• Jun 29th 2006, 10:19 AM
jessicash
a trainer in an athletic department was asked to arrange the towels in the locker room in stacks of equal size. When she separated the towels into stacks of 4, one was left over. When she tried stacks of 5, one was left over. The same was true for stacks of 6. However, she was successful in arranging the towels in stacks of 7 each. What is the smallest possible number of towels in the locker room?
• Jun 29th 2006, 12:47 PM
Quick
Quote:

Originally Posted by jessicash
a trainer in an athletic department was asked to arrange the towels in the locker room in stacks of equal size. When she separated the towels into stacks of 4, one was left over. When she tried stacks of 5, one was left over. The same was true for stacks of 6. However, she was successful in arranging the towels in stacks of 7 each. What is the smallest possible number of towels in the locker room?

The answer is 301. I really couldn't find a good strong way to find the answer, just a series of logical clues that gave me a boost.

clues:
The number of towels, $x$, is divisible by 4, 5 and 6 when subtracted by 1, it is also divisible by 7 without any subtraction.

numbers divisible by 5 end in either 5 or 0, that means that $x$ ends in 6 or 1.

4 and 6 can only go into even numbers, so $x$ ends in 1 (because subtract 1 by 1 and you'll get an even number).
• Jun 29th 2006, 01:14 PM
Soroban
Hello, Jessica!

My method is of only slightly higher-level than Quick's.
. . Nice job, Q!

Quote:

A trainer in an athletic department was asked to arrange
the towels in the locker room in stacks of equal size.
When she separated the towels into stacks of 4, one was left over.
When she tried stacks of 5, one was left over.
The same was true for stacks of 6.
However, she was successful in arranging the towels in stacks of 7 each.
What is the smallest possible number of towels in the locker room?

Let $N$ = the number of towels.

$N \div 4$ had a remainder of $1$ . . . $N$ is of the form: $4a + 1$

$N \div 5$ had a remainder of $1$ . . . $N$ is of the form: $5b + 1$

$N \div 6$ had a remainder of $1$ . . . $N$ is of the form: $6c + 1$

Hence, $N$ is of the form: $60k + 1$ .[1]
. . (60 is the LCM of 4, 5, 6.)

Since $N$ is divisible by 7: $N \,= \,7n$

So we have: . $60k + 1 \,= \,7n$

Solve for $n:\;\;n \:= \:\frac{60k+1}{7} \:=\:8k + \frac{4k+1}{7}$

Since $n$ is an integer, $4k + 1$ must be divisible by $7$.

The first time this occurs is when $k = 5$.

Substitute into [1]: . $N \:=\:60(5) + 1 \:= \:\boxed{301}$

• Jun 30th 2006, 06:02 AM
jessicash
Thank you all. I ended up figuring it out on my own. Of course, my answer didn't look quite as nice as yours. :)