Question

If the number 357×25× is divisible by both 3 and 5, then the
missing digits in the unit's place and the thousandth place
respectively are​

Answer 1

Answer:

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)

Answer 2

Answer:

(5,0),(5,3),(5,6),(5,9)

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