if 41y is a multiple of 3,where y is a digit what is the value of y
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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
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