Question

Please slove the problems

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Answer 1

Answer:

1) value of y=0

2)value of z=0

Answer 2

Answer:

Answer 1

Since the number 21y5 is a multiple of 9.

So, the sum of its digits 2+1+y+5=8+y is a multiple of 9.

∴(8+y) is either 0 or 9 or 18 or ...

Since y is a digit, so (8+y) must be equal to 9.

i.e., 8+y=9

⇒y=9−8=1

Answer 2

see attachment

answer 3

in second attachment

answer 4

Answer

Since 31z5 is a multiple of 3 ,the sum of its digits  must be a multiple of 3 i.e. 3+1+z+5=9+z.

Hence 9+z is a multiple of 3

Since z is a single digit which can value from 0−9,

Possible values of z will be,

9+0=9

9+3=12

9+6=15

9+9=18

Thus, there can be four possible values i.e. 0,3,6 or 9