how can you solve:5a+10b+20c+50d=1995 and a>b,c,d and a+b+c+d=100. also a=3c. Sorry for the wordiness!?

Question

I think diophantine equations is the way to go, but I can't figure them out with 4 unkown values. anyway anyone who answers - thanks in advance!!!

I mean that 'a' is larger than any other value... is that the confusing bit?

sorry, c may be larger than d (but it isn;t a set fact so it may not be)

Answer 1

ok.. lemme see..

divide 5a+10b+20c+50d=1995 by 5 and get
a + 2b + 4c + 10d = 399 ==> since even numbers multiplied by any number produce even numbers.. and even numbers added equal even numbers.. then.. a MUST be an ODD number

and .. since a is an odd number.. then 3c is odd, therefore c must be odd also. Also, if a > b, c, d.. then a must be positive and c must be positive.

now subtracting equations:

..a + 2b + 4c + 10d = 399
-(a + b + c + d = 100 )
--------------------------------
0 + b + 3c + 9d = 299

but 3c = a.. so we can substitute this in and get

b + a + 9d = 299

a + b + 9d = 299
-(a+b+c+d=100 )
----------------------
+ 0+0 -c +8d = 199
or... 8d - 199 = c
and since a = 3c.. a = 3(8d - 199) = 24d - 597... and since a is odd.. then d must be even and if d is even then b is even.

a = odd = 3c = 24d - 597 = positive number
b = even
c = odd = 8d - 199 = positive number
d = even

take a = 3c and substitute the 3c for a in
5a +10b+20c+50d=1995 ..
5(3c) + 10b + 20c + 50d = 1995
or... 15c + 10b + 20c + 50d = 1995
combining the c's and get 10b + 35c + 50d = 1995

but also .. if you take 5(a + b + c + d = 100) and subtract it from the first equation.. you get..

..5a+10b+20c+50d=1995
-(5a +5b + 5c + 5d = 100)
-------------------------------
..........5b + 15c + 45d = 1495
subtract 5 of the equation..
a+b+c+d=100 ..o..r 5a + 5b + 5c + 5d = 500

and you get 5b + 15c + 45d = 1495

hmmm... ok.. a = 3c.. so .. 15c = 5a = 5(24d - 597)
or... 5b + 5(24d - 597) + 45d = 1495
multiply out and combine.. get.. 5b + 165d - 2985 = 1495
or 5b + 165d = 4480
divide by 5... and get b + 33d = 896
or... b = 896 - 33d

a = odd = 3c = 24d - 597 = positive number
b = even = 896 - 33d
c = odd = 8d - 199 = positive number
d = even

now 24d - 597 > 0
so 24d > 597
d > 597/24 ... or.. d > 24.875
check 8d - 199 > 0
or 8d > 199
or d > 199/8..or.. d > 24.875 ok

guess d = 26
a = odd = 24(26) - 597 = positive number = 27
b = even =896 - 33(26) = 38 [no good, a > b, c, d]
hmmmm.. let's try
24d - 597 > 896 - 33d
57d > 1493
d > 26.19
ok.. try d = 28

a = odd = 24(28) - 597 = positive number = 75
b = even =896 - 33(28) = -28
c = odd = 8(28) - 199 = positive number = 25
d = even = 28
a + b + c + d = 100 becomes.. 75 -28 + 25 + 28 = 100 ok
try 5a+10b+20c+50d=1995
5(75) + 10(-28) + 20(25) + 50(28) = 1995 OK!!

a = 75
b = -28
c = 25
d = 28

Whew!

Answer 2

It's like making chnage for $19.95 using nickels, dimes, twenty cent peices :-) and half dollars.

Since a = 3c we can simplify to

5*(3c) + 10b + 20c + 50d = 1995 and 3c + b + c + d = 100 or

35c + 10b + 50d = 1995 and 4c + b + d = 100

The first simplifies some more to 7c + 2b + 10d = 399

Double the = 100 one and subtract from the = 399 one to get
8d - c = 199. The next larger multiple of 8 above 199 is 200 = 8 * 25 so c is 1 more than a multiple of 8: c = 1, 9, 17, 25, ...

c = 1 => a = 3, 8d = 200 or d = 25 is greater than a

c = 9 => a = 27, 8d = 208 or d = 26
b = 100 - (9 + 27 + 26) = 100 - 62 = 38 is greater than a

c = 17 => a = 51, 8d = 216 or d = 27
b = 100 - (17 + 51 + 27) = 100 - 95 = 5 so a is the max

So:

a = 51
b = 5
c = 17
d = 27

a + b + c + d = 51 + 5 + 17 + 27 = 56 + 17 + 27 = 73 + 27 = 100 ok

5a + 10b + 20c + 50d = 5*51 + 10*5 + 20*17 + 50*27 = 255 + 50 + 340 + 1350 = 305 + 340 + 1350 = 645 + 1350 = 1995 ok

So one possible solution is:

a = 51
b = 5
c = 17
d = 27

Lesser values of c violate the a is the max constraint. Greater values of c make b negative, which may or may not be okay.

Answer 3

In short. Solve
5a+10b+20c + 50d =1995
a=3c
a+b+c+d=100
a is the biggest of all

By eliminating d and c you still have an equation with a and b
while a>b. In a graph it is a half line.
It is impossible to solve a and b. For that you need two more equations.

Answer 4

see ur question. i think it should be c Sorry

I have solved it by Trial and Error method.
33,29.75,11,26.25
36,25.625,12,26.375

Answer 5

thats a hard question!