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

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

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New post 13 Nov 2019, 02:54
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A
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C
D
E

Difficulty:

  65% (hard)

Question Stats:

29% (02:18) correct 71% (02:12) wrong based on 24 sessions

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

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New post 14 Nov 2019, 09:00
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1
Bunuel 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

Are You Up For the Challenge: 700 Level Questions


We will divide it into three parts 750-799, 800-899 and 900-999

Two ways almost similar to the ones explained above..

(I) Find the numbers with unique digits..


750-799 - 1st digit is 7, tens digit can be any of 5, 6, 8 or 9--so 4 ways, and the ones digit will be any of the digits except first and tens digit so 10-2=8 ways..
Numbers = 1*4*8=32 ways
800-899 - 1st digit is 8, tens digit can be any of 10 digits except first/hundreds digit--so 9 ways, and the ones digit will be any of the digits except first and tens digit so 10-2=8 ways..
Numbers = 1*9*8=72 ways
900-999 - Same as above 1*9*8=72 ways
Total numbers = 999-750+1=250 and numbers with different digits above 750 = 32+72+72=176.
Thus, numbers with at least 2 digits similar = 250-176=74 numbers

(II) Find the numbers straight -


There are 3 ways this can happen - Hundreds and tens digit are similar OR Hundreds and ones digit are similar OR ones and tens digit are similar
750-799 -
1st digit and tens is 7, and the ones digit will be any of the digits so 1*1*10= 10 ways..
1st digit and ones is 7, and the tens digit will be any of the digits- 5, 6, 8 and 9 ,so 1*4*1=4 ways..
1st digit is 7, and the tens digit and ones digit will be any of the digits- 5, 6, 8 and 9 ,so 1*4*1=4 ways..
Total with 7 in hundreds place = 10+4+4=18 numbers
800-899 -
1st digit and tens is 8, and the ones digit will be any of the digits so 1*1*10= 10 ways..
1st digit and ones is 8, and the tens digit will be any of the digits except 8 ,so 1*9*1=9 ways..
1st digit is 8, and the tens digit and ones digit will be any of the digits except 8 ,so 1*9*1=9 ways..
Total with 8 in hundreds place = 10+9+9=28 numbers
900-999 - Same as above 28 numbers

[b]Total numbers =18+28+28=74

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

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New post 13 Nov 2019, 03:36
1
Bunuel 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

Are You Up For the Challenge: 700 Level Questions


Answer is D

Method 1 :

Total 3 digit numbers greater than 750 = 249
Total number of numbers having unique digits = Total no. of unique digits between 800 to 999 + Total no. of unique digits between 750 and 799
= 2*9*8 + 1*4*8 = 144 + 32 = 176 (This contains 750 as well)
Hence the numbers which have at least two digits equal = 249 - (176 -1) = 74.

Method 2 :

The numbers which have all the 3 digits equal (777,888,999) = 3
Lets calculate the numbers which have two digits equal.

First for the range (800-999) :
XXY 2*1*9 = 18
XYX 2*9*1 = 18
XYY 2*9*1 = 18

Total for this range = 3*18 = 54

For the range (750-799)
XXY 1*1*9 = 9
XYX 1*4*1 = 4
XYY 1*4*1 = 4

Total for this range = 17
Hence the numbers which have at least two digits equal = 54 + 17 + 3 = 74.

Please give Kudos, If you find my answer explanation good :thumbup: :thumbup:
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Re: Of the three-digit integers greater than 750, how many have at least   [#permalink] 13 Nov 2019, 03:36
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