Category: Arithmetic
"Published in Newark, California, USA"
Find the missing digit so that it becomes divisible by 6 for
a. 45?673
b. 34562?
Solution:
a. Consider the given number
Since the last digit of a given number is an odd number, then it is not divisible by 6. A number is divisible by 6 if it is both divisible by 2 and 3. All even numbers are divisible by 2. There's
nothing that we can do in
order to become divisible by 6 since the last digit of a given number is
not an even number. You can assign any number to the missing digit but
still, the given number will never become divisible by 6.
b. Consider the given number
A number is divisible by 6 if it is both divisible by 2 and 3. In short, an even number that is divisible by 3. Add all the digits of the given number as follows
Since 20 is not a multiple of 3, then we need to add a number so that it becomes a multiple of 3. So, 20 + 1 = 21.
We can add also 4 (1 + 3) so that 20 + 4 = 24. We can add also 7 (1 + 3
+ 3) so that 20 + 7 = 27. 7 is the highest digit that we can use
because 7 + 3 = 10 will be a two digit number and we need to use only
one digit to fill up the missing digit. The numbers 1, 4, and 7 are the right digits to fill up the missing digit in order to become the given number divisible by 3.
Since we want a given number to be divisible by 6, then we have to choose 4 as a digit because the missing digit is the last digit. The last digit must be an even number so that the given number becomes divisible by 6. Therefore, the possible number is 345624.
Category: Arithmetic
"Published in Newark, California, USA"
Find the missing digit so that it becomes divisible by 5 for
a. 94?763
b. 32178?
Solution:
a. Consider the given number
Since the last digit of the given number is 3, then it is not divisible by 5. A number is divisible by 5 if the last digit is 5 or 0. There's
nothing that we can do in
order to become divisible by 5 since the last digit of a given number is
not 5 or 0. You can assign any number to the missing digit but
still, the given number will never become divisible by 5.
b. Consider the given number
A number is divisible by 5 if the last digit is 5 or 0. Since the missing digit is the last digit, then we can assign 5 and 0 so that the given number becomes divisible by 5. Therefore, the possible numbers are 321785 and 321780.
Category: Arithmetic
"Published in Newark, California, USA"
Find the missing digit so that it becomes divisible by 4 for
a. 5?627
b. 6721?
Solution:
a. Consider the given number
Since
the last two digit of a given number which is 27 is not a multiple of 4, then the
given number is not divisible by 4. There's nothing that we can do in
order to become divisible by 4 since the last digit of a given number is not a multiple of 4. You can assign any number to the missing digit but
still, the given number will never become divisible by 4.
b. Consider the given number
A number is divisible by 4 if the last two digit is a multiple of 4. Since 1 is located at the second to the last digit, then we can assign 2 and 6 to the last digit so that 1 becomes 12 and 16 which are the multiples of 4. Therefore, the possible numbers are 67212 and 67216.
Category: Arithmetic
"Published in Newark, California, USA"
Find the missing digit so that it becomes divisible by 3 for
a. 35?83
b. 7895?
Solution:
a. Consider the given number
A number is divisible by 3 if the sum of the digits is a multiple of 3. If you add the rest of the digits, the sum will be equal to
Since 19 is not a multiple of 3, then we need to add a number so that it becomes a multiple of 3. So, 19 + 2 = 21. We can add also 5 (2 + 3) so that 19 + 5 = 24. We can add also 8 (2 + 3 + 3) so that 19 + 8 = 27. 8 is the highest digit that we can use because 8 + 3 = 11 will be a two digit number and we need to use only one digit to fill up the missing digit. Therefore, the possible numbers are 35283, 35583, and 35883.
b. Consider the given number
A
number is divisible by 3 if the sum of the digits is a multiple of 3.
If you add the rest of the digits, the sum will be equal to
Since 29 is not a multiple of 3, then we need to add a number so that it becomes a multiple of 3. So, 29 + 1 = 30. We can add also 4 (1 + 3) so that 29 + 4 = 33. We can add also 7 (1 + 3 + 3) so that 29 + 7 = 36. 7 is the highest digit that we can use because 7 + 3
= 10 will be a two digit number and we need to use only one digit to
fill up the missing digit. Therefore, the possible numbers are 78951, 78954, and 78957.
Category: Arithmetic
"Published in Vacaville, California, USA"
Find the missing digit so that it becomes divisible by 2 for
a. 245?7
b. 3259?
Solution:
a. Consider the given number
Since the last digit of a given number which is 7 is an odd number, then the given number is not divisible by 2. There's nothing that we can do in order to become divisible by 2 since the last digit of a given number is an odd number. You can assign any number to the missing digit but still, the given number will never become divisible by 2.
b. Consider the given number
The given number is divisible by 2 if the last digit is an even number. Since the missing digit is the last digit, then we can assign 0, 2, 4, 6, and 8 in order to become divisible by 2. Therefore, the possible numbers are 32590, 32592, 32594, 32596, and 32598.
Category: Arithmetic
"Published in Vacaville, California, USA"
Perform the divisibility test of a number by 1 to 15 for
Solution:
Consider the given number above
Without using a calculator, we can perform the divisibility test by using divisibility rules.
Divisibility by 1:
All numbers are divisible by 1 because 1 is a universal factor. Any number divided by 1 is always the same number.
Divisibility by 2:
Since the given number is an odd number, then it is not divisible by 2. All even numbers are divisible by 2.
Divisibility by 3:
To test a given number, add all the digits of a given number as follows
Since 6 is a multiple of 3, then the given number is divisible by 3.
Divisibility by 4:
Consider the given number
Since the last two digits of a given number is an odd number, then the given number is not divisible by 4. The multiples of 4 are all even numbers.
Divisibility by 5:
Since
the last digit of a given number is 5, then the given number is
divisible by 5. A number that ends with 5 or 0 is divisible by 5.
Divisibility by 6:
Since the given number is an odd number, then it is not divisible by 6. If an even number that is divisible by 3, then it is divisible by 6.
Divisibility by 7:
To test a given number, double the last digit and then subtract it to the rest of the digits as follows
Since the result of this process is 0, then the given number is divisible by 7. If the result of the process is a multiple of 7 or 0, then the given number is divisible by 7.
Divisibility by 8:
Consider the given number
Since
the last three digits of a given number is an odd number, then the given
number is not divisible by 8. The multiples of 8 are all even numbers.
Divisibility by 9:
To test a given number, add all the digits of a given number as follows
Since 6 is not a multiple of 9, then the given number is not divisible by 9.
Divisibility by 10:
Since
the last digit of a given number ends with 5, then the given number is not
divisible by 10. A number that ends with 0 is divisible by 10.
Divisibility by 11:
To test a given number, create two groups of the sum of alternating digits and then get their difference as follows
Since -1 is not a multiple of 11, then the given number is not divisible by 11.
Divisibility by 12:
Since the given number is divisible by 3 only and not divisible by 4 because it is an odd number, then the given number is not divisible by 12.
Divisibility by 13:
To test a given number, multiply the last digit by 4 and then add it to the rest of the digits as follows
Since 39 is a multiple of 13, then the given number is divisible by 13.
Divisibility by 14:
Although
the given number is divisible by 7 but it is not divisible by 2 because it is an odd number.
Because of this, the given number is not divisible by 14.
Divisibility by 15:
Since the given number is both divisible by 3 and 5, then it is divisible by 15.
Therefore,
1365 is divisible by 1, 3, 5, 7, 13, and 15.
