Find a number so that when the digit 3 is put in front of it, the new number is six times the original number. Can you find more than one number
2007-02-24 06:33:55 · 6 answers · asked by asdfasdffdas 1 in Science & Mathematics ➔ Mathematics
Start with a single digit number, say 1 through 9, call it N.
Now appending a 3 in front of N yields N+30 ( and example makes this clear. Suppose N=2, then appending the 3 gives 32 ).
So you are looking for N with 6*N = 30 + N, which solves to N=6.
Now try a 2 digit number NN. Appending the 3 adds 300, so the equation to solve is 6*NN = NN+300 which gives NN = 60.
The pattern is clear: any number of the form 6, 60, 600, and so forth will work.
2007-02-24 06:46:25 · answer #1 · answered by Anonymous · 0⤊ 1⤋
Heinz is right for the first half of the question.
Now can you find more numbers.
What happens with 60 instead of 6.
6^2=36 (or 6 * 6)
How about 600 * 6?
How many more on the same pattern
2007-02-24 14:53:50 · answer #2 · answered by U-98 6 · 0⤊ 0⤋
6 (36)
no.... six is unique in that is it the only EVEN number that when multiplied by itslef ends with a 6
2007-02-24 14:43:42 · answer #3 · answered by A H 1 · 0⤊ 0⤋
6
36
2007-02-24 14:42:21 · answer #4 · answered by Anonymous · 0⤊ 0⤋
assuming it is between 1 and 10,
6x = 30+x
x = 30/6 + x/6
x - x/6 = 5
5x/6 = 5
x = 6
assuming it is between 10 and 100,
6x = 300+x
x = 300/6 + x/6
x - x/6 = 50
5x/6 = 50
x = 60
etc.
2007-02-24 14:46:38 · answer #5 · answered by math freak 3 · 0⤊ 0⤋
Put 3 in front of 100,000 to make 3,100,000. Put a 3 in front of 3,100,000 to make 33,100,00. That's two and you can make thousands more.
2007-02-24 14:37:50 · answer #6 · answered by spencermurraygaunt 2 · 0⤊ 2⤋