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Insert a digit in place of each “…” to make numbers that are divisible by 6 if it
is possible:
4…6
?


Sagot :

Answer: Either 2, 5, or 8

This means the number 426 is divisible by 6. So are 456 and 486

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Explanation:

A number is divisible by 6 if both of the following are true

  • The number is divisible by 2
  • The number is divisible by 3

This is simply because 6 = 2*3. So if 6 is a factor of a number, then 2 and 3 must be factors.

  • To have 2 be a factor, the units digit must be in the set {0,2,4,6,8} which is the case here (the units digit is 6 in this case). Therefore we know the number is a multiple of 2 regardless of what the other digits are.
  • To have 3 be a factor, the digits must add up to a multiple of 3. Through trial and error, we see that 0 doesn't work because 4+0+6 = 10 which is not a multiple of 3. Same goes for 4+1+6 = 11, but 4+2+6 = 12 is a multiple of 3.

Therefore, 426 is a multiple of 6

Increment that middle digit 2 by 3 and we jump from 426 to 456. Those three digits add to a multiple of 3 as well (4+5+6 = 15). Following that line of logic, we go from 456 to 486 as the last possible three digit number that has these conditions of having 4 first and 6 last, and the number is a multiple of 6.

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In short,

The numbers 426 and 456 and 486 are all multiples of 6 since they are multiples of 2 and 3 at the same time.

So we could replace that middle digit with either 2, 5 or 8.