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22 Sep 2020, 07:03
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rnemani
We can solve the problem in two easy ways;
First:
Using the number divisible formula
[(Last Divisible number - First Divisible number)/Divisor]+1
The first divisible number after 310 (exclusive) is 312 and the last divisible number within 400 is is 399
so, [(399-312)/3]+1
= (87)/3]+1=30
The other way is
We know the last number and first number, the second number will be 315; thus the gap between the numbers will be 3,
so we can use the nth number formula:
nth number = a1+(n-1)d
399= 312+(n-1)3
399 = 312 + 3n -3
399 = 309 + 3n
3n = 90
n= 30
Ans is D.
27 Oct 2021, 14:31
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Here's my thought process: if the number if divisible by 3, flipping digits wont matter. So finding the number that are divisible by 3 is sufficient.
Looking from 310 to 320: we have 312, 315 and 318 are divisible by 3 since they add up to 6, 9 and 12 respectively.
So a total of 3 multiples every 10 numbers. From 310 to 400 we have 9 sequences of 10 numbers.
Finally, 9 x 3 = 27 , plus the three extras imo which are 330, 360 and 390. so a total of 30 multiples of 3.
It might look long but it took me like 30 seconds to solve. (also this could be a terrible non mathematical way of approaching this i'm just sharing my thoughts here)
Looking from 310 to 320: we have 312, 315 and 318 are divisible by 3 since they add up to 6, 9 and 12 respectively.
So a total of 3 multiples every 10 numbers. From 310 to 400 we have 9 sequences of 10 numbers.
Finally, 9 x 3 = 27 , plus the three extras imo which are 330, 360 and 390. so a total of 30 multiples of 3.
It might look long but it took me like 30 seconds to solve. (also this could be a terrible non mathematical way of approaching this i'm just sharing my thoughts here)
30 Sep 2023, 05:05
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