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Hi Mayur,
I think you need to elaborate your question a bit. Is it taking all the digits or any of the digits. If it is taking any of the digits then One needs to take into account all the single digit numbers, two digit numbers, three digit numbers and so on.
BTW, This is my first post in the Math forum and if the threads spin around the assumption that always all digits are considered then please excuse me.
The answer is 5199960. I think you have overlooked those starting with Zero. They do not count in the solution. Keep trying. I know the answer but am not able to get the solution. _________________
Hi Mayur, I think you need to elaborate your question a bit. Is it taking all the digits or any of the digits. If it is taking any of the digits then One needs to take into account all the single digit numbers, two digit numbers, three digit numbers and so on.
BTW, This is my first post in the Math forum and if the threads spin around the assumption that always all digits are considered then please excuse me.
Thanks Venkat
The question is clear, the numbers are > 10,000. You have to find the sum of these. So, you need not worry about single digit numbers, double digit numbers and three digit numbers. The only hitch in this problem is the presence of zero as you cannot start a number with zero. _________________
I figured out the solution!!! [#permalink]
18 May 2004, 09:55
Here is my solution and why I did what I did. I've never tried to explain something like this before so please let me know if you have any trouble understanding my explanation. It is rather long because I wanted to fully explain my reasoing for each step.
1. Since it is an addition problem I decided to look at the sum of the digits in the ones place first. If I could figure out how many times each number appeared in the ones place I could find the sum of the ones place, which would be a good start to finding the total sum.
2. I wrote one of the possibilities, 20468, and focused on figuring out how many times the 8 will appear. Now, because 0 can not be the first digit there are 3 other possibilities for the 1st position(2,4,6).
3. For each of the 3 possibilities where 8 is in the ones place there are 3!, or 6 ways of arranging the other 3 numbers. For example, 20468, 20648, etc.
4. Using the multiplication principle I multiplied 8*3*6 to get 144 as the sum for all possible 8s in the ones position.
5. I used the same reasoning for the other 3 digits(2,4,6) and added their sums with the sum of all the 8 digits.
6*3*6=108
2*3*6=36
4*3*6=72
8*3*6=144
sum: 360
6. Since the same digits appear in every position except the 1st one, the sum will be 360 for each digit position.
7. The last digit of the sum will be 0. Now you have to carry the 36 (from 360) to the 10's place.
8. The sum for this position is 360+36= 396. The next digit of the sum will be 6. Now carry the 39 to the 100's place.
9. The sum for this position is 360+39=399. The next digit of the sum will be 9. Now carry the 39 to the 1000's place.
10. The sum for this position is 360+39=399. The next digit of the sum will be 9. Now carry the 39 to the 1,000's place.
11. So far we have 9960 as the sum of the numbers > 2,000. The sum of the digits in the 1,000's position will be different than the other positions.
12. For each of the non 0 digits, there are 4! or 24 ways of arranging the other 4 places. This means that the 2 will appear 24 times, the 4 will appear 24 times, etc.
13. To get the sum of this position multiply each digit by 24 and add.
2*24=48
4*24=96
6*24=144
8*24=192
sum: 480
14. Add 480 to the 39 carried from step 10. This sum=519.
15. Finally, we put the 519 in with the 9960 from the other digits to get the solution: 5,199,960
Here is another approach to solve the same problem.
Q: find the sum of all the numbers greater than 10,000 using the numbers 0,2,4,6,8 no digit repeated.
There can be 96 such numbers : 5! - 4! = 120 -24= 96 (Alt as explained by hallelujah1234: 4*4! = 4*24 = 96).
Approach : Find the sum of all the 5 digit numbers using 0,2,4,6,8 and then subtract the sum of all the 5 digit numbers that start with 0 (Basically get rid of the sum of all the 4 digit numbers using 2,4,6,8 )
Solution :
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
Hope this helps,
Venkat
Last edited by go2venkat on 18 May 2004, 10:34, edited 2 times in total.
1. Where does this question come from? Is it a real GMAT Q? 2. Hallelujah, how many minutes did it take to solve?
P.S. I wouldn't even try to solve it. Unless it is at the end of the test and I have about 5 spare minutes. Should I?
i will say what i say to everyone. if you aim for a 48+ quant, i will assure you will be tested on some very good problems.
Hard questions are meant to improve your concepts. DO not look to solve ONLY gmat type questions if you even dream of more than 48 in quant.
I have taken the GMAT and inspite of working hard on math, i saw some different questions on the test and that is enough to rattle most guys.
GMAT will throw incredibly intelligent questions at you. Unless you understand everything about a concept, it will be very hard for you to even start 50Q type problems.
If you have taken the time to study the courses we designed for you all,
I would recommend solving 50 tough tough problems on each concept that make your life miserable than solving 500 medium difficulty problems that you would have solved correctly anyway.
As akamai would say, practice hitting with a heavier bat, so that it is easier on game day.
See what works for you. Just my advice, nothing scientific about it.