what is the sum of all possible digits that could fill the blank in 47_21 so that the resulting five-digit number is divisible by 3?
2006-09-02 02:08:32 · 8 answers · asked by Ninik 3 in Science & Mathematics ➔ Mathematics
If the sum of the digits in any number is divisible by 3, then the whole number is also divisible by 3.
The sum of the digits is already 14, so we need numbers which, added to 14, will be divisible by 3...
0 - no
1 - yes
2 - no
3 - no
4 - yes
5 - no
6 - no
7 - yes
8 - no
9 - no
so 1 + 4 + 7 = 12
2006-09-02 02:10:09 · answer #1 · answered by hawk22 3 · 0⤊ 0⤋
1+4+7 = 12
2006-09-02 10:34:24 · answer #2 · answered by bob h 3 · 0⤊ 0⤋
12
For a number to be divisible be 3 the sum of its digits should be divisible by 3.
2006-09-02 09:18:52 · answer #3 · answered by nayanmange 4 · 0⤊ 0⤋
1,4,7
Here is the rule for "divisible by 3"--add up the digits in the number. If that number is divisible by 3, then the big number is as well.
47121
47421
47721
Cool, huh? Hey, hawk22, I see you figured it out too! Nice work.
2006-09-02 09:13:20 · answer #4 · answered by EXPO 3 · 0⤊ 1⤋
1, 4, and 7 can be plugged in because they would all make it divisible by 3
2006-09-02 11:55:11 · answer #5 · answered by Terryn M 3 · 0⤊ 0⤋
47721
47421
47121
are all div by 3
7+4+1=12
12
2006-09-02 09:18:35 · answer #6 · answered by Zero 1 · 0⤊ 0⤋
I just draw a blank on this one. I just wanted to add my two cents and get two points but also to remember this question to think about it but the answers seem legit.
2006-09-02 09:51:21 · answer #7 · answered by radtadstar 2 · 0⤊ 0⤋
7,4,1
2006-09-02 09:17:54 · answer #8 · answered by lktt71 2 · 0⤊ 0⤋