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How many even 3 digit integers greater than 700 with distinct non zero

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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 04 Jul 2015, 10:14
twixt wrote:
How many even 3 digit integers greater than 700 with distinct non zero digits are there ?

A. 729
B. 243
C. 108
D. 88
E. 77


There are only Three possible cases

In each case Hundreds places is fixed so we fill the unit's place first in order to make the number even and at last we fill the Ten's digit place

Case 1: 7 __ __ When Hundreds digit is 7
7 _7 choices (remaining 7 out of 9 non-zero digits)_ * _4 choices (2,4,6,8)_ = 28 Numbers[/b]

Case 2: 8 __ __ When Hundreds digit is 8
8 _7 choices (remaining 7 out of 9 non-zero digits)_ * _3 choices (2,4,6)_ = 21 Numbers[/b]

Case 3: 9 __ __ When Hundreds digit is 9
9 _7 choices (remaining 7 out of 9 non-zero digits)_ * _4 choices (2,4,6,8)_ = 28 Numbers[/b]

Total Such Numbers = 28 + 21 + 28 = 77 Numbers

Answer: Option E
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 12 Apr 2017, 18:10
Quickly decide to use brute force. Count the number of matching cases in each group of ten numbers.

70 => 0
71 => 4
72 => 3
73 => 4
74 => 3
75 => 4
76 => 3
77 => 0
78 => 3
79 => 4
80 => 0
81 => 3
82 => 2
83 => 3
84 => 2
85 => 3
86 => 2
87 => 3
88 => 0
89 => 3
90 => 0
91 => 4
92 => 3
93 => 4
94 => 3
95 => 4
96 => 3
97 => 4
98 => 3
99 => 0

4*8 + 3*13 + 2*3 = 77

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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 20 May 2017, 21:37
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twixt wrote:
How many even 3 digit integers greater than 700 with distinct non zero digits are there ?

A. 729
B. 243
C. 108
D. 88
E. 77

These types can be handled by considering the number of possibilities for each digit. Start with the one with the most constraints.
1. Last digit can be 2,4,6,8 when the first digit is 7 or 9 and 2,4,6 when the first digit is 8. So there are 4 possibilities in the first case and 3 possibilities in the second case for the last digit.
2. Middle digit cannot be one of the first and last digits and cannot be 0. So there are 7 possibilities
3. When first digit is 7 or 9, then we have (4*7) + (4*7)=56
4. When first digit is 8, we have 3*7=21
5. Total is (3)+(4)= 77
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 09 Jun 2017, 23:35
[E]



Lets go digit wise:

A B C, A is the hundredth digit and C is the ones digit.

As ABC is a three digit greater than 700, A can have values 7,8,9 ( 700<ABC<999)
As ABC is a even number then, C can take values 2,4,6,8

Now Imagine A is 7, and C can take 2,4,6,8 for each of these possibilities of C there are 7 different values B can take up (except 7,0,and the one in C as the digits are unique)

Therefore for A = 7, there are 4*7 = 28 numbers
for A = 8 there are 3*7 (as 8 is already used and C can take only 2,4,6 as its even) = 21
for A = 9 there are 4*7 = 28 numbers

Adding, 28+21+28 = 77
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How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 27 Jul 2018, 11:42
Bunuel

712 756 812 856 912 956
714 758 814 858 914 958
716 762 816 862 916 962
718 764 818 864 918 964
722 766 822 866 922 966
724 768 824 868 924 968
726 772 826 872 926 972
728 774 828 874 928 974
732 776 832 876 932 976
734 778 834 878 934 978
736 782 836 882 936 982
738 784 838 884 938 984
742 786 842 886 942 986
744 788 844 888 944 988
746 792 846 892 946 992
748 794 848 894 948 994
752 796 852 896 952 996
754 798 854 898 954 998

Count: 108 [Horizontal*Vertical = 6*18]
Answer is C
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 27 Jul 2018, 12:10
1
shal1 wrote:
Bunuel

712 756 812 856 912 956
714 758 814 858 914 958
716 762 816 862 916 962
718 764 818 864 918 964
722 766 822 866 922 966
724 768 824 868 924 968
726 772 826 872 926 972
728 774 828 874 928 974
732 776 832 876 932 976
734 778 834 878 934 978
736 782 836 882 936 982
738 784 838 884 938 984
742 786 842 886 942 986
744 788 844 888 944 988
746 792 846 892 946 992
748 794 848 894 948 994
752 796 852 896 952 996
754 798 854 898 954 998

Count: 108 [Horizontal*Vertical = 6*18]
Answer is C


Please read the question and solutions carefully:

How many even 3 digit integers greater than 700 with distinct non zero digits are there ? (I marked couple of incorrect numbers)
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 27 Jul 2018, 20:16
Poor me.. Thank you Bunuel
Re: How many even 3 digit integers greater than 700 with distinct non zero &nbs [#permalink] 27 Jul 2018, 20:16

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