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The first whole number divisible by 3 is 102 and the last one 399.

Let n be the number of whole numbers between 102 and 399

102 + (n - 1)x3 = 399 (this is an arithmetic progression)

Solving n, n-1 = (399 - 102)/3 = 99

n = 100

Since even whole numbers among these 100 will be divisible by 6, the number not divisible is half of 100, i.e. 50.

regards,

lpokbeng

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Q: How many whole number between 100 and 400 are divisible by 3 but not divisible by 6?
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