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luda_lava [24]
3 years ago
5

If 6!+7!+8!=(n)(6!), then the value of n is A) 15 B)16 C)63 D)64 E)15!

Mathematics
1 answer:
gladu [14]3 years ago
6 0
6!+7!+8!=(n)(6!)
calculate the individual values first:
6!=720
7!=5040
8!=40320

plug them into the equation:
720+5040+40320=n720
solve for n
5760+40320=n720
46080=n720
divide both sides by 720 to isolate n
46080/720=n720/720
64=n
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  • Call the first integer n,
  • Then the second integer will be n + 1,
  • The third will be n + 2,
  • The fourth will be n + 3,
  • And the fifth will be n + 4.

Now, let's break down the equation in English. We'll use #1, #2, #3, #4, and #5 as stand-ins for the five integers just so we can get things set up, then we'll replace them with the expressions we came up with in the first part. Starting from the beginning of the statement, let's translate words to symbols:

  • "The sum of the first and 4 times the third" → #1 + 4 × #3
  • "is equal to" → = (maybe obvious, but still!)
  • "60 less than 3 times the sum of the second, fourth, and fifth" → 3 × (#2 + #4 + #5) - 60

Putting that all together, we get the equation

#1 + 4 × #3 = 3 × (#2 + #4 + #5) - 60.

Replacing our stand-ins for the expressions we came up with, it becomes

n + 4(n + 2) = 3[(n + 1) + (n + 3) + (n + 4)] - 60

We can now start to simplify our equation to find n (the first integer) and the rest will follow easily from that. I'll start by focusing on the left side. We can distribute and combine like terms like so:

n + 4(n+2) = n + 4n + 8 = 5n + 8

Dealing with the right, we can get rid of some redundant parentheses and carry out the same process:

3[(n + 1) + (n + 3) + (n + 4)] - 60 = 3(n + 1 + n + 3 + n + 4) - 60

= 3(3n + 8) - 60 = 9n + 24 - 60 = 9n - 36

Our simplified equation now becomes

5n + 8 = 9n - 36.

Subtracting 5n from either side:

8 = 4n - 36

Adding 36 to either side:

44 = 4n

And dividing either side by 4:

11 = n.

We now know that our first integer is 11, which means that our 5 consecutive integers are 11, 12, 13, 14, and 15.

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