# Some problems in Number theory

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?

• How about 7 - 15^{81}? – Qiaochu Yuan Oct 29 '10 at 11:06
• Hmm not getting what you meant ?? – Quixotic Oct 29 '10 at 11:08
• I mean that 15^{81} + (7 - 15^{81}) is divisible by 7. – Qiaochu Yuan Oct 29 '10 at 11:26
• Wow, that's intuitive but unfortunately that's not the solution set :) – Quixotic Oct 29 '10 at 11:29
• If any of the answers you get help you, you could accept them by clicking the ghostly check mark beside them. – Ross Millikan Dec 22 '10 at 14:45

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.

• A good approach is to note that the order of a subgroup is a divisor of the order of the group. – awllower Apr 2 '11 at 7:25
• I am just kidding. – awllower Apr 2 '11 at 7:25

HINT: For the first one use the fact that $15^{2} \equiv 1 \ (\text{mod} \ 7)$

• Is it really $15^{2} \equiv 1 \ (\text{mod} \ 7)$ ? or you meant $15 \equiv 1 \ (\text{mod} \ 7)$ ? – Quixotic Oct 29 '10 at 11:23
• However the results is same. – Quixotic Oct 29 '10 at 11:24
• Okay I can see it's 6,but may be I am not getting the correct way to .. since what abt "What will be added to $29^2$ so that it will be a divided by 5 , here $29 \equiv 4 \text{ mod 5 }$ , How to get the result here ? – Quixotic Oct 29 '10 at 11:27
• @Trewick Marian: Correct – anonymous Oct 29 '10 at 11:30
• In short I am getting ... what should be subtracted to get it divisible but not what should be added. – Quixotic Oct 29 '10 at 11:31