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12 Feb 2021, 08:00
Kudos
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
42% (02:23) correct
58%
(02:15)
wrong
based on 138
sessions
History
Date
Time
Result
Not Attempted Yet
What is the sum of all numbers greater than 10000 formed by using the digits 0, 1, 2, 4, 5 no digit being repeated in any number ?
A. 3119972
B. 3119973
C. 3119974
D. 3119975
E. 3119976
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
A. 3119972
B. 3119973
C. 3119974
D. 3119975
E. 3119976
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
13 Feb 2021, 00:18
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Bunuel
We don't require to calculate the entire answer, as we can see that the UNIT's digit is different in each option..
If abcde is the number, then taking e as 1, we get 3 positions for a ( less 0 and the digit at e), and 3*2*1 for remaining three.
So, 3*3*2*1*1=18
Thus each of the digits 1, 2, 4 and 5 will come 18 times in the units digit.
The units digit of sum = 18*(1+2+4+5)=18*12...8*2=16
E is the answer.
If you want to find the exact answer.
abcde.
e : e will be the units digit of 18*12=216, so 6
d : Now 21 will get carried over to tens place and it will become 216+21=237, so 7
c : Again 23 will get carried over to hundreds place and it will become 216+23=239, so 9
b : Again 23 will get carried over to hundreds place and it will become 216+23=239, so 9
a : For a, the sum of digits will change, as it is restricted by only non-zero integers. Hence if 1 is at a position, other arrangements will be 4*3*2*1=24. So 24*(1+2+4+5)=24*12=288. But we have a carry over of 23. So 288+23=311
Number = 3119976
10 Feb 2022, 05:27
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Bunuel
Let the number be in the form ABCDE
Sum of all the numbers that can be formed by using the n digits without any digit repeated is: (n−1)!∗(sum of the digits)∗(111... n times)
First let us take all ways without any restriction
=>(5−1)!(0+1+2+4+5)∗11111=24∗12∗11111=3199968
Now, to get numbers less than 10000, let us subtract ways in which A is 0, so the number becomes BCDE
=>(4−1)!(1+2+4+5)∗1111=6∗12∗1111=79992
Our answer = 3199968-79992= 3119976
E
