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Q.

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


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a

0,6

b

5,6

c

5,4

d

None of these 

answer is B.

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Detailed Solution

Concept- By looking at the options available, one can say that given no. is divisible by 5 and 3 both. So, applying divisibility rules of 5 and 3 in the given no. will yield a real answer.
Divisibility rule of 5 says that unit digit place shall have 0 or 5.
Divisibility rule of 3 says that the sum of all digits must be a multiple of 3.
Let’s apply these rules in the given no. 357y25x
For it to divisible by 5, x=0 or x=5
Now we have two options for this assumption. We will now apply the divisibility rule of 3 on filtered options.
Trial 1 : For 357y250 to be divisible by 3,
Sum of digits,  3+5+7+y+2+5+0=22+y
For this to be divisible by 3,
0y9
y=2 or 5 or 8
Possible combinations of (x,y):
0,2;0,5;(0,8)
Trial 2: For 357y255 to be divisible by 3,
Sum of digits,3+5+7+y+2+5+5=27+y
For this to be divisible by 3,
0y9
y=0 or 3 or 6 or 9
Possible combinations of (x,y):
5,0;5,3;5,6;(5,9)
Therefore, from all possible combinations we can conclude that (0,6) is correct.
Hence,. the correct option is (2).
 
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