1.What should be added to $15^{81}$ to make it divisible by 7 ? Suggest a general technique to counter problems like this.
2.For how many two-digit numbers, the sum of the digits is greater than the product of the digits?
1.What should be added to $15^{81}$ to make it divisible by 7 ? Suggest a general technique to counter problems like this.
2.For how many two-digit numbers, the sum of the digits is greater than the product of the digits?
For the second question, for single digits $a+b\geq ab$ when a=0 or 1, b=0 or 1, or a=b=2.
A good approach to these is to reduce your base number according to the modulus. So since $15\equiv 1\pmod{7}$, you would have $15^{89}\equiv 1^{89}\equiv 1\pmod{7}$. In other words, $15^{89}$ is 1 greater than a multiple of $7$, or 6 less than a multiple of $7$. So to get a complete solution set, you could add any integer of the form $6+7k$, or $-1+7k$, and these two sets coincide. So you can take either what must be subtracted or added to get the number to a multiple of your modulus. For example, for $29^2=841$, you could either add 4 to get 845, or subtract 1 to get 840, both of which are divisible by 5.
For the second, as was noted in the comments, $29^2\equiv(-1)^2\equiv 1\pmod{5}$, and you should be able to get a complete solution set by the same reasoning. I just mentioned $29\equiv -1$ since $(-1)^2$ is nicer to deal with than $4^2$, and you would have to reduce 16 modulo 5 anyway.
HINT: For the first one use the fact that $15^{2} \equiv 1 \ (\text{mod} \ 7)$