Application of the Binomial Theorem-remainder I am having a confusion in this question-
What is the remainder when $7^{103}$ is divided by 24?
I attempted it as follows - 
It can be written as
$(7^2)^{51} \cdot 7$
Which can be written as
$(24*2+1)^{51} \cdot 7$
Now using Binomial Theorem clearly the remainder should be $1^1 \cdot 7=7$
But the answer in most places is 18 and whereas in some sites it is 7!!  
Which one is correct and why am I wrong if I am? 
PS-for those who think i am correct please prove that they are wrong:) 
http://www.askiitians.com/iit-jee-algebra/binomial-theorem-for-a-positive-integral-index/application-of-binomial-expression.aspx
Check this one out also... you have to download the pdf and scroll down to almost the middle to find the question! https://www.google.co.in/url?sa=t&source=web&rct=j&ei=dVwUVeKtDoKyuATQ6IH4DA&url=http://www.arbindsingh.com/wp-content/uploads/2012/07/Introduction-To-Binomial-Theorem.pdf&ved=0CC8QFjAG&usg=AFQjCNGErAyk6Qh_fO2nlD-XvlDyuMFblQ&sig2=HiEovIrDxhZy4DPIgGHA_w
 A: Your reasoning is correct. The remainder is $7$.
If you want another way to check, we can use modular arithmetic. This is where in arithmetic you replace a number by its remainder (with respect to $24$ in this case).
As you have shown, $7^2$ has remainder 1 modulo 24. This means $$7^{103} \equiv 7^{102} \cdot 7 \equiv 1^{51} \cdot 7 \equiv 7 \mod 24.$$

If you want, I can spell out the binomial expansion.
When you expand $(2\cdot 24 + 1)^{51}$, the result is: $$_{51}C_0 (2\cdot 24)^{51} +\ _{51}C_{1} (2 \cdot 24)^{50} + \cdots +\ _{51}C_{50} (2\cdot 24) +\ _{51}C_{51}$$
Since $24$ is attached to all but the last term, and $_{51}C_{51} = 1$, we have $$(2\cdot 24 + 1)^{51} = 24 \cdot M + 1$$
This means $$7 \cdot (2\cdot 24 + 1)^{51} = (7M)\cdot 24 + 7$$ and so the remainder after division by $24$ is $7$.
A: Your answer is correct. But both sources compute the remainder modulo $25,\,$ not $\,24,\,$ i.e.  
$\,{\rm mod}\ 25\!:\ \color{#c00}{7^2\equiv 1}\,\Rightarrow\,7^{103}\equiv 7(\color{#c00}{7^2})^{51}\equiv 7(\color{#c00}{-1})^{51}\equiv -7\equiv  18\  $ by $ $ Congruence Rules. 
Therefore the $24$ is a typo for $25$ in both sources.
A: $7\cdot \left(48+1)^{51} = 7(48^{51} + \binom{51}{1}48^{50} +\cdots + \binom{51}{50}48 + 1\right) = 7 \pmod {24}$. Can you take it from here?
