4) If the number 357*25*is divisible by 3 and 5, find the missing digits in the unit's place and the thousand's place respectively.
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Let the required number be 357y25x.
Then, for divisibility by 5, we must have x = 0 or x = 5.
Case I. When x = 0
Then, sum of digits = (22 + y). For divisibility by 3, (22 + y) must be divisible by 3.
∴y=2or5or8
∴Numbersare(0,2)or(0,5)or(0,8)
Case II. When x = 5.
Then, Sum of digits = (27 + y). For divisibility by 3, we must have y = 0 or 3 or 6 or 9.
∴Numbersare(5,0)or(5,3)or(5,6)or(5,9)
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