FInd a number so that when the digit 1 is put in front of it, the new number is 8 times the original number. explain your answer
2007-02-24 06:25:01 · 5 answers · asked by asdfasdffdas 1 in Science & Mathematics ➔ Mathematics
assuming it is between 1 and 10,
8x = 10+x
x = 10/8 + x/8
x - x/8 =1.25
7x/8 = 1.25
x = 1.4285714285714...
assuming it is between 1 and 10,
8x = 100 + x
x = 100/8 + x/8
x - x/8 =12.5
7x/8 = 12.5
x = 14.285714285714...
2007-02-24 06:38:58 · answer #1 · answered by math freak 3 · 0⤊ 0⤋
10/7 and 80/7 (1.428571429... and 11.428571429...)
100/7 and 800/7 (14.28571429... and 114.28571429...)
1000/7 and 8000/7 (142.8571429... and 1142.8571429...)
etc.
I came up with a system of equations.
If x is the smaller number and y is the bigger, then
8x = y
and based on place value
x + 10 = y
x + 100 = y
x + 1000 = y
etc.
and found the intersection of 5x = y and the place value graph on my caluclator, which indicates a pattern that the answer is always a multiple of 10/7 and 80/7.
I used substitution to come up with the fraction equivalents since TI calcs aren't smart enough to do all the work!
Fun, thanks!
2007-02-24 06:40:05 · answer #2 · answered by nicolosi81 2 · 0⤊ 0⤋
original 133
number 064
Put one in front 1064
calcultor 8+8+8+8 till you get to 1064-8=133
I don't know if it is right but it is a try. I hope i helped...
2007-02-24 06:33:58 · answer #3 · answered by cheeky_little_monkey! 2 · 0⤊ 1⤋
â the equation should look: 10^n +x =8*x, where x and n both integers to be found.
Thus 7x = 10^n, hence x=10^n /7 is not integer with any n; no solution!
2007-02-24 06:51:19 · answer #4 · answered by Anonymous · 0⤊ 0⤋
There is no such whole number possible.
2007-02-24 06:33:53 · answer #5 · answered by LEPTON 3 · 0⤊ 1⤋