I know its hard but if you can figure it out and show me how, i would really appreciate it
If 1 x 10^A + 2 x 10^B + 3 x 10^c + 4 x 10^D = 24,130 and
A ≠ B ≠ C ≠ D then what does A/2 + B/4 + C/8 + d/16 equal?
2006-11-21 14:08:07 · 10 answers · asked by VHS123 2 in Science & Mathematics ➔ Mathematics
it twenty four thousand
2006-11-21 14:17:02 · update #1
Cool. Just recall what our place value number system means.
0 is in the 1s col, so it's absent
3 is in the 10s col, so c = 1
1 is in 100s, so a =2
4 is in 1000s, so d =3
2 is in 10,000s so b=4
2006-11-21 14:14:49 · answer #1 · answered by modulo_function 7 · 1⤊ 0⤋
I assume that A, B, C, and D are single digits. Then they are the power indices corresponding to the nonzero digits of 24130, that is, A = 2, B=4, C=1, and D=3.
So A/2 + B/4 + C/8 + D/16 = 2/2 + 4/4 + 1/8 + 3/16 = 2+(5/16).
2006-11-21 14:29:49 · answer #2 · answered by Anonymous · 0⤊ 0⤋
This is a digits problem.
Perhaps this will lead you in the right direction.
20000
4000
100
30
Adding those numbers up gives you 24130
So since 1 x 10^A has to be a number such as 100, 1000, 10000... it depends on what A is raised to... in this case A MUST be 2 (A = 2)
Similar logic applies to the other cases 2 * 10^B. 20000 is 2 * 10000. So B must be 4 (B = 4).
The other cases are C = 1 and D = 3
So...
2/2+4/4+2/16+3/16=37/16
2006-11-21 14:27:06 · answer #3 · answered by Anonymous · 0⤊ 0⤋
2 X 10^4 = 20,000
4 X 10^3 = 4,000
1 X 10^2 = 100
3 X 10^1 = 30
-------------------------
24,130
A = 2, B = 4, C = 1, D = 3
2/2 + 4/4 + 1/8 + 3/16 = 2.3125
As for HOW I did it? Hmmmmm....I just kind of looked at it, and noticed the mantissas of the numbers in scientific notation matched the digits 2, 4, 1, and 3.
2006-11-21 14:26:02 · answer #4 · answered by pack_rat2 3 · 0⤊ 0⤋
the base 10 #'s give away the value of ....
24,130 = 2*10^4 + 4*10^3 + 1*10^2 + 3*10^1
D=3
A=2
C=1
B=4
a/2 + b/4 + c/8 + d/16 =
1 + 1 + .125 + .1875 = 2.3125
(2/2) + (4/4) + (1/8) = (3/16) = 2 (5/16)
2006-11-21 14:19:02 · answer #5 · answered by Brian D 5 · 1⤊ 0⤋
a million. permit S be a series such that for each element x of S there exists a special element x' of S. 2. there is an element in S, we will call it a million, such that for each element x of S, a million isn't equivalent to x'. 3. If x and y are factors of S such that x' = y', then x = y. 4. If M is any subset of S such that a million is an area of M, and for each element x of M, the element x' is likewise an area of M, then M = S. purely as a be counted of notation, we write a million' = 2, 2' = 3, etc. We define addition in S as follows: (a1) x + a million = x' (a2) x + y' = (x + y)' The element x + y is termed the sum of x and y. Now to instruct that a million + a million = 2. From (a1), with x = a million, we see that a million + a million = a million' = 2.
2016-10-17 08:53:18 · answer #6 · answered by Anonymous · 0⤊ 0⤋
A=2
B=4
C=1
D=3
So , A/2 + B/4 + C/8 + d/16 = 1+1 +1/8 +3/16 = 2.313
PS forget matrices, that's just needless complications. It's simple, once you see it it'll become obvious.
2006-11-21 14:14:34 · answer #7 · answered by Anonymous · 0⤊ 0⤋
Dude- there's a 1 in the hundreds digit, a 2 in the ten-thousands digit, a 3 in the tens digit, and a 4 in the thousands digit. How is this hard?
2006-11-21 14:15:39 · answer #8 · answered by moronreaper 2 · 2⤊ 0⤋
Use a matrix to solve this. You'll have to substitute all the number parts in and then use that to solve. If you have a TI-83 or higher, just plug it in to a matrix there.
2006-11-21 14:14:46 · answer #9 · answered by candy 2 · 0⤊ 1⤋
how can the same equation be equal to both 24 and 130 ?
sorry -- could you please clarify
-- this problem sounds interesting
2006-11-21 14:15:11 · answer #10 · answered by ilovemath_pi 2 · 0⤊ 1⤋