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vivado [14]
3 years ago
6

Cheryl Falkowski has a collection of silver spoons from all over the world. She finds that she can arrange her spoons in sets of

7 with 1 left​ over, sets of 8 with 7 left​ over, or sets of 15 with 3 left over. If Cheryl has fewer than 200​ spoons, how many are​ there?
Mathematics
1 answer:
Ad libitum [116K]3 years ago
3 0
Since she can arrange it in sets
let N=total number of spoons so
N mod7 1 ---->7a+1=n where a is a natural num
N mod 8 7---->8b+7=n where b is a natural num
N mod 15 3----->15c+3=n where c is a natural n
for the first set {8,15,22,29,36,43,50,57,65,71....}
for the second set {15,23,31,39,47,55,63,71,79..}
for the third set {18,33,48,63,78,....}
and find them manually but what is easier is to solve them using algebra and logic (college level)
N=840m+183 where m is zero or a postive integer
so the only N below 200 is 183
let us check
183/7=26 R 1
183/8=22 R 7
183/15=12 R 3
which means our soultion is correct
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Answer:

-13/6 or (in mixed number form) -2 1/6

Step-by-step explanation:

Rewriting our equation with parts separated

=10+512−12−712

Solving the whole number parts

10−12=−2

Solving the fraction parts

512−712=−212

Reducing the fraction part, 2/12,

−212=−16

Combining the whole and fraction parts

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Another way to do it:

Convert any mixed numbers to fractions.

Then your initial equation becomes:

125/12−151/12

Applying the fractions formula for subtraction,

=(125×12)−(151×12) / 12×12

=1500 − 1812/ 144

=−312/144

Simplifying -312/144, the answer is

=−2 1/6

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mote1985 [20]

Answer:

30 seconds

Step-by-step explanation:

Find the least common multiple of 6 and 15.

First, write the prime factorizations.

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4 years ago
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Let us say that the number of people in the row is = x

We can also assume that the people arranged in the row are counted from one end.

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hence, we will have

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