2007-09-05 04:30:07 · 20 answers · asked by arnawaz b 1 in Science & Mathematics ➔ Mathematics
to be divisible by 2, it need to end in an even number (0, 2, 4, 6, or 8)
To be divisible by 3, the sum of the digits must be divisible by 3.
So, start with the largest 4 digit number and work backwards.
9999 = not div by 2
9998 = not div by 3
9997 = not div by 2 or 3
9996 = Yes, div by 2 and 3.
So, 9996 is the answer
2007-09-05 04:35:22 · answer #1 · answered by Mathematica 7 · 4⤊ 0⤋
If a number should be divisible by 2 and 3, it should be duvisible by 6 (LCM of 2 and 3).
Multiples of 6 ends with numbers 0,2,4, 6, 8 only in unit place where 8 is highest
largest 4 digit number ending with 8 is 9998
the sum of digits in 9998 = (9 + 9+ 9 +8 = 35 = 8), which is not divisible by 3
so next largest numbers`is 9996
the sum of digits is 6 (9996 = 9+9+9+6 = 33 = 6) , whic is divisible by 3.
So the largest 4 digit number divisible by 2 nd 3 is 9996
2007-09-05 05:02:04 · answer #2 · answered by mohanrao d 7 · 0⤊ 0⤋
The number is divisible by 2 and 3 means it is divisible by the number which is LCM of 2 and 3 i.e. 2 × 3 = 6
To find largest 4 digit number
The largest 4 digit number is 9999
Divide this number by the required divisor, in this case it is 6
9999/6 quotient is 1666, so the required number is 1666 × 6 = 9996.
By this method you can find any number with any condition.
2007-09-06 03:46:44 · answer #3 · answered by Pranil 7 · 0⤊ 0⤋
write any 4 digit number which must have o,2,4,6,8.
& now add all 4 digits if sum is a multiple of 3 then that number will be divisible by 2 & 3 both.
eg., 9996
in this case last number is even & sum is
9+9+9+6 = 33.
2007-09-05 18:08:59 · answer #4 · answered by ajay_undefeated 1 · 0⤊ 0⤋
9999 is the largest 4 digit number divisible by 3.
9998 is the largest 4 digit number divisible by 2.
9996 is the largest 4 digit number that is divisible by 2 and 3.
2007-09-05 05:04:24 · answer #5 · answered by Swamy 7 · 1⤊ 0⤋
Find a number which has a digit ending with an even number 0,2,4,6,8 and has a sum of the 4 digits divisible by 3 (eg. 9999 but since it does not end with an even number its wrong). My answer? 9996.
2007-09-05 04:35:50 · answer #6 · answered by nothingtodo007 2 · 2⤊ 0⤋
10000/6 = 1666.67;
So, the largest number divisible by 2 & 3 is
6*1666 = 9996 --- (ans)
2007-09-05 05:02:30 · answer #7 · answered by Devarat 7 · 1⤊ 0⤋
9996 divisible by both 2 and 3
2007-09-05 22:30:48 · answer #8 · answered by sprite 6 · 0⤊ 0⤋
9999 by 3
9998 by 2
9996 by both 2 and 3
2007-09-05 20:49:59 · answer #9 · answered by Anonymous · 0⤊ 0⤋
the number which is divisible by both 2 and 3 is divisible by 6......
larget 4 digit num ---- 9999
9999/6
remainder = 3
Thus, the num is 9996.......
2007-09-05 05:42:09 · answer #10 · answered by Ann 3 · 1⤊ 0⤋