2007-08-16 03:30:51 · 6 answers · asked by nick1995 1 in Science & Mathematics ➔ Mathematics
The divisibility rule is to take the last 3digits and minus from the rest. If the answer is divisible by 7,11 and 13, then it is divisible. abc-abc=0.
rule: any number divided by 0 = 0. but if 0divides0 then u get infinity.
i think i know you..... u GEP?
2007-08-16 03:45:45 · answer #1 · answered by Az 3 · 0⤊ 0⤋
Abcabc
2016-11-09 09:45:07 · answer #2 · answered by nader 4 · 0⤊ 0⤋
abcabc=abc(1001)
1001 is divisible by 7
1001=7(143)
1001 is divisible by 11
1001=11 (91)
Hence abcabc is divisible by 7 and 11.
Check it for 13 by yourself
2007-08-16 03:40:29 · answer #3 · answered by iyiogrenci 6 · 0⤊ 1⤋
note 1001 is divisiable by 7, 11 and 13.
abcabc = 1001 * abc
and we are done.
2007-08-16 03:39:41 · answer #4 · answered by doctor risk 3 · 2⤊ 1⤋
a = 7
b = 11
c = 13
AxBxCxAxBxC = 7x11x13x7x11x13 = 1002001
1002001/7 = 143143
1002001/11 = 91091
1002001/13 = 77077
That should be the answer to your question. If you have a choice, make a = 7, b=11, and c=13. Your work should look something like this.
2007-08-16 03:42:29 · answer #5 · answered by William H 1 · 0⤊ 1⤋
7x11x13 = 1001. The smallest 6 digit multiple of this is 100100. Multiply this by ab.c, and = abcabc
2007-08-16 03:42:04 · answer #6 · answered by John V 6 · 0⤊ 2⤋