A 4 digit no. is formed using the digits 0,2,4,6,8 without repeating any 1 of them. What is the sum of all such possible no.s?
1)519960 2)402096 3)133320 4)4321302
If you fix 8 as the last digit, you see that there are 4*3*2 ways to complete the number. Thus, 8 appears 24 times as the last digit. By the same logic, if we enumerate all possible numbers using these 5 digits, each number appears 24 times in each of the 4 positions. That is, the digit 8 contributes (24*8+240*8+2400*8+24000*8). In total, we have
(0+2+4+6+8)(24+240+2400+24000)=533280
as our total sum.
In case 4-digit numbers cannot start with 0, then we have over-counted. Now we have to subtract the amount by which we over-counted, which is found by answering: "What is the sum of all 3-digit numbers formed by using digits 2,4,6, and 8?" Now if 8 appears as the last digit, then there are 6 ways to complete the number, so 8 contributes (6*8+60*8+600*8). In total, we have
(2+4+6+8)(6+60+600)=13320.
Subtracting this from the above gives us 519960.
(0+2+4+6+8)(24+240+2400+24000)=533280
as our total sum.
In case 4-digit numbers cannot start with 0, then we have over-counted. Now we have to subtract the amount by which we over-counted, which is found by answering: "What is the sum of all 3-digit numbers formed by using digits 2,4,6, and 8?" Now if 8 appears as the last digit, then there are 6 ways to complete the number, so 8 contributes (6*8+60*8+600*8). In total, we have
(2+4+6+8)(6+60+600)=13320.
Subtracting this from the above gives us 519960.
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