:( its soo lengthy and tough

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sana.noor: Legendary Member; Posts: 512; Joined: Mon Jun 18, 2012 11:31 pmFollowed by: 20 members; Thanked: 42 times

by sana.noor » Fri Oct 04, 2013 9:58 pm

1,257

1,275

1,527

.........

.........

+ 7,521

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1, 2, 5, and 7 exactly once in each integer. What is the sum of these 24 integers?

a) 26,996
b) 44,404
c) 60,444
d) 66,660
e) 99,990

OA is E

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Source: Beat The GMAT — Problem Solving | Fri Oct 04, 2013 9:58 pm

theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Fri Oct 04, 2013 10:22 pm

Four digits : 1,2,5,7

Number starting from 1's (i.e. 1 _ _ _ )
Total combination = 1 * 3! [arranging the last three digits] = 6

Similarly, for all we will have 6 combinations each

Now,
1000*6 + 2000*6 + 5000*6 + 7000*6 [this would be the minimumiest possible sum, even lesser than the possible sum made from the combination of numbers]
Because 1000*6 < 1257 + 1527 + ....
= 6 * ( 1+2+5+7) = 90,000 --> this is the minimum SUM will would be infact lesser than the minimum possible sum made using these number

No other answer choice is above 90,000 ..

So surely {E}

R A H U L

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rakeshd347: Master | Next Rank: 500 Posts; Posts: 391; Joined: Sat Mar 02, 2013 5:13 amFollowed by: 4 members; Thanked: 50 times

by rakeshd347 » Fri Oct 04, 2013 10:30 pm

sana.noor wrote:1,257

1,275

1,527

.........

.........

+ 7,521

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1, 2, 5, and 7 exactly once in each integer. What is the sum of these 24 integers?

a) 26,996
b) 44,404
c) 60,444
d) 66,660
e) 99,990

OA is E

Its not that tough or lengthy as it looks. In fact you can solve this question under 1 minute.
Here is my approach: whenever I see such a lengthy addition or multiplication first thing comes into my mind it that GMAT never expects you to do the whole addition or multiplication.
Now if you notice there will be 24 numbers with those 4 digits. Out of those 24....each number will appear in the unit place 4 times...Correct....because each number get the same priority. So each number will appear only 6 times.

So lets calculate the unit digit.
1 will apear 4 times and will give unit digit=1*6=6
2 will apear 4 times and will give unit digit=2*6=2
5 will apear 4 times and will give unit digit=5*6=0
7 will apear 4 times and will give unit digit=7*6=2

Now if you add all the unit digits you will get 6+2+0+2=10 or unit digit is again 0.
So you can eliminate A B and C.

The above part is optional just in case if the answers are too close to eliminate.

Now here it is even easier. Suppose if you were to add all the numbers starting with 1,2,5 and 7 with just 000 at the end you will get 90,000
1000*6=6000
2000*6=12000
5000*6=30,000
7000*6=42,000

So the total sum is 90,000. Of course there is no answer is 90000 but the only answer that is more than 90,000 is
E.
The reason the sum will be more than 90,000 is because you will have all number more than 1000 (1257,1527,1725 and so on) same is with 7000 (7521,7251 and so on) so the final answer will be more than 90,000. Hence answer is E.

it looks lengthy as I explained it here but I solved it in less than a minute. if you want you don't have to even do the unit digit part as the answers are so wide apart. I would do the Unit digit part only if the answer were like 98000 and 99568 so they are too close and Gmat will give other hints in these cases.

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vinay1983: Legendary Member; Posts: 643; Joined: Wed Aug 14, 2013 4:27 amFollowed by: 7 members; Thanked: 48 times

by vinay1983 » Fri Oct 04, 2013 10:45 pm

Now,
1000*6 + 2000*6 + 5000*6 + 7000*6 [this would be the minimumiest possible sum] = 6 * ( 1+2+5+7) = 90,000 --> this is the minimum SUM

Almost near your answer. Hmmm. Strange question though. But i like this about your answer

You can, for example never foretell what any one man will do, but you can say with precision what an average number will be up to!

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sana.noor: Legendary Member; Posts: 512; Joined: Mon Jun 18, 2012 11:31 pmFollowed by: 20 members; Thanked: 42 times

by sana.noor » Fri Oct 04, 2013 11:59 pm

rakesh347, i used the same approach for this question but Veritas expert says that some times answer are so close together that it wont help you to reach the right answer choice. veritas used different approach which was hard for me to digest.

Last edited by sana.noor on Sat Oct 05, 2013 3:04 am, edited 1 time in total.

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rakeshd347: Master | Next Rank: 500 Posts; Posts: 391; Joined: Sat Mar 02, 2013 5:13 amFollowed by: 4 members; Thanked: 50 times

by rakeshd347 » Sat Oct 05, 2013 1:21 am

sana.noor wrote:rakesh347, i used the same approach for this question but Veritas experts says that some times answer are so close together that it wont help you to reach the right answer choice. they sed different approach which was hard for me to digest.

I have mentioned 2 approaches for this question:
1. to check the unit digit
2. to get the estimate of the sum. If you do these two I can bet you can solve these question in less than a minute.

I don't see anything wrong with this approach. Even Vinay and Rahul have used the same approach. I elaborated it to make it easy for you to understand otherwise it is very short. You can do these calculations in your head. This is around 600 difficulty level question if you can do this question in less than a minute then its good. I don't think there can be some other approach unless you want to learn lot of formulas and memorise them but these sort of question is better solved in less time if you think conceptually.

Moreover, there is no point looking at an approach if that confuses you and hard to understand. It better to go back to basics and think conceptually.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Oct 05, 2013 3:03 am

sana.noor wrote:1,257

1,275

1,527

.........

.........

+ 7,521

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1, 2, 5, and 7 exactly once in each integer. What is the sum of these 24 integers?

a) 26,996
b) 44,404
c) 60,444
d) 66,660
e) 99,990

OA is E

Estimating the sum is a great approach.
Here's one way to quickly calculate the sum if we need to be more exact:

The sum is composed of 24 integers.
In the UNITS place, 1, 2, 5 and 7 each will appear the SAME NUMBER OF TIMES.
Thus, each digit will appear in the units place a total of 6 TIMES.
The same will be true for the THOUSANDS place, the HUNDREDS place, and the TENS place.
Thus, the SUM of the digits in each position = 6 * (1+2+5+7) = 90.

Each of the integers can be represented as follows:
1000A + 100B + 10C + D.
Since the sum of the digits in each position = 90, we get:
90(1000) + 90(100) + 90(10) + 90 = 90(1000+100+10+1) = 90(1111) = 99,990.

The correct answer is E.

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ducefun: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Sun Sep 29, 2013 5:55 pm

by ducefun » Sat Oct 05, 2013 4:48 pm

My solution:
The sum is abcde.
e = 0 [6*(1+2+5+7)=90] and you have to add 9 to the tens digit.
But before adding 9, what do you have for the tens digit? Still 0, because of the same combination of {1, 2, 5, 7}.
So add 9: 0+9 = 9 = d
The sum is now abc90. Check all answers, choose one.

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