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Counting problem

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Counting problem

by STEVEN SPIELBERG » Sun Dec 01, 2013 10:07 am
Q:The total number of seven digit numbers,the sum of whose digits is even are:

(a)250000

(b)4500000

(c)3500000

(d)6500000

(e)7500000

OA[spoiler](b)[/spoiler]

Although the answer given is (b) I am not getting the answer (b). I think there is something wrong with the question or maybe I am missing something.
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by Brent@GMATPrepNow » Sun Dec 01, 2013 10:44 am
STEVEN SPIELBERG wrote:Q:The total number of seven digit numbers,the sum of whose digits is even are:

(a)250000
(b)4500000
(c)3500000
(d)6500000
(e)7500000
IMPORTANT: For HALF of the 7-digit numbers, the sum of the digits will be odd, and for the other half, the sum will be even.

To see why, let's make some observations using 2-digit numbers:
10: sum is odd
11: sum is even
12: sum is odd
13: sum is even
14: sum is odd
etc.
The same will apply to 7-digit integers.

The question involves 7-digit numbers.
In other words, we're examining the integers from 1,000,000 to 9,999,999 inclusive

We have a rule that says: the number of integers from x to y inclusive equals y - x + 1
So, the number of integers from 1,000,000 to 9,999,999 inclusive = 9,999,999 - 1,000,000 + 1 = 9,000,000

Half of these integers will be such that the sum of their digits will be even.
(1/2) of 9,000,000 is 4,500,000

Answer: B

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by Mathsbuddy » Mon Dec 02, 2013 8:34 am
Can anyone see the mistake?

From 0 to 9999999, the quantity of 7 digit integers is 10^7 combinations minus: the [6, 5, 4, 3, 2, 1 and 0 digit combinations] , which equals 10^7 - (10^6 + 10^5 + 10^4 + 10^3 + 10^2 + 10^1) = 8888890
altogether.

Half of these have an even sum of digits, giving 8888890/2 = 4444445

What's the flaw in this logic?
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by Brent@GMATPrepNow » Mon Dec 02, 2013 8:46 am
Mathsbuddy wrote:Can anyone see the mistake?

From 0 to 9999999, the quantity of 7 digit integers is 10^7 combinations minus: the [6, 5, 4, 3, 2, 1 and 0 digit combinations] , which equals 10^7 - (10^6 + 10^5 + 10^4 + 10^3 + 10^2 + 10^1) = 8888890
altogether.

Half of these have an even sum of digits, giving 8888890/2 = 4444445

What's the flaw in this logic?
The flaw is that you're subtracting the same amount more than once.

For example, there are 100 integers from 0 to 99 inclusive. INCLUDED in those 100 integers are 10 1-digit integers (0, 1, 2, ...,8, 9).
So, when you subtract 100 AND subtract 10 from 10^7, you're subtracting the quantity of 1-digit numbers TWICE.
This occurs all the way up to 10^6.

If you want to find the number of 7-digit numbers take 10^7 and subtract 10^6 (the number of digits that don't have fewer than 7 digits) to get 9,000,000

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by STEVEN SPIELBERG » Tue Dec 03, 2013 11:55 am
The solution I have is with the use of an Anagram: |9|10|10|10|10|10|5|=4500000.If we fix first digit then unit digit will determine the odd/even nature of numbers so formed. So unit digit can assume only 5 digits which are even. This implies that whatever the first six digits are the sum will be even if the unit's digit will be even. But say- 1212122 is odd. Maybe I missed the meaning of the highlighted statement. Or maybe I misunderstood something else. How we interpret this question using the Anagram. Please explain.
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by [email protected] » Wed Dec 04, 2013 12:16 am
Hi Steven,

This question DOES NOT ask how many 7-digit numbers are even; it asks how many 7-digit numbers are there whose DIGITS SUM to an even number.

As examples:

The number 2,111,111 has digits that add up to an EVEN number
The number 2,111,112 has digits that add up to an ODD number

Brent provides a logical way to solve this problem by pointing out a "math pattern" that exists in sequential numbers. By definition, every OTHER positive integer will have digits that SUM to an even number.

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by Mathsbuddy » Wed Dec 04, 2013 8:24 am
7 digits including leading zeros -> 10^7 combinations (including 0000000)

Number of 7 digit combinations that include leading zeros = 10^6 (not including 0000000)

So 10^7 - 10^6 = 9000000

Half of these have an even sum of digits:

9000000/2 = 4500000

I's all clear to me now!
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by STEVEN SPIELBERG » Wed Dec 04, 2013 11:06 am
[email protected] wrote:Hi Steven,

This question DOES NOT ask how many 7-digit numbers are even; it asks how many 7-digit numbers are there whose DIGITS SUM to an even number.

As examples:

The number 2,111,111 has digits that add up to an EVEN number
The number 2,111,112 has digits that add up to an ODD number

Brent provides a logical way to solve this problem by pointing out a "math pattern" that exists in sequential numbers. By definition, every OTHER positive integer will have digits that SUM to an even number.

GMAT assassins aren't born, they're made,
Rich
Hey Rich,

I am not arguing Brent's method. That's a supercool method. And I am also talking about sum of digits to be even. I am just asking why the anagram method isn't working, or maybe I'm not interpreting the anagram correctly. Let me write the solution in detail: Anagram as above- since if we fix the first digit then unit digit will determine the odd/even nature of the numbers so formed. But unit place can assume half even and half odd. Hence unit place can assume just 5 digits which makes the sum of previous six digit even.

Now that's the actual solution I was given. So if the above is true then the first six digits can assume any digits and the unit's digit if even will determine whether the sum is even or odd. So if we take a number say 1212122 the unit's digit is even but still we are not getting an even sum. So maybe I am not interpreting the Anagram correctly.
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by Mathsbuddy » Thu Dec 05, 2013 2:16 am
STEVEN SPIELBERG wrote:
[email protected] wrote:Hi Steven,

This question DOES NOT ask how many 7-digit numbers are even; it asks how many 7-digit numbers are there whose DIGITS SUM to an even number.

As examples:

The number 2,111,111 has digits that add up to an EVEN number
The number 2,111,112 has digits that add up to an ODD number

Brent provides a logical way to solve this problem by pointing out a "math pattern" that exists in sequential numbers. By definition, every OTHER positive integer will have digits that SUM to an even number.

GMAT assassins aren't born, they're made,
Rich
Hey Rich,

I am not arguing Brent's method. That's a supercool method. And I am also talking about sum of digits to be even. I am just asking why the anagram method isn't working, or maybe I'm not interpreting the anagram correctly. Let me write the solution in detail: Anagram as above- since if we fix the first digit then unit digit will determine the odd/even nature of the numbers so formed. But unit place can assume half even and half odd. Hence unit place can assume just 5 digits which makes the sum of previous six digit even.

Now that's the actual solution I was given. So if the above is true then the first six digits can assume any digits and the unit's digit if even will determine whether the sum is even or odd. So if we take a number say 1212122 the unit's digit is even but still we are not getting an even sum. So maybe I am not interpreting the Anagram correctly.
Hi there,

I believe the key to your question is as follows:

ODD + ODD = EVEN
EVEN + EVEN = EVEN
ODD + EVEN = ODD

So if the sum of the 1st 6 digits is EVEN then the last digit needs to be EVEN to give an even sum
and if the sum of the 1st 6 digits is ODD then the last digit needs to be ODD to give an even sum.

If you go through all the combinations from 1000000 to 9999999 in numerical order, you get:
1000000, 1000001, 1000002, 1000003, .... etc... to 9999998, 9999999
which is a sequence, the sum of each term of which goes:
Odd, even, odd, even, odd, even, etc...
Altogether there are 9000000 terms in 4500000 pairs of (odd, even).
Hence there are 4500000 evens.
I hope that helps in some way.
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by STEVEN SPIELBERG » Thu Dec 05, 2013 6:22 am
Hey there,

Yep, I was misinterpreting the Anagram. The Anagram is actually a method for counting the number of possible numbers that meet the criteria and not for checking what the actual numbers will be i.e the actual details say 1212122. So the 5 in the unit's digit position is actually 10/2 which is in line with Brent's method. Thus I guess we can calculate half of any digit number using the Anagram just by taking the half of unit's digit in the Anagram and the rest will remain same- |9|10|10|10|.......and so on with last digit |10/2|.
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