## random math problem time

November 10th, 2009

What is the remainder when 356 is divided by 7?

## Representations of Number as Sum of Consecutive Integers

October 2nd, 2009

How many ways are there of writing XXXX as a sum of positive consecutive integers?

I found this discussion very interesting (after coming upon it after a search due to encountering one of these problems on FTW and failing horribly).

Basically, the result is that the number of ways of writing a number as a sum of consecutive integers is the number of odd divisors of that number, minus one. This is an awesome conclusion.

How many ways are there of writing 123123 as a sum of positive consecutive integers?

$123123 = 3 \cdot 7 \cdot 11 \cdot 13 \cdot 41$

Therefore, the number of factors (all of which must be odd, because there is no factor of $2$ in $123123$) are:

$2^5 = 32 \text{ factors}$

Since there are 32 odd factors of 123123, the number 123123 can be written as a sum of positive consecutive integers in 31 ways.

Each of these ways is one of the odd divisors of 123123 (we subtract 1 because we can’t count the divisor 1 as that would yield only 1 integer, 123123 itself, and we need plural integers anyways).

If we count the even divisors, that will result in some of the divisors being negative. (See thread.)

Ok, lame post, sorry. Guys, do some of the earlier problems, like that staircase one ^_^

## Variation of a previous number theory problem

September 19th, 2009

Lol, just saw this on the math team forum, was posted by like last last year, but anyways…

In the prime factorization of the expression $100! \cdot 99! \cdot 98! \cdot \ldots \cdot 3! \cdot 2! \cdot 1!$ there is a $7^x$ term. Find x.

## Number Theory – Solutions

September 18th, 2009

1) Written out in base 10, the number $345!$ ends in how many zeroes?

Each zero in a number represents a factor of 10, or a factor each of five and two.

Therefore, to find the number of zeros after a number, we need to find the number of factors of five and the number of factors of two in $345!$.

How many factors of five are in $345!$?

Simply divide 345 by five.

$\dfrac{345}{5} = 69$

However, we need to count the factors of five squared (25) twice, so we divide again by 25.

$\dfrac{345}{25} = 13$

And finally, let’s not forget the factors of five cubed which need to be counted three times.

$\dfrac{345}{125} = 2$

Summing these together:

$69 + 13 + 2 = \boxed{84}$

Thus there are 84 zeroes after the number $345!$.

2) The product of the 12-digit number $493,\!224,\!762,\!945$ and the 12-digit number $471,\!910,\!894,\!925$ is the 24-digit number $232,\!758,\!139,\!280,\!5A5,\!928,\!554,\!125$. Find the digit $A$.

Did you try to find the digits that affected the value of A and try multiplying things out?

A smarter solution, do this calculation mod 9. It’s called “casting out nines”. Cross out all the nines and all the numbers that add up to nine, etc, and add up the remaining numbers, subtracting nine when needed.

$493,\!224,\!762,\!945 \equiv 3 \mod 9$ $471,\!910,\!894,\!925 \equiv 5 \mod 9$

Therefore,

$493,\!224,\!762,\!945 \cdot 471,\!910,\!894,\!925 \equiv 15 \mod 9$

And

$15 \equiv 6 \mod 9$

The product mod 9 is:

$232,\!758,\!139,\!280,\!5A5,\!928,\!554,\!125 \equiv 2 + A \mod 9$

And since the products are supposed to be equal, they will also be equal mod 9.

$2 + A \equiv 6 \mod 9$ $\boxed{A \equiv 4 \mod 9}$

It’s as simple as that!

## Number Theory

September 18th, 2009

Exercises for Hanchan:
(note to self, move this to private/password protected tomorrow)

1) Written out in base 10, the number $345!$ ends in how many zeroes?

2) The product of the 12-digit number $493,\!224,\!762,\!945$ and the 12-digit number $471,\!910,\!894,\!925$ is the 24-digit number $232,\!758,\!139,\!280,\!5A5,\!928,\!554,\!125$. Find the digit $A$.

Hey, Ben, this is K. Apparently I can edit this. I can also see all your private posts. You should probably fix this.

Fixed. Thanks.