Just Some Division

We see, $10N\equiv70(mod100)$

$10N=100k+70$ [ $k\in(0,\infty)$ ]

$N=10k+7$

So we aren't sure what k is actually and the remainders may be (7,17,27,37,47,57,67,77,87,97) respectively.

Note that both $N = 7$ and $N = 17$ are values that satisfy the congruence above. However, they leave different remainders when divided by 100. A modular arithmetic error may lead one to believe that since $10\times N \equiv 10 \times 7 \bmod{100} \text{ then it must be that } N \equiv 7 \bmod{100}.$

However, in order to divide the equivalence $10N \equiv 70 \bmod{100}$ by 10, we must also divide our modulus (in this case 100) by 10. In reality, the correct division looks like this:

$10\times N \equiv 10 \times 7 \bmod{100} \longrightarrow N \equiv 7 \bmod{10}.$

One Line solution :

From the given options , both 17 and 7 are correct and it is clear that we cannot choose two so there should be more information provided about $N$ .

We realize that 10N is basically a number of the form 100x +70,or 10(10x+7).Doing basic algebra(10n=10(10x+7),n=10x+7) we find that N has multiple(infinitely many)values(it depends on your choice for x).Therefore we need more information to determine which one is the right one for the question.

So is it 7,17,27,37,47,57,67,77,87 or 97?We have no idea.