# When will Raj and Vikram meet?

Here's a fun little question I encountered along one of my many mathematical adventures:

Raj and Vikram are two travelers in ancient India, walking along the Silk road. On the first day of travel, Raj travels 5 yojanas, and on each successive day travels 3 yojanas more than the previous day. Vikram started at the same place and travels the same path, but started 5 days earlier, and travels 7 yojanas a day. On which day of Raj's travel will the two of them meet?

Details and assumptions: A yojana is an ancient Indian unit of measure. The silk road does not cross itself. Raj and Vikram will meet only if they have travelled the same distance. If they meet at any point on the 123rd day (say at 12:34pm), then your answer is that they meet on the 123rd day.

My approach:

Let us first consider Raj.
From the question, Raj covers an initial 5 yojanas and covers 3 yojanas per each succeeding day.
The total distance covered by Raj can be considered as an Arithmetic Progression, $$5,\space 8,\space 11,\space 14,\space 17,\space 20,\space 23,\space ...$$ Let the the initial distance be $a$, the constant addition of distance be $d$ and let total distance be $a_n$ on the $n^{th}$ day, $$\implies a_n = a + (n-1)\cdot d$$ Since $a = 5$ and $d = 3$, $$\implies a_n= 5+(n-1)\cdot 3 = 3n + 2$$

So, on the $n^{th}$ day Raj meets Vikram at a distance of $3n+2$ yojanas.

Now let's consider Vikram.
Vikram started 5 days prior to when Raj started his journey and on each day he traveled 7 yojanas. Let the total distance traveled by Vikram be $x$. $$\implies x = \frac{7 \text{ yojanas}}{1\text{ days}} * 5 \text{days} = 35 \text{ yojanas}$$

So, for Raj to get to that distance, $$\implies 3n+2 = 35$$ $$\implies n = \frac{35 - 2}{3} = 11$$

Thus, Raj and Vikram meet on the $11^{th} \text{ day}$.

But the provided answer is the $7^{th}$ day. Where have I gone wrong? Is there an error in my logic? and most importantly, is there a faster way of doing this question?

You forgot to consider their total distance traveled, and also interpreted the statement "Vikram started ... 5 days earlier" differently than the author intended.

You are correct that the distance Raj travels each day can be represented as an arithmetic progression, namely $a_n = 2 + 3 n$.

However, the total distance Raj has covered at the end of the $n$-th day is $$R_n = \sum_{k = 1}^n a_k = 2 \sum_{k = 1}^n 1 + 3 \sum_{k = 1}^n k = 2 n + 3 \cdot \frac{1}{2} n (n + 1) = \frac{1}{2} n (7 + 3 n).$$

On the contrary, at the end of the first day of Raj's travels, Vikram has already traveled $7 \cdot 6 = 42$ yojana; then $7$ each day afterwards. The total distance Vikram has traveled on the $n$-th day of Raj's travels is thus $$V_n = 42 + 7 \cdot (n - 1) = 35 + 7 n.$$

These can be listed as

$$R_n = 5, 13, 24, 38, 55, 75, 98, \ldots$$ $$V_n = 42, 49, 56, 63, 70, 77, 84, \ldots$$

Thus we can see that at the end of the $6$th day of Raj's travels, Raj has traveled $75$ yojanas, while Vikram has traveled $77$ yojanas. Thus they will meet up on the $7$th day.

• Actually, Kyle, I was already considering the total distance by Raj as an A.P. Was I misinterpreting 3 yojanas more than the previous day? Because if travel 3 yojanas tomorrow, I would have travelled 3 yojanas more than the yojanas I traveled today. Also, yojanas is a really fun word to say.
– Nick
Commented Dec 12, 2013 at 2:24
• I interpret Raj travels 5 yojanas, and on each successive day travels 3 yojanas more than the previous day as he travels 5 yojanas on the first day, then 3 more (i.e. 8) on the second day, then 8 + 3 = 11 on the third day, and so on. ... Now that you bring up your interpretation, I agree that the problem statement isn't clear.
– Kyle
Commented Dec 12, 2013 at 2:30
• Yeah, I guess so too but thank you for making me realize Vikram doesn't stop moving. ..hey wait, shouldn't we consider that these fellows sleep as well.
– Nick
Commented Dec 12, 2013 at 8:17

Just to put another approach out there...

Simply consider the relative pace of Raj and Vikram. After the 0th day, Vikram has traveled 35, which is the total distance Raj will have to make up. This is when Raj starts traveling, and his relative pace to Vikram can be written:

$-2, 1, 4, 7, 10, 13, 16, ...$


The distance that Raj makes up can be written as $\sum_{n=1}^{N}{-2 + 3(n-1)} = -2N + 3N(N-1)$. Setting this equal to 35, we get $N = \frac{7+\sqrt{49 + 3*4*70}}{6} \approx 6.13$. And if we round up, we get 7. I think this is correct.