SOLUTION: A clerk in a clothing department is arranging t-shirts in a display in stacks of equal size. When he separated the t-shirts into stacks of 4, there was one left over. When he tried

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Question 1059511: A clerk in a clothing department is arranging t-shirts in a display in stacks of equal size. When he separated the t-shirts into stacks of 4, there was one left over. When he tried stacks of 5, there was still one left over. The same was true for stacks of six. However, he was able to arrange the shirts evenly in stacks of 7. How many t-shirts were in the display that he was arranging?
Found 2 solutions by ankor@dixie-net.com, ikleyn:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A clerk in a clothing department is arranging t-shirts in a display in stacks of equal size.
When he separated the t-shirts into stacks of 4, there was one left over.
When he tried stacks of 5, there was still one left over.
The same was true for stacks of six.
However, he was able to arrange the shirts evenly in stacks of 7.
How many t-shirts were in the display that he was arranging?
:
The last statement tells us it has to be a multiple of 7 and because division
by 5 having a remainder of 1, the number will end in a 1 or a 6,
A list of these numbers
21, 56, 91, 126, 161, 196, 231, 266, 301, 336, 371
Divide these by 4 and 6, find a quotient with a remainder of 1.
Only 301 shirts satisfies these requirements (in 43 stacks)

Answer by ikleyn(52800) About Me  (Show Source):
You can put this solution on YOUR website!
.
Was solved a month ago under the link
>https://www.algebra.com/algebra/homework/coordinate/word/Linear_Equations_And_Systems_Word_Problems.faq.question.1058380.html

https://www.algebra.com/algebra/homework/coordinate/word/Linear_Equations_And_Systems_Word_Problems.faq.question.1058380.html

For your convenience I reproduce this solution here again:
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The traditional solution for such kind of problems is THIS:

Let N be an unknown number of t-shirts.
Let us take one shirt aside for a moment (mentally) and consider the number N-1.

Then this number is divided by 4, by 5 and by 6 without a remainder.
The smallest such integer is 4*5*3 = 60.

Consider the numbers (a sequence) 60+1=61, 2*60+1=121, 3*60+1=181, 4*60+1=241, 5*60+1=301, . . . and check if these numbers are multiples of 7.

The 301 is (301 = 7*43).

So, your answer is 301.

For similar problem see 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.