Veritas Test

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Veritas Test

by CSASHISHPANDAY » Sun Nov 03, 2013 9:54 am
The sum of the digits used to write the sum 10 + 11 + 12 + 13 is 10. What is the sum of the digits used to write the sum of the integers from 1 to 110, inclusive?

900

911

955

957

1001
OA D
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by theCodeToGMAT » Sun Nov 03, 2013 10:18 am
10 + 11 + 12 + 13 = 46 => 10 => 1

Sum to 110 ==> (110)(111)/2 = 6105 => 12 => 3

{A} - 900 = 9 NO
{B} - 911 = 11 = 2 NO
{C} - 955 = 19 = 1 NO
{D} - 957 = 21 = 3 YES
{E} - 1001 = 2 NO
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by [email protected] » Sun Nov 03, 2013 1:05 pm
Hi CSASHISHPANDAY,

These types of questions will almost always involve some type of pattern in the numbers (or in this case, the digits of the numbers). Here's one way to solve this problem (based on the patterns in the digits):

We're asked to deal with the numbers 1 to 110, inclusive. I'm going to break this into two groups: 1 - 99 and 100 - 110

The numbers from 1 to 99 follow an interesting pattern: each digit appears 10 times in the "tens spot" and 10 times in the "units spot"

For example, the digit 1:

10, 11, 12, .....19 ---> 10 times in the "tens spot"
1, 11, 21, 31....91 ---> 10 times in the "units spot"

So, we have to multiply each digit by 20 to get it's total sum. There's a faster way to do it though:

1+2+3+4+5+6+7+8+9 = 45

45 x 20 = 900

So the digits of the numbers from 1 to 99 is 900

Now, for the numbers from 100 to 110

We have 11 "ones" in the "hundreds spot" = 11
We have the digits 1 to 9 in the "ones spot" = 45
We have one more 1 in the "tens spot" = 1

11+ 45 + 1 + 900 = 957

Final Answer: D

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by Brent@GMATPrepNow » Sun Nov 03, 2013 3:58 pm
CSASHISHPANDAY wrote:The sum of the digits used to write the sum 10 + 11 + 12 + 13 is 10. What is the sum of the digits used to write the sum of the integers from 1 to 110, inclusive?

A) 900
B) 911
C) 955
D) 957
E) 1001
IMPORTANT: Always check the answer choices before beginning any solution. The answer choices may hint at an approach and/or suggest that you can skip tedious calculations.

Here, the units digits in the answer choices are all different, which suggests that I may be able to avoid some "grunt" work.

First, let's examine the numbers from 00 to 99 inclusive (i.e., 00, 01, 02, .... 97, 98, 99) [note: adding 00 to the mix doesn't change the final answer]
Notice that there are 100 digits from 00 to 99 inclusive.
Also notice that the digits in the tens and units position are equally distributed

So, in the UNITS position, there will be ten 0's, ten 1's, ten 2's . . . ten 8's and ten 9's
In the TENS position, there will be ten 0's, ten 1's, ten 2's . . . ten 8's and ten 9's
So, the sum of ALL DIGITS from 00 to 99 will equal 20(1+2+3+...7+8+9)

IMPORTANT: We don't need to calculate 20(1+2+3+...7+8+9). We need only recognize that the units digit will equal 0. That is 20(1+2+3+...7+8+9) = ??0

From here, we need to add the digits in 100 to 110 inclusive.
To do so, we can use Rich's approach, or we might even list the values and add them in our head, since there aren't many to add here.
When we add the digits in 100 to 110 inclusive, we get 57

So, the sum of the digits from 00 to 110 inclusive = ??0 + 57 = ??7

Since only D has a 7 in the units position, this is the correct answer.

Cheers,
Brent
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by Mathsbuddy » Sun Nov 17, 2013 11:19 am
Here's a super-quick way of securing the answer:

The sum of the integers from 1 to 110 = 55 * 111 = 6105 (using formula S = (n + 1) * n/2)
The sum of the digits of this is 6 + 1 + 0 + 5 = 12, which in turn reduces to 1 + 2 = 3.

Place value can be ignored as multiplying by 1, 10, 100 or 1000, etc. does not change the sum. Therefore this procedure produces the same result as adding all the digits and reducing similarly.

Of the 5 answer options, only one of them reduces to 3:

9+5+7 = 21
2+1 = 3

Therefore the answer is D

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by namnomnom » Fri Dec 13, 2013 10:46 am
I don't get the question..."The sum of the digits used to write the sum 10 + 11 + 12 + 13 is 10".

So 10+11+12+13 = 46. It takes 10 digits to write the sum 46?

So how many digits does it take to write "the sum of 1-110 inclusive"?

I'm confused.

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by GMATGuruNY » Fri Dec 13, 2013 10:51 am
namnomnom wrote:I don't get the question..."The sum of the digits used to write the sum 10 + 11 + 12 + 13 is 10".

So 10+11+12+13 = 46. It takes 10 digits to write the sum 46?

So how many digits does it take to write "the sum of 1-110 inclusive"?

I'm confused.
10 + 11 + 12 + 13.
The expression above is composed of the following digits:
1,0...1,1...1,2...1,3.
Sum of the digits = 1+0+1+1+1+2+1+3 = 10.
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