Find sum of all the number greater than 10000 formed by the digits 0,2,4,6,8 no digit being repeted?
Raj kumar , 7 Years ago
Grade 11
Magical Mathematics[Interesting Approach]> Find sum of all the number greater than 1...
1 Answers
Arun
Last Activity: 7 Years ago
Dear Raj
1. Fix the units digit of the 5 digit number . say _ _ _ _ 2
2. The total of such numbers with 2 in the units place is 4! i,e 24
3. In the same manner there are 24 such numbers that end with 0,4,6 & 8
4. Now the sum of all the numbers in the Units place is
24 * (0+2+4+6+8) = 24 * 20 = 480
5. The same is true for the 10th place, 100th place, 1000th place & 10,000th place.
Hence the sum of all the 5 digit numbers using 0,2,4,6,8 is
480 (1+10+100+1000+10000)
= 480 (11111)
= 5333280-----------------------------(A)
6. Following the same approach as mentioned in steps 1 to 5,
the sum of all the 4 digit numbers (5 digit numbers starting with 0 )
is
= 6*(2+4+6+8)* (1+10+100+1000)
= 6* 20 * 1111
= 133320--------------------------------(B)
7. Hence the sum of all the 5 digit numbers using 0,2,4,6,8 > 10000
= (A) - (B)
= 5333280 - 133320
= 5199960
1. Fix the units digit of the 5 digit number . say _ _ _ _ 2
2. The total of such numbers with 2 in the units place is 4! i,e 24
3. In the same manner there are 24 such numbers that end with 0,4,6 & 8
4. Now the sum of all the numbers in the Units place is
24 * (0+2+4+6+8) = 24 * 20 = 480
5. The same is true for the 10th place, 100th place, 1000th place & 10,000th place.
Hence the sum of all the 5 digit numbers using 0,2,4,6,8 is
480 (1+10+100+1000+10000)
= 480 (11111)
= 5333280-----------------------------(A)
6. Following the same approach as mentioned in steps 1 to 5,
the sum of all the 4 digit numbers (5 digit numbers starting with 0 )
is
= 6*(2+4+6+8)* (1+10+100+1000)
= 6* 20 * 1111
= 133320--------------------------------(B)
7. Hence the sum of all the 5 digit numbers using 0,2,4,6,8 > 10000
= (A) - (B)
= 5333280 - 133320
= 5199960
Regards
Arun(askIITians forum expert)
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