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if t = 137486x225, where x is a single digit. Determine the value(s) of x such that 3|t. Which values of x make t divisible by 9?
My soluctions
x = 0 to make it divisible by 3
x = 4 to make it divisible by 9
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In this case you could just brute-force it, but in general the trick to check whether something is divisible by 3 is to add all its digits together. If the sum of its digits is divisible by 3, then so is the number. (Same goes for the sum being divisible by 9)
Here, the sum is 38+x.
This is divisible by 3 when x = 1, 4 or 7 and divisible by 9 when x = 7.
Why did the vector cross the road?
It wanted to be normal.
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