Bunuel wrote:
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 QuestionsWe 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
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