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Re: Sum of 3-digit numbers [#permalink]
29 Jun 2009, 03:40

Is it C .. This is how I did it .. Keeping 1 as Hundred digit .. 158+185=343 -----1 Keeping 5 as Hundred digit .. 518+581=1099-------2 Keeping 8 as Hundred digit .. 815+851=1666-------------3 Adding 1+2+3 = 3108 ..
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Re: Sum of 3-digit numbers [#permalink]
29 Jun 2009, 03:49

6

This post received KUDOS

There can be 27 numbers. Every digit has three possibilities because question permits repetition. So it becomes a very good summation problem. If there are 27 digits. There are 27 hundred digits, 27 tens digit and 27 ones digit to be summed. Of 27 hundred digits, 9 of them are 1, 9 of them are 5 and the last 9 are 8. Same is true for the other two digits. Thus the sum is: 9x100+9x500+9x800+9x10+9x50+9x80+9x1+9x5+9x8= 999x(1+5+8)=999x14=13986 E

Re: Sum of 3-digit numbers [#permalink]
29 Jun 2009, 06:01

Since problem permits repetition. There are 27 numbers that satisfies. e.g.: Lets say: first digit is 1, then numbers can be: 111 115 118 151 155 158 181 185 188 Same is true when first digits are 5 and 8. As you can see, there are nine 1 in first digit. Nine 5 in first digit. And nine 8 in first digit. Same is true for the other digits.

Re: Sum of 3-digit numbers [#permalink]
29 Jun 2009, 13:08

maliyeci wrote:

Since problem permits repetition. There are 27 numbers that satisfies. e.g.: Lets say: first digit is 1, then numbers can be: 111 115 118 151 155 158 181 185 188 Same is true when first digits are 5 and 8. As you can see, there are nine 1 in first digit. Nine 5 in first digit. And nine 8 in first digit. Same is true for the other digits.

One more tip: when we add up all the numbers, we can start with the hundreds. We know each number (1, 5, & 8) will appear in the hundreds place a total of 9 times. So let's see how many hundreds we have. 1X9=9 5X9=45 8X9=72 Add this up we have a total of 126 hundreds, or also expressed as 12,600. We see that only one answer could possibly match the size of this sum, which is (E), so without calculating the exact sum, we already know (E) is the only possible choice. On a real test however, the writers could make life difficult by adding a few answer choices that are close to this sum (i.e. 11,950, or 14,088, etc).

Re: Sum of 3-digit numbers [#permalink]
29 Jun 2009, 21:30

maliyeci wrote:

There can be 27 numbers. Every digit has three possibilities because question permits repetition. So it becomes a very good summation problem. If there are 27 digits. There are 27 hundred digits, 27 tens digit and 27 ones digit to be summed. Of 27 hundred digits, 9 of them are 1, 9 of them are 5 and the last 9 are 8. Same is true for the other two digits. Thus the sum is: 9x100+9x500+9x800+9x10+9x50+9x80+9x1+9x5+9x8= 999x(1+5+8)=999x14=13986 E

Now these are my early days here. I even have some problems to use this site When I saw this squestion, I thought how is it possible to do that, to add twenty seven numbers, but thanks to maliyeci!!

I learned a new approach today to add numbers!! +1
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Re: Sum of 3-digit numbers [#permalink]
29 Jun 2009, 22:29

1

This post received KUDOS

maliyeci wrote:

There can be 27 numbers. Every digit has three possibilities because question permits repetition. So it becomes a very good summation problem. If there are 27 digits. There are 27 hundred digits, 27 tens digit and 27 ones digit to be summed. Of 27 hundred digits, 9 of them are 1, 9 of them are 5 and the last 9 are 8. Same is true for the other two digits. Thus the sum is: 9x100+9x500+9x800+9x10+9x50+9x80+9x1+9x5+9x8= 999x(1+5+8)=999x14=13986 E

Great ! Kudos to you.

Another approach is intelligent guess, based on which I would have opted E. Explanation: Total possibilities = 3*3*3 =27 Now, taking examples of numbers starting with 8. Sum of any four 3-digit numbers starting with 8 > 3200,

We know that there are 9 possible nos starting with 8 (apart form other 18 numbers), so sum would certainly be much much greater then 3200.

All other options, except E is less then 3200. (Btw, one can eliminate A, B and D on the 1st glance itself)
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Re: Sum of 3-digit numbers [#permalink]
14 Jul 2009, 12:27

Expert's post

skim wrote:

Could you please explain how did you get 111?

Thanks.

Of course,

(1+5+8)/3 - "average" digit. (1+5+8)/3 * 111 - another way to write 3-digit number formed from "average digit": xyz = (1+5+8)/3 (1+5+8)/3 (1+5+8)/3 or (1+5+8)/3 * 111
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Re: Sum of 3-digit numbers [#permalink]
24 Nov 2011, 10:36

maliyeci wrote:

There can be 27 numbers. Every digit has three possibilities because question permits repetition. So it becomes a very good summation problem. If there are 27 digits. There are 27 hundred digits, 27 tens digit and 27 ones digit to be summed. Of 27 hundred digits, 9 of them are 1, 9 of them are 5 and the last 9 are 8. Same is true for the other two digits. Thus the sum is: 9x100+9x500+9x800+9x10+9x50+9x80+9x1+9x5+9x8= 999x(1+5+8)=999x14=13986 E

Great explanation .. +1 kudos.
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