It is embarrassing to see how for long I have not posted anything on this blog! But the summer break broke my momentum, and too many events intervened subsequently. Thankfully I am able to make an entry today, after a break of close to ten weeks.
A reader - Mr Vinod Kumar, lecturer in mathematics, of Payyanur College, Kannur District, Kerala - has sent in the following problem:
Consider all four digit numbers with the property that the sum of the first two digits is equal to the sum of the last two digits. Examples of such numbers are $4123$, $6372$ and $0413$. (In this problem we shall include the numbers with one, two or three digits among the four digit numbers - we simply consider their leading digits to be $0$. So we shall write $311$ as $0311$, $12$ as $0012$, and so on.)
Here is the problem: Show that the sum of all such four digit numbers is divisible by $11$.
A second question is: If we consider instead all the six digit numbers with the corresponding property - that is, the sum of the first three digits equals the sum of the last three digits - what would be the number that divides the sum of all such numbers? Once again, we pad the number with enough zeros from the left side to ensure that it has six digits.
Readers are invited to solve these problems!
THe property is sum of odd numbers and sum of even numbers is same!
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