Bank Exam Government Jobs Exams Numerical Ability LCM and HCF

X is the greatest number which divides 1305, 4665 and 6905 and gives the same remainder in each case.
What is the sum of the digits in X?

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displaystyle{text{Number of pairs * Sum of each pair} = (frac{n}{2})(n+1) = frac{n(n+1)}{2}}
which is the formula above.

Wait — what about an odd number of items?
Ah, I’m glad you brought it up. What if we are adding up the numbers 1 to 9? We don’t have an even number of items to pair up. Many explanations will just give the explanation above and leave it at that. I won’t.

Let’s add the numbers 1 to 9, but instead of starting from 1, let’s count from 0 instead:

0 1 2 3 4
9 8 7 6 5
By counting from 0, we get an “extra item” (10 in total) so we can have an even number of rows. However, our formula will look a bit different.

Notice that each column has a sum of n (not n+1, like before), since 0 and 9 are grouped. And instead of having exactly n items in 2 rows (for n/2 pairs total), we have n + 1 items in 2 rows (for (n + 1)/2 pairs total). If you plug these numbers in you get:

displaystyle{text{Number of pairs * Sum of each pair} = (frac{n + 1}{2})(n) = frac{n(n+1)}{2}}

which is the same formula as before. It always bugged me that the same formula worked for both odd and even numbers – won’t you get a fraction? Yep, you get the same formula, but for different reasons.

Technique 2: Use Two Rows
The above method works, but you handle odd and even numbers differently. Isn’t there a better way? Yes.

Instead of looping the numbers around, let’s write them in two rows:

1 2 3 4 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Notice that we have 10 pairs, and each pair adds up to 10+1.

The total of all the numbers above is

displaystyle{text{Total = pairs * size of each pair} = n(n + 1)}

But we only want the sum of one row, not both. So we divide the formula above by 2 and get:

displaystyle{frac{n(n + 1)}{2}}

Now this is cool (as cool as rows of numbers can be). It works for an odd or even number of items the same!
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