Pls. explain. Thanks!
2006-11-06 16:15:17 · 7 answers · asked by wolfwood 2 in Science & Mathematics ➔ Mathematics
There are exactly 100
108 is the smallest and is 12*9
999 is the largest and is 111*9
Therefore there are 111 - 12 + 1 or 100 such numbers.
(the number of numbers between a and b inclusive is a-b+1)
Nice try heartsensei but that wasnt your original answer. copying others is low.
2006-11-06 16:19:35 · answer #1 · answered by Scott R 6 · 1⤊ 0⤋
The smallest 3 digit number divisible by 9 is 108. The largest is 999. There are 100 numbers including these two. The easiest way to see this is to construct a formula matching the numbers 108 and 999 to 1 thru 100. In this case, you divide by 9 and subtract 11...
108 /9 - 11 = 1
999 /9 - 11 = 100
2006-11-06 16:18:29 · answer #2 · answered by heartsensei 4 · 0⤊ 1⤋
The answer is 100
Y? for each 100 you have 11 three digit numbers that can be divided by nine but (11*9 = 99) BUT 100-99=1 which means during the ninth set you'll have added up 9 1's (1+1+1+1+1+1+1+1+1) which will give your 12 for that last set. So....you have 11 set for 100's through 800's which = 8*11 = 88 plus 12 for set 900's becuase of the ones issue so 88+12 = 100 :)
2006-11-06 16:23:05 · answer #3 · answered by bourqueno77 4 · 0⤊ 1⤋
About 100
2006-11-06 16:17:36 · answer #4 · answered by Anonymous · 0⤊ 1⤋
Take the largest 3 digit number divisible by 9 and divide it. The answer should tell you how many numbers in between that are divisible by 9 as well.
2006-11-06 16:19:24 · answer #5 · answered by Anonymous · 0⤊ 1⤋
I got 99 numbers. I figure the first is 108 and the last is 999, so I took 111 and subtracted 12 from it. I checked it by figuring that there are 11 multibles per 100. 11 times 9 equals 99.
2006-11-06 16:23:30 · answer #6 · answered by Michael T 1 · 0⤊ 2⤋
Just FYI, numbers divisible by nine have digits that add up to multiples of nine.
2006-11-06 16:19:49 · answer #7 · answered by iblori58 2 · 0⤊ 1⤋