Question

If the number 357y25x is divisible by both 3 and 5 then find the missing digit in the unit’s place and the thousand place respectively are
(A). 0,6
(B). 5,6
(C). 5,4
(D). None of these

Answer 1

Hint: Take the given number in terms of x,y. Now look at the divisibility rules of 5,3. By each rule conclude the value of each variable. Write all the possible cases you get from divisibility rules and then verify it yourself. Apply the divisibility rule of 5 first, as it is easy. Now, you set the condition on one variable. Then use the divisibility rule of 3 to derive a condition on another variable, which will in turn convert as your result.

Complete step-by-step solution -
Divisibility rule of 5: The last digit that is the units digit of the number must be either 0 or 5 for the condition that the number is divisible by 5.
Divisibility rule of 3: The sum of all the digits of the given number must be a multiple of 3. It must be any integral multiple for it to be divisible by 3.
Given number in the question, is given be the form of:
$357y25x$.
By using the divisibility rule of 5, we can say that the unit digits must be 0 or 5. By the number we get units digits $ = x$.
So, $x = 0$ (or) $x = 5$. We get two cases of x.
Now, we will apply divisibility rule of 3 on 2 cases respectively:
Case I: Let the value of x be 0 in this case.
By above condition, the number given in the question is:
$357y250$
The sum of digits of the above number can be written as:
$3 + 5 + 7 + y + 2 + 5 + 0$.
By simplifying it, we get the value of sum of digits as:
$22 + y$.
For $\left( {22 + y} \right)$ being multiple of 3 and as y is digit it must be $0 \le y \le 9$. So we get y values as 2 or 5 or 8.
So, $\left( {x,y} \right)$ can be written as $\left( {0,2} \right);\left( {0,5} \right);\left( {0,8} \right)$ ………………..(1)

Case II: Let the value of x be 5 in this case.
By above condition, the number in the suggestion is written in form:
$357y255$
By adding all the digits of the above number, we get it as:
$27 + y$.
For $\left( {27 + y} \right)$ being multiple of 3 and as y is digit it must be $0 \le y \le 9$. So we get y values as 0 or 3 or 6 or 9.
So, $\left( {x,y} \right)$ can be written as $\left( {5,0} \right);\left( {5,3} \right);\left( {5,6} \right);\left( {5,9} \right)$.
So, writing altogether we get \[\left( {0,2} \right);\left( {0,5} \right);\left( {0,7} \right);\left( {5,0} \right);\left( {5,3} \right);\left( {5,6} \right);\left( {5,9} \right)\].
Therefore, option (b) is the correct answer.

Note: Be careful after considering the sum of digits you must take care of the value of y must be in $\left[ {0,9} \right]$ (as it is digit). Students generally forget this condition. All the other values are also correct. We selected the option which is only one of possible answers. But don’t forget to verify all the cases just by dividing 3 and 5.