers where 7 is twice
7(5 to 9)7 so 1*5*1=5
77(0 to 9)=10
Subtract 1 as 777 is common in both so 10+5-1=14
Numbers where other than 7 is used twice
7(5,6,8,9)(1) so 1*4*1=4
Total 14+4=18

Combined 28+28+18=74

D
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: Of the three-digit integers greater than 750, how many have at least t  [#permalink]

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New post 29 Jul 2018, 11:12
OA: D

Three digit number can have only following 3 patterns:
A. all digits are distinct.
B. two digits are alike and the third is different.
C. all three digits are alike.

Total =A+B+C
As per the question, We have to find out B+C or Total-A.
Total =999 - 750 = 249

Now for A, Case when all digits are distinct
Between 750-759: 7 (Unit digit can be 1,2,3,4,5,6,7,8,9)
Between 760-799: 24 (tens digit have 3 options i.e 6,8 or 9, units digits have 8 option (0,1,2,3,4,5,7,two out of 6,8 and 9)
Between 800-899: 72 (tens digit have 9 options , units digit have 8 options)
Between 900-999: 72 (tens digit have 9 options , units digit have 8 options)
A= 7+24+72+72=175

B+C = Total-A = 249-175=74
Re: Of the three-digit integers greater than 750, how many have at least t &nbs [#permalink] 29 Jul 2018, 11:12
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