I am a 3 digit number divisible by 7 but not 2, sum of digit is 4 received 301 for answer?

Question

My son is in the 6th grade and received this homework without any knowlege of this using divisibilty before. I thank you for the answer but would like the answer to be given more detail on how he could explain how the answer was given. I would the answer to be step by step so that if he had to do another questioning similar he would be able to use extensive response with his answer. Thank you for answering this question so promptly.

Answer 1

You know it's an odd number...Because it's not divisible by 2.

7 * 15 = 105 (adds up to 6)
7 * 16 = 112 (divisible by 2)
7 * 17 = 119 (adds up to 11)
7 * 18 = 126 (divisible by 2)

So you know you don't have to worry about 7 * even number, because that results in an even number.

So just keep adding 14 to previous numbers..see what you get...

119
133
147
161
175
189
203
217
231
245
259
273
287
301 - first one that adds up to 4

Answer 2

Given that the number cannot be divided by 2, the last digit must be an odd number and given that the sum of the digits is 4, this rules out 4+0 and 2+2, leaving 3+1 as the only odd numbers adding up to 4. So, the only possible 3 digit numbers using 3 and 1 are 130, 103, 301 and 310. Furthermore, 310 and 130 can be divided by 2, so these can be eliminated, leaving 301 and 103. If you divide each by 7 the only number which does not give a decimal answer is 301. Hope this helps.

Answer 3

There are only a few 3-digit numbers whose digits sum to 4:
400, 301, 310, 220, 211, 202, 130, 121, 112, 103

Since the number here is not divisible by 2, it must be odd, so eliminate all the even numbers from the above list:
301, 211, 121, 103

Trial and error quickly shows that only 301 is divisible by 7.

Answer 4

If the sum of the digits is 4 and the number is not divisible by 2 (in other words, odd), the last digit must be either 1 or 3.

If the last digit was 3, the only number with digits that add to 4 would be 103, which is not divisible by 7. Therefore, the last digit must be 1.

The only possibilities with a last digit of 1 are 121, 211, and 301. If you try to divide each by 7, you soon figure out that only 301 divides evenly.

Answer 5

I would explain the process like this: It's a 3 digit number, so you know it's between 100 and 999. Since the sum of the digits is 4, the number can't be higher than 400, lower than 103, or have any digit higher than 3 in it. Since the number can't be divided by 2, you can eliminate all even numbers. This leaves you with numbers between 103 and 301 with a 0, 1, 2, or 3 in them. Just check one by one to see which one is divisible by 7. Hope this is what you were looking for.

Answer 6

First figure out the combinations of digits which sum to 4...

4+0+0
3+1+0
2+2+0
2+1+1

Since the number must be odd, it must contain an odd digit, therefore we can exclude 4+0+0 and 2+2+0

The only odd numbers we can construct with 2+1+1 and 3+1+0 are 211, 121, 103, 301,
Dividing by 7, we see only 301 leaves no remainder.

Answer 7

there are only so many combinations you can have to solve this equation. Start with the 3 digit number that has a sum of 4. Your possibilities are 004, 013, 022, 031, 121, 112, 202, 211, 220, 310, 301, 400.

Then take these numbers and remove all even numbered candidates. (they are divisible by 2)

Then divide the remainder by 7 and 301 is your answer.

Answer 8

Alright, we can break it down to easier things. Like that it's three digit, that's obvious.
The sum of the digits is 4. so 112, 121, 103, 130, 211, 202, 220, 301, 310. I think that's all, but don't yell if It's not, it's the idea of it :).

But not divisible by 2, so it rules out any even number. That cuts the possible answers to half of what it was:
121, 103, 211, 301

Then it has to be divisible by 7, the only one left is 301.


I'm pretty much just saying that my method of choice with a question like this would be just working it out. I went backwards, but that was just because it seemed easiest to me. I don't think it really matters what you start with, since it's not really order of operations or anything.

Answer 9

Assume x,y,z are the three numbers.
Then the number is
100*x + 10*y + z
x,y,z belongs to {0,1,2,3,4} // sum of numbers is 4
Since the number is not divisible by 2
z belongs to {1, 3}
if z = 1
x + y = 3
(x,y) = (3,0), (2,1), (1,2)
number divisible by 7 is 301
if z= 3
x+y = 1
(x,y) = (1,0)
Thus number is 301

Answer 10

301 is correct