Integer Quotient Problem
Brenda Keimig
larger divided by the smaller, the integer part of the quotient is 1 and the
remainder is 16. Find the original integers. OK, I have found them (21 &
37), but not using a pretty algebraic equation. Can anyone help me with
setting it up?
THANKS!!
Ray Cribbs
news:76233778FDECAC9C.8B10F992...@lp.airnews.net...
Let's call the greater of the two positive integers 'g' and the smaller 's.'
Two positive integers have a sum of 58:
g + s = 58
If 3 is added to each and then the larger divided by the smaller, the
integer part of the quotient is 1 and the remainder is 16:
(g+3)/(s+3) = 1 + 16/(s+3)
(g+3) = (s+3) + 16
g = s + 16.
Now, substituting back into the original equation:
g + s = 58
(s+16) + s = 58
2s = 42
s = 21
g = s + 16
g = 21 + 16
g = 37
Thus, your two numbers must be 21 and 37.
Greg Ball
Let the numbers be x and y with y>x. Then,
x + y = 58
That's the easy one. Also,
(y+3)/(x+3) = 1 + 16/(x+3)
See how the 1 and the remainder of 16 have been built into the equation.
This gives two equations with two variables. I would multiply the second one
by (x+3) to get rid of the denominators and then use the first equation to
substitute for one of the variables.
Greg
Brenda Keimig wrote in message
<76233778FDECAC9C.8B10F992...@lp.airnews.net>...
Stan Brown
Please follow up in the newsgroup.]
Said kei...@pdq.net (Brenda Keimig) in alt.algebra.help:
>Two positive integers have a sum of 58.
So one of them is x, and the other is 58-x. It will help you A LOT
to recognize this pattern: if two quantities add to 40, one of them
can be x and the other will be 40-x. There's no magic about it: x +
(40-x) = x + 40 - x = 40, and similarly for x and 58-x.
> If 3 is added to each
x+3, 61-x
>and then the
>larger divided by the smaller,
We don't know which is larger, so let's arbitrarily assume that x
is smaller. Therefore dividing larger by smaller is
(61-x) / (x+3)
>the integer part of the quotient is 1 and the
>remainder is 16. Find the original integers.
Since we said 58-x > x, and therefore 61-x > x+3, we know that the
fraction above is greater than 1. And the problem tells us that the
fraction is less than 2, because the integer part is exactly 1. But
(x+3)/(x+3) = 1, so the above fraction decomposes as
(x+3 + 58-2x)/(x+3)
(x+3)/(x+3) + (58-2x)/(x+3)
Now we are told that the remainder is 16, so
16 = 58 - 2x
2x = 42
x = 21
One number is 21, and the other is 58-21 = 37.
Check: 21+3 = 24, 37+3 = 40. 40 div 24 goes once with remainder 16,
as required.
--
Stan Brown, Oak Road Systems, Cleveland, Ohio, USA
http://oakroadsystems.com
alt.algebra.help FAQ: http://www.vex.net/~trebla/stuff/aah-faq.txt
sci.math FAQ: http://www.cs.unb.ca/~alopez-o/math-faq/mathtext/math-faq.html
Eric's Treasure Trove: http://mathworld.wolfram.com/letters/
trigonometry without tears: http://www.mindspring.com/~brahms/trig.htm
more FAQs: http://www.mindspring.com/~brahms/faqget.htm
Darrell Ryan
Let the larger number be 58-x
Add 3 to each and divide the larger by the smaller:
(61-x) / (x+3)
To form an equation that relates the given information, recall that
Dividend/Divisor = Quotient+Remainder, and what it means to say, "The
remainder is 16." For example, if you divide 5 (dividend) by 2
(quotient) you get 2 (quotient) + 1/2 (don't forget the actual
"remainder" is the integer part DIVIDED BY the dividend. In our
problem, the equation would be:
(61-x) / (x+3) = 1 + 16/(x+3)
DIVIDEND DIVISOR = QUOTIENT + REMAINDER
This is a fairly simple linear equation that can be solved pretty
easily:
61 - x = x + 3 + 16 ...clear fractions by multiplying each term by
x+3
x = 21
so the two numbers are 21 and (58-21)=37
--
Regards,
Darrell
Brenda Keimig wrote:
>
> Two positive integers have a sum of 58. If 3 is added to each and then the
> larger divided by the smaller, the integer part of the quotient is 1 and the