A Notorious Number Theory Problem

Using the formula $\sum_{k=1}^n k^2 = \frac16 n(n+1)(2n+1)$ we require $\frac16 \left( c(c+1)(2c+1)-(b-1)b(2b-1) \right) \equiv 3 \pmod5$

where $c=a+b$ . Since $5$ and $6$ are coprime, we can just multiply over to get $c(c+1)(2c+1)-(b-1)b(2b-1) \equiv 18 \equiv 3$

Since both terms have the same form, the easiest approach here is just to check cases, as per the table below:

where the results are taken modulo $5$ . It's clear that the only way we can get a difference of $3$ is by taking $b-1 \equiv 1$ and $c \equiv 3$ . So the units digit of $c$ is either $3$ or $8$ ; we're told that $c$ is odd, so this leaves $\boxed3$ as the answer.

Define $s(b,a) = \displaystyle \sum_{k=b}^{b+a} k^2$ and consider modulo $5$ of $s(b,a)$ .

$\begin{aligned} s(1,a) & \equiv 1 + 4 + 9 + 16+25 + 36 + 49+64+81+100 + \cdots + (1+a)^2 \text{ (mod 5)} \\ & \equiv 1 + 4 + 4 + 1+ 0 + 1 + 4 + 4+ 1+0 + \cdots + (1+a)^2 \text{ (mod 5)} \end{aligned}$

We note, for $b \bmod 5 = 1$ , $s(b_1,a) \bmod 5 = \begin{cases} 1 & \text{if }a \bmod 5 = 0 \\ 4 & \text{if }a \bmod 5 = 2 \\ 0 & \text{otherwise} \end{cases}$ and $(s(b_1,a) - 3) \bmod 5 \ne 0$ . So we have no solution for $b \bmod 5 = 1$ .

Now consider $b \bmod 5 = 2$ . Then $s(b_2,a) \equiv 4+4+1+0+1+\cdots \text{ (mod 5)}$ . $\implies s(b_2,a) \bmod 5 = \begin{cases} 0 & \text{if }a \bmod 5 = 0 \\ 3 & \text{if }a \bmod 5 = 1 \\ 4 & \text{otherwise} \end{cases}$ .

Therefore $(s(b,a) - 3) \bmod 5 = 0$ and solutions exist when $b \bmod 5 = 2$ and $a \bmod 5 = 1$ . Then $a$ and $b$ are of the form $a=5m+1$ and $b=5n+2$ , where $m$ and $n$ are integers. $a+b$ is odd when both $m$ and $n$ are odd or both are even and the units digit is $\boxed 3$ .

Let $a+b=2c+1\implies b=2c+1-a$ , where $c$ is a positive integer. Then total number of terms in the series is $a+1, i^{th}$ term of the series is $(b-1)^2+2(b-1)i+i^2=(2c-a)^2+2(2c-a)i+i^2\implies S-3=(a+1)(2c-a)^2+(a+1)(a+2)(2c-a)+\dfrac{(a+1)(a+2)(2a+3)}{6}-3=\dfrac{a+1}{6}(24c^2-12ca+2a^2+24c-5a+6)-3$ , where $S$ is the sum of the series. For $a=1$ , $S-3=8c^2+4c-2=2[2c(2c+1)-1]$ . Since $2c(2c+1)$ is even, therefore unit's place digit of $2c(2c+1)$ must be $6$ and hence that of $c(2c+1)$ must be $3$ or $8$ . By direct inspection we see that $c=1$ is a feasible candidate. So $a=1, a+b=2c+1=\boxed 3$

We notice that,square numbers modulo 5 follow this pattern:0,1,-1,-1,1.We need to only consider at most last 4 terms,since other terms cancel out. And the consecutive squares have to sum equal to 3 mod 5.The only way that can happen is when a+b is 3 modulo 5 (follows from the pattern) hence 3 modulo 10 .

I did it by Hit and Trial :-

I observed that $[7^2+(7+1)^2+(7+2)^2+...+(7+6)^2]-3$ is divisible by 5

$a+b=13$

Hence unit digit came out to be $\boxed{3}$