Problem Solving

gmatter2012 wrote:What is the remainder, after division by 100, of 7^10 ?

7^10 = (7²)^5 = 49^5
Odd powers of 9 have 9 as the unit's digit and even powers of 9 has 1 as the unit's digit.

Now, unit's digit of 7^10 = unit's digit of 49^5 = 9

Hence, the unit's digit of the remainder when 7^10 s divided by 100 will be 9. Only such option is 49.

The correct answer is D.

aakgoel wrote:Is there any way to find the digit in tens place also.
If exists, please tell.
Thank you.

In general to find out the ten's digit of numbers of the form x^n you need to know binomial expansion.

We can expand (a + b)^n as follows,


Now, the trick is to write x as sum or difference of the nearest multiple of 10. For example, say we need to find out the last two digits of 121^21. Then we will write 121 = (120 + 1)

So, 121^21 = (120 + 1)^21 = (1 + 120)^21
Now, we will consider a = 1, b = 120, and n = 21, and accordingly we will expand (1 + 120)^21. Note that, every term in the expansion after the second term will be a multiple of 100. Hence, we just need to determine the first two terms.

First term = a^n = (1)^21 = 1
Second term = n*[a^(n - 1)]*b = 21*[1^20]*120 = 21*120 = 2520

Hence, last two digits of 121^21 are last two digits of (2520 + 1) = 2521, i.e. 2 and 1.

This method works great when we can write x as sum or difference of nearest multiple of 10 with respect small numbers like 1 or 2. Even for 3, it gets complicated.

For example, in this case 7^10 = (-3 + 10)^10
First term = (-3)^10 = 3^10 ---> Needs lot of calculation.

This method will work for last three digits also. In that case, we need to write x as sum or difference of nearest multiple of 100.

Hope that helps.