If the number 357*25* is divisible by both 3 and 5, then the missing digits in the unit's place and the thousandth's place respectively are :
A. 0, 6
B. 5, 1
C. 5, 4
D. None of these
Answer: Option D
Solution(By Examveda Team)
Let the unit's place be x and the thousand's place be y.Then, 357y25x is divisible by 5 only when x = 0 or x = 5
Also, this number is divisible by 3 only when sum of its digits is divisible by 3
So, (22 + x + y) must be divisible by 3
∴ x + y = 2
Taking x = 0, we get y = 2
So, the unit place = 0 and thousand's place = 2
Related Questions on Number System
Three numbers are in ratio 1 : 2 : 3 and HCF is 12. The numbers are:
A. 12, 24, 36
B. 11, 22, 33
C. 12, 24, 32
D. 5, 10, 15
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