## Divisibility & Prime Number

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### Divisibility & Prime Number

by relaxin99 » Wed Mar 11, 2009 8:16 pm
A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?
86

53

49

46

33

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by DanaJ » Wed Mar 11, 2009 9:31 pm
I'm thinking: go for the multiples of 4, add 1 and then check to see if that number fits the second condition:
4 + 1 = 5 - NO
8 + 1 = 9 - NO
12 + 1 = 13 - YES - first one
16 + 1 = 17 - NO
20 + 1 = 21 - NO
24 + 1 = 25 - NO
28 + 1 = 29 - NO
32 + 1 = 33 - YES - second one

This makes the sum, IMHO, 46.

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by franciskyle » Thu Mar 12, 2009 1:33 am
For this one, I would start with deciding which answers gave a remainder of 3 when divided by 5 (the units digit would have to be either 3 or 8 because to be divisible by the units would have to be 0 or 5... add 3 to get 3 and 8 respectively). That leaves us with B & E.

A quick look and you know that 4 goes into 33 eight times with a remainder of 1. Therefore E.

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by cramya » Thu Mar 12, 2009 2:13 pm
Good solutions!

n = 4k+1 and n=5q+3

Find the smallest number that satisfies this i.e. 13

Therefore the family of numbers will be

a*LCM(4,5) + 13

20a+13

When a=0 the number is 13 which u obtained above
When a = 1 the number will be 33

So 33+13 = 46

Hope this helps!

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CR

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by Ian Stewart » Thu Mar 12, 2009 2:58 pm
franciskyle wrote:For this one, I would start with deciding which answers gave a remainder of 3 when divided by 5 (the units digit would have to be either 3 or 8 because to be divisible by the units would have to be 0 or 5... add 3 to get 3 and 8 respectively). That leaves us with B & E.

A quick look and you know that 4 goes into 33 eight times with a remainder of 1. Therefore E.
Ah, note that the correct answer shouldn't give the same remainders that are mentioned in the question. The question asks for the sum of two different numbers with certain remainders. We know each of those two numbers will have a remainder of 3 when divided by 5; when we add those two numbers we should now have a remainder of 1 when we divide by 5 (because 3+3 has a remainder of 1). We also want a number with a remainder of 2 when divided by 4, since the two numbers we're adding give a remainder of 1 when divided by 4.

Still, two answer choices remain, unfortunately, so some other method would be needed to choose between 46 and 86.
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