how many are numbers up to 2000 divisible by 5 and 9?
Answers
Answer:
A n-digit number is written as such: a(n)a(n-1)….a(3)a(2)a(1). We can represent it as a(n)x10^(n-1)+a(n-1)x10^(n-2)+….+a(3)x10^2+a(2)x10+a(1).
Or a(n)x999….9 {n-2 times of 9} +a(n)
+a(n-1)x999….9 {n-3 times of 9} + a(n-1)
…….….
a(3)x99 +a(3)
a(2)x9 +a(2)
+a(1)
We will notice that all the product of 9 are already multiple of 9, hence the rest has to be a multiple of 9 so that the n-digit number be divisible by 9… We can say that a number is divisible by 9 if the sum of its digits is divisible by 9. Example, 423 is divisible by 9 because 4+2+3=9, which is divisible by 9 of course.. In fact 423=47x9…
2) On the other hand, a number is divisible by 5 if it ends with 5 or 0.
Now, you can use 1) and 2) to come up with the final solution…
Step-by-step explanation:
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Answer:
40
Step-by-step explanation:
5 and 9 are co-prime number so the product of the numbers is 45 and 45 have 2 divisions in 100 and there are 20 100 in 2000 so 20 ×2 = 40
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