SOLUTION: A band master wished to arrange the band members in pairs for a marching patterns. He found that he was one person short. He tried to arrange by 5s and 7s and he was still one pers

Algebra ->  Divisibility and Prime Numbers -> SOLUTION: A band master wished to arrange the band members in pairs for a marching patterns. He found that he was one person short. He tried to arrange by 5s and 7s and he was still one pers      Log On


   

Question 1144398: A band master wished to arrange the band members in pairs for a marching patterns. He found that he was one person short. He tried to arrange by 5s and 7s and he was still one person short. What is the least number of people in the marching band?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


(corrected response, after reading the problem correctly)

The number of band members is 1 cross%28more%29 less than a multiple of 5 and 1 cross%28more%29 less than a multiple of 7. Since 5 and 7 are relatively prime (have no common factor), the number of band members is 1 cross%28more%29 less than a multiple of 5*7=35.

Then, since the number of band members is odd (1 cross%28more%29 less than a multiple of 2), the smallest possible number of band members is the smallest odd number that is 1 cross%28more%29 less than a multiple of 35.

ANSWER: cross%282%2835%29%2B1%29 2(35)-1 = 69

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

Add (mentally) one person to the band.


Then the number of persons is a multiple of 2, is a multiple of 5 and is a multiple of 7.


The smallest integer positive number with such properties is 2*5*7 = 70.


Now subtract (detach) this additional person, whom you added at the very beginning.


Thus you will get the ANSWER : least number of people in the marching band is 70-1 = 69.


CHECK.  69 is 1 short to be arranged in pairs;

           is 1 short to be arranged by 5s,

           and is 1 short to be arranged by 7s.

Solved and completed.

-----------------

The tutor @greenestamps misread the problem and gave wrong answer.


/\/\/\/\/\/\/\/

To see many other similar solved problems, look into the lesson
    - The number that leaves a remainder 1 when divided by 2, by 3, by 4, by 5 and so on until 9
in this site.

The  Problem 2  of that lesson is an exact analogue to your problem in this post.