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18 Feb 2021, 09:45
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C
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Difficulty:
85%
(hard)
Question Stats:
58% (03:25) correct
42%
(02:53)
wrong
based on 101
sessions
History
Date
Time
Result
Not Attempted Yet
What is the number of odd numbers between 3000 to 6300 that have all different digits?
A. 59
B. 413
C. 826
D. 911
E. 991
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
A. 59
B. 413
C. 826
D. 911
E. 991
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
18 Feb 2021, 12:32
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Bunuel
Let us start with the slot method, we need to divide between many cases however as this question is more complicated than it seems.
3000~4000 case: 3 _ _ _ . We should start with the last slot because it must be an odd number. The last slot only has 4 options since the digit three is taken. Then the other slots have 8*7 options as we took two digits away. Thus in total 8*7*4 options.
4000~5000 case: 4 _ _ _. The last slot has 5 options this time because the digit four is even. The other slots are still 8*7 so we have 8*7*5 options.
5000~6000 case: Same as the 3 _ _ _ case, 8*7*4 options.
6000~6300 case: 6 _ _ _. The hundreds digit can only be 0, 1, or 2. Let us start with the 60_ _ case. The unit digit can have 5 options and the tens digit can have 7 options, 5*7 for this case. The 61_ _ case has only 4 options for the unit digit, but still 7 for the tens digit. The 62_ _ case is the same as the 60_ _ case. Thus in total \(5*7 + 4*7 + 5*7 = 14*7 = 98\) cases.
In total we have \(8*7*(4+5+4) + 98 = 56*13 + 98\). The last digit for this calculation is 6*3+8 -> 6 and only C ends in 6.
Ans: C
RanonBanerjee
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28 Jun 2023, 07:13
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I am getting 801 somehow.
My approach is for any number that starts with 3 should have next 3 digits be filled with 9*8*7 ways.
Similarly for 4 and 5 (1×9×8×7) ways.
Then for any number less than 6300, we can fill out thousands digit in 1 way, hundreds digit in 3 ways(0,1,2), tens digit in 6 ways (3,4,5,7,8,9) and the unit digit in 5 ways. Therefore, (3*6*5) ways.
So total ways would be (3×9×8×7)+(3×6×5)=1602.
Odd numbers should be half of it I.e., 801.
What am I doing wrong here, can anyone please help?
Posted from my mobile device
My approach is for any number that starts with 3 should have next 3 digits be filled with 9*8*7 ways.
Similarly for 4 and 5 (1×9×8×7) ways.
Then for any number less than 6300, we can fill out thousands digit in 1 way, hundreds digit in 3 ways(0,1,2), tens digit in 6 ways (3,4,5,7,8,9) and the unit digit in 5 ways. Therefore, (3*6*5) ways.
So total ways would be (3×9×8×7)+(3×6×5)=1602.
Odd numbers should be half of it I.e., 801.
What am I doing wrong here, can anyone please help?
Posted from my mobile device
