# Find how many solutions (divisibility question)

• Jul 5th 2010, 08:18 AM
Mukilab
Find how many solutions (divisibility question)
There is an integer 27*36*5 where the asterisks are also numbers, you just don't know them. This number is divisible by 225. How many integer solutions does it have (not negative)

This is a follow up question to another one which was find a number such that *676* can be divided by 72. Found 2 different solution by brute force (there were only 45 things to pick from)

I don't understand how it is a follow up method. Is there a technique or method involved because I probably missed it in using brute force.

• Jul 5th 2010, 08:28 AM
undefined
Quote:

Originally Posted by Mukilab
There is an integer 27*36*5 where the asterisks are also numbers, you just don't know them. This number is divisible by 225. How many integer solutions does it have (not negative)

This is a follow up question to another one which was find a number such that *676* can be divided by 72. Found 2 different solution by brute force (there were only 45 things to pick from)

I don't understand how it is a follow up method. Is there a technique or method involved because I probably missed it in using brute force.

So in *676* you were able to narrow down from 90 to 45 possibilities because you knew the number had to be even.

Here, you can tell that the second blank in 27*36*5 has to be a 2 or a 7 because the last two digits of multiples of 225 are either 00, 25, 50, or 75.
• Jul 5th 2010, 09:23 AM
eumyang
Quote:

Originally Posted by Mukilab
This is a follow up question to another one which was find a number such that *676* can be divided by 72. Found 2 different solution by brute force (there were only 45 things to pick from)

Quote:

Originally Posted by undefined
So in *676* you were able to narrow down from 90 to 45 possibilities because you knew the number had to be even.

I could narrow *676* down to 18 possibilities at the start. A number is divisible by 72 if it is divisible by both 8 and 9. A number is divisible by 8 if the last 3 digits are divisible by 8. I happen to know that 768 is divisible by 8:
$\displaystyle 768 = 2^8 \times 3$
(you see the number 768 in the resolution for XGA: 1024 x 768),
so that means that the only digits in the ones place could be 0 or 8.

A number is divisible by 9 if the sum of the number's digits is divisible by 9. Use that to find which numbers among *6760 and *6768 work.
• Jul 5th 2010, 01:13 PM
Soroban
Hello, Mukilab!

Quote:

An integer $\displaystyle N = 27a36b5$, where $\displaystyle a$ and $\displaystyle b$ are digits, is divisible by 225.

How many positive integer solutions does it have?

Since the number is divisible by 225, it is divisible by 25 and 9.

To be divisible by 25, the number must end in 00, 25, 50, or 75.
. . Since the last digit is 5, the only choices are: .$\displaystyle b = 2\text{ or }7$

To be divisible by 9, the sum of the digits must be divisible by 9.
. . $\displaystyle 2 + 7 + a + 3 + 6 + b + 5 \:=\:9k\:\text{ for some positive integer }k.$
We have: .$\displaystyle a \:=\:9k - 23 - b$

If $\displaystyle {\bf b = 2}$ we have: .$\displaystyle a \:=\:9k-25$
. . Then: .$\displaystyle k = 3,\;{\bf a = 2}$

If $\displaystyle {\bf b = 7}$, we have: .$\displaystyle a \:=\:9k-30$
. . Then: .$\displaystyle k = 4,\;{\bf a = 6}$

There are two solutions.

. . $\displaystyle a = 2,\:b=2\!:\;\;27{\bf2}36{\bf2}5$

. . $\displaystyle a = 6,\:b = 7\!:\;\;27{\bf6}36{\bf7}5$