Question

what is the sum of all positive 1-digit integers that 4221462 is divisible by?

Answer 1

Hello,

n=4221462
n is divisible by 1 as any integer.
n ends by 2==>n is divisible by 2

4+2+2+1+4+6+2=21 is a multiple of 3 ==> n is divisible by 3

62 is not n is divisible by 4 and nor n.

n don't end with 0 or 5 ==>n is not divisible by 5

n  is divisible by 2 and n is divisible by 3==n is divisible by 6

divisibility by 7:

1|231|231
4|221|462

(1*2+3*6+2*4 ) mod 7=(2+4+1) mod 7=0
1*4=4                                                                0+4=4
----------------------------------------------------------
(1*1+3*2+2*2) mod 7= (1+6+4) mod 7= 4

4-4=0 so n is divisible by 7

462 is not divisible by 8; nor n.

4+2+2+1+4+6+2=21 is not divisible by 9, nor n

==>1+2+3+6+7=19

Answer 2

Answer:

The sum of all positive 1-digit integers that 4221462 is divisible by is 19.

Step-by-step explanation:

All integers are divisible by 1, so this is also divisible by 1.

4221462 is an even number. So, this is divisible by 2.

The sum of the digits is $4+2+2+1+4+6+2=21$.

21 is divisible by 3 and 7, so this is divisible by 3 and 7 also.

The last two digits of this number are 62. 62 is not divisible by 4, so it is not divisible by 8 either.

This number is also divisible by 6, because it is divisible by two and three.

It is not divisible by 9. It is not divisible by 5.

Hence, the sum of all positive 1-digit integers will be : $1+2+3+6+7=19$

Therefore,

The sum of all positive 1-digit integers that 4221462 is divisible by is 19.