How do you evaluate #65\times 5\times 42\times 542\times 58\times 556\div 78963#?

2 Answers
EZ as pi
Aug 5, 2017

#3,021,419.885#

Explanation:

It often makes simplifying easier if you break the numbers down into the factors first.

#(65xx5xx42xx542xx58xx556)/(78,963)#

#=(65xx5xx42xx542xx58xx556)/(3xx26,321)#

However in this case, the only factor of #78963# which I could find was #3#

#(65xx5xxcancel42^14xx542xx58xx556)/(cancel3xx26321)#

This leaves us with an unthinkable manual calculation or using a calculator.

#3,021,419.885#

Jim H
Aug 7, 2017

I assume you want to know how to simplify the evaluation if you are not using a calculator. Here are some tips.

Explanation:

Although I notice that both the numerator and denominator are divisible by #3#, we'll leave the simplfication for later.

Look at the product first.

#65xx5xx42xx542xx58xx556#

# = 65xx5xx(42xx58)xx(542xx556)#

# = 325 xx[(50-8)(50+8)] xx [[549-7)(549+7)]#

In the first brackets:

#(50-8)(50+8) = 50^2-8^2 = 2500-64 = 2450-14 = 2436#

(Note that #50^2 = 5^2*10^2 = 25*100#)

In the second brackets we have #[549-7)(549+7) = 549^2-49#

To find #549^2# either multiply it out long hand or use

#549^2 = (550-1)^2 = 550^2 - 2 xx 550 +1#

#550^2 = (5xx11xx10)^2 = 25 xx 11xx11 xx 100 = 275xx11xx100 = 302500#

Returning, we have

#549^2 = (550-1)^2 = 302500 - 1100 +1 = 301400+1=301401#

So in the second brackets we have

#[549-7)(549+7) = 549^2-49 = 301401-49 = 301401-50+1 = 301351+1 = 301352#

The total product is

#65xx5xx42xx542xx58xx556#

# = 325 xx[(50-8)(50+8)] xx [[549-7)(549+7)]#

# = 325 xx [2436]xx[301352]#

Now we will reduce by the factor of #3# (using #2436 = 3xx812#) and we'll drop the brackets.

#(65xx5xx42xx542xx58xx556)/78963 = (325 xx 2436xx301352)/(3xx26321)#

# = (325 xx 812xx301352)/26321#

I would then move a factor of #4# from the #812# to multiply #325# and get a nice multiple of #100.

# = (325 xx 4 xx 203 xx301352)/26321#

# = (1300 xx203 xx301352)/26321#

Now it gets a bit tedious. We can either

multiply #203 xx301352# and then multiply that product be #13# and tack on the two #0#'s or

Make it # = (100 xx (13xx203) xx301352)/26321#

which is # = (100 xx 2639xx301352)/26321#

and now do #2639xx301352# longhand.

(If you prefer, you could do #2640 xx 301352 - 301352#

Yes, it is tedious, but we get

#(100 xx 795,267,928) /26321= (79,526,792,800) /26321#

Now either approximate the division or go through it long hand.

I used a calculator to get

#3,021,419(23,301)/(26,321) ~~ 3,021,419.8852627179818# (and so on until it goes into a cycle)

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