Remainders

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Remainders

by MBA.Aspirant » Mon Jul 04, 2011 11:55 pm
What's the remainder when (4444)^(4444) is divided by 9?


Remainder when 7^0+7^1+7^2......+7^25 is divided by 14?

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by top_business_2011 » Tue Jul 05, 2011 2:44 am
MBA.Aspirant wrote:What's the remainder when (4444)^(4444) is divided by 9?


Remainder when 7^0+7^1+7^2......+7^25 is divided by 14?
1]This is for the first question.
For such questions, I usually want to see if there is any pattern. Remainder of[(4444^1)/9] =7
Remainder of[ (4444^2)/9]=4
Remainder of [(4444^3)/9]= 1
Remainder of [(4444^4)/9]= 7
see that? after the 3rd, it follows a similar pattern. Though finding the remainders appears too cumbersome, it shall take you a minute or less to do so if you really understand the properties of remainders. The following formula is useful in this regard:
Remainder[(m * n)/d] = Remainder[m/d] * Remainder[n/d] [I call this the multiplication rule of remainders, this is only for my purpose] N.B.Here if you get a value greater than d keep on dividing it by d till you get a value less than d, and that value is the remainder.

Getting back to the solution. See the pattern above closely. Excluding the first, the remainder is 7 for every 3rd. value. So getting back to our original question, we need to find for the 4444th. pattern. 4443 is divisible by 3; so the remainder for the 4443th pattern is 7. Therefore, looking at the pattern, the reminder for the 4444th. pattern shall be 4.

So the answer is 4.

2] This is my solution to the second question.
When working with remainder problems, the following rule may, at times, be handy.
Remainder[(X + Y)/z] = Remainder(X/z)+ Remainder (Y/z)[ I call this the addition rule of remainders, that is, once again, because I don't know the name of the rule ]
In like vein, your question can be expressed as:
Remainder[(7^0 +7^1 +...)/14] = Remainder(1/14)+ Rema.[(7^1)/14]...
= 1 +7 +7 +7...[ you have 25 such sevens]
Here it is helpful to note that for any integer x>1 and where a is dividend, d is a divisor and a>d, (a^x)/d leaves the same remainder.
Getting back to the solution, 1+7+7+7...[25 such sevens] =176[ Had this been a number less than 14, we would have taken it as an answer;however as in this case it is greater than 176, we have to divide 176 by 14 and find the remainder. That way we get 8 as a remainder.
Hope this helps.

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by Ian Stewart » Tue Jul 05, 2011 3:00 am
These questions test principles of modular arithmetic, which is beyond the scope of the GMAT, so test takers reading this - you can move on.

The solutions above are perfect, except for the last step in the first problem:
YAMLAKSIRA wrote:
MBA.Aspirant wrote:What's the remainder when (4444)^(4444) is divided by 9?


Remainder when 7^0+7^1+7^2......+7^25 is divided by 14?
1]This is for the first question.
For such questions, I usually want to see if there is any pattern. Remainder of[(4444^1)/9] =7
Remainder of[ (4444^2)/9]=4
Remainder of [(4444^3)/9]= 1
Remainder of [(4444^4)/9]= 7
see that? after the 3rd, it follows a similar pattern. Though finding the remainders appears too cumbersome, it shall take you a minute or less to do so if you really understand the properties of remainders. The following formula is useful in this regard:
Remainder[(m * n)/d] = Remainder[m/d] * Remainder[n/d] [I call this the multiplication rule of remainders, this is only for my purpose] N.B.Here if you get a value greater than d keep on dividing it by d till you get a value less than d, and that value is the remainder.

Getting back to the solution. See the pattern above closely. Excluding the first, the remainder is 7 for every 3rd. value. So getting back to our original question, we need to find for the 4444th. pattern. 4443 is divisible by 3; so the remainder for the 4443th pattern is 7. Therefore, looking at the pattern, the reminder for the 4444th. pattern shall be 4.

So the answer is 4.
Since the pattern repeats in 'blocks of 3', 4444^3, 4444^6, 4444^9, and so on, will all have the same remainder of 1 when divided by 9. Thus when k is an integer, 4444^(3k) has a remainder of 1 when divided by 9. So 4444^4443 has a remainder of 1 when divided by 9 (and not a remainder of 7), and 4444^4444 will thus have a remainder of 7, since it must be the start of a 'block'.
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