There are 45multiples of 3 between 85 and integer b. What is the largest possible value of b.
$(x-87)/3 + 1 = 45 \implies x=219$
Why is the answer 221?
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4Since neither $220$ nor $221$ are divisible by $3$ you can go a bit further. – lulu Dec 12 '22 at 21:58
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The way to calculate $b$ it is to re-state the question as
There are $45$ multiples of 3 between $\color{red}{84}$ and integer b.
The reason is that $84$ is a multiple of $3$. So any multiple of $3$ that is greater than $84$ will also be greater than $85$.
So, the subsequent $\color{red}{46}$-th multiple of $3$ occurs at
$$84 + (46 \times 3) = 222.$$
So, $b$ must be the largest integer that is strictly less than $222$.
That is, there will be exactly $45$ multiples of $3$ between $85$ and $b$ if and only if $b$ is any element in $\{219, 220, 221\}.$
user2661923
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I tell you why, because the question asks for the highest possible value of b, therefore since the answer is 219, and we use 219 only because it is dividable by 3. However, the true number may be not 219 it can be greater than this and we choose only 219 because it is the first number bellow it that is dividable.
to make it clear, how many numbers that are not dividable by 3 greater than 219? the answer is 220, 221, and that's it. 222 is dividable by 3
so the answer is 221 because it is the largest number that is not dividable by 3 after the 219
