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
Source:OG
OA:E
How many even 3 digit integers greater than 700 with
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Case 1: Hundreds digit is 7 or 9NandishSS 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
Number of options for the hundreds digit = 2. (7 or 9)
Number of options for the units digit = 4. (2, 4, 6, or 8)
Number of options for the tens digit = 7. (Of the 9 remaining nonzero digits, any but the 2 already used)
To combine these options, we multiply:
2*4*7 = 56.
Case 2: Hundreds digit is 8
Number of options for the hundreds digit = 1. (Must be 8)
Number of options for the units digit = 3. (2, 4, or 6)
Number of options for the tens digit = 7. (Of the 9 remaining nonzero digits, any but the 2 already used)
To combine these options, we multiply:
1*3*7 = 21.
Total ways = Case 1 + Case 2 = 56+21 = 77.
The correct answer is E.
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Hi Mitch,GMATGuruNY wrote:Case 1: Hundreds digit is 7 or 9NandishSS 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
Number of options for the hundreds digit = 2. (7 or 9)
Number of options for the units digit = 4. (2, 4, 6, or 8)
Number of options for the tens digit = 7. (Of the 9 remaining nonzero digits, any but the 2 already used)
To combine these options, we multiply:
2*4*7 = 56.
Case 2: Hundreds digit is 8
Number of options for the hundreds digit = 1. (Must be 8)
Number of options for the units digit = 3. (2, 4, or 6)
Number of options for the tens digit = 7. (Of the 9 remaining nonzero digits, any but the 2 already used)
To combine these options, we multiply:
1*3*7 = 21.
Total ways = Case 1 + Case 2 = 56+21 = 77.
The correct answer is E.
I tried to solve it as follows:
Number of options for the hundreds digit = 3. (7 or 8 or 9)
Number of options for the units digit = 3. (2, 4, or 6) - excluding 8
Number of options for the tens digit = 7. (Of the 9 remaining nonzero digits, any but the 2 already used)
To combine these options, we multiply: 3*3*7= 63
Where did I go wrong?
Thanks
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The red portion miscounts the number of options for the units digit.Mo2men wrote:Hi Mitch,
I tried to solve it as follows:
Number of options for the hundreds digit = 3. (7 or 8 or 9)
Number of options for the units digit = 3. (2, 4, or 6) - excluding 8
Number of options for the tens digit = 7. (Of the 9 remaining nonzero digits, any but the 2 already used)
To combine these options, we multiply: 3*3*7= 63
Where did I go wrong?
Thanks
The red portion is valid only if the hundreds digit is 8, in which case the units digit can be 2, 4, or 6, yielding a total of 3 options.
But if the hundreds digit is 7 or 9, then the units digit can be 2, 4, 6, or 8, yielding a total of 4 options.
Thus, it is incorrect to say that -- in every case -- the number of options for the units digit = 3.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
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