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sveticcg [70]
3 years ago
11

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 o

ver. What is the sum of the two smallest possible values of n?
Mathematics
1 answer:
mel-nik [20]3 years ago
7 0

Answer:

  46

Step-by-step explanation:

The smallest value of n (found by trial and error) is 13. The next larger value is 4·5 = 20 more than that, 33.

The sum of these values is 13 +33 = 46.

_____

Actually, we looked for solutions to ...

  n = 4k +1 = 5m +3

  4k -5m = 2 . . . . . . subtract 5m+1

We found the smallest solution to be k=3, m=2 giving 12 -10 = 2. Then further solutions are of the form ...

  k = 5p+3, m = 4p+2 . . . . . for any integer p

The corresponding values of n are ...

  n = 4(5p+3)+1 = 20p+13

The two smallest values correspond to p=0 and p=1, so are 13, 33.

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Answer: There are 4 students in the group.

Step-by-step explanation:

Let the number of students in the group be n

If one of them scored 9 more points , their average score would be 81 points.

So, our equation (1) will be

\frac{\sum x+9}{n}=81\\\\\sum x+9=81n

Similarly,

If one of them scored 3 points less, their average score would be 78 points.

So, our equation (2) will be

\frac{\sum x-3}{n}=78\\\\\sum x-3=78n

So, from equation (1) and (2) we get:

81n-9=\sum x=78n+3\\\\81n-9=78n+3\\\\81n-78n=3+9\\\\3n=12\\\\n=\frac{12}{3}\\\\n=4

Hence, there are 4 students in the group.

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