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.
