A certain positive integer has the interesting property that to multiply it by 7, all that is needed is to take the 7 from its right end and place it at the beginning (far left). What is the number?
I know that the last digit is 7 and the digit before 7 is 9
2007-06-07 13:12:58 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I think the answer is endless, but I'm not sure
2007-06-07 13:17:30 · update #1
The answer doesn't exist.
7 at the far right end can come in two ways either the number has the rightmost digit 1 with 0 carry or the just previous to the right most digit be 9 with carry 7. but the second option can't be fulfilled because we can't have a carry of 7 when multiplying a number with 7.hence the second option is ruled out.
Now we are left with the first choise that is the rightmost digit be 1 with 0 carry. that can only be obtained when the second right most digit be 0. Since according to question we need 1 in place of 0 when multiplied by 7 means the carry must be 1 at that place. That means the number left to the 0 must be 2 which can't be possible satisfying 1 in place of 0 and 0 in place of 2 after being multiplied.
Hope that clarifies your answer.
2007-06-07 13:55:04 · answer #1 · answered by om j 2 · 1⤊ 0⤋
7
2007-06-07 20:18:28 · answer #2 · answered by psychoffspring 2 · 0⤊ 2⤋
Wow
I feel bad for you and your class!
2007-06-07 20:17:04 · answer #3 · answered by Luis B 2 · 0⤊ 1⤋
Are you kidding?
there's no answer to this question.
2007-06-07 20:36:33 · answer #4 · answered by Alberd 4 · 1⤊ 0⤋