## Quants doubt

Please tell me how to approach this... What is the digit at the ten`s place of 7^99. Units place can be done by cyclicity but what about ten`s place.

## The answer is same as for

The answer is same as for the last digit i.e. unit place The last 2 digits also repeat with a definite period For example 7 to the power 1 – 07 7 to the power 2 – 49 7 to the power 3 – 343 7 to the power 4 – 2401 7 to the power 5 – 16807 7 to the power 6 – 117649 7 to the power 7 – 823543 So for 7 the last 2 digits repeat after every 4 i.e. 1st , 5th , 9th . . . all have same last 2 digit numbers So 99th power‘ll have the same ten’s place as 3rd term and that’s 4.

## Thanks Can it be applied for

Thanks Can it be applied for numbers other that 7.

## Thanks Can it be applied for

Thanks Can it be applied for numbers other than 7.

ya this cyclicity concept can be used for any natural number for any place of digit.

## Though mathematically above

Though mathematically above statement is correct you 'll hardly find any other number except 7 which repeats both the unit and tens place with a fixed periodicity.

## this method works for all

this method works for all integers the only difficult part being that of finding the least nonzero power that leaves a reaminder of 1
we have to keep on finding the power till we get a number with remainder 1. But still Rajorshi's method is advisable and better than the way i solved it initially bcz as per that way we had to find the powers till we get the numbers repeating itself 7 to the power 1 – 07 7 to the power 2 – 49 7 to the power 3 – 343 7 to the power 4 – 2401 7 to the power 5 – 16807 7 to the power 6 – 117649 7 to the power 7 – 823543 but as per rajorshi calculating 7 to the power 1 – 07 7 to the power 2 – 49 7 to the power 3 – 343 7 to the power 4 – 2401 'll be sufficient now i am thinking how should I approach if the question had asked for the hundredths place ??

## i have another solution :

i have another solution : the above problem may be stated as : what is the reaminder when 7^99 is divided by 100. now the least nonzero power of 7 that leaves a reaminder of 1 when divided by 100 is 4. thus 7^(4k+c) will leave the same reaminder as 7^c when divided by 100. now 99 = 4*24 + 3. thus 7^(4*24+3) will leave the same reaminder as 7^3 when divided by 100. = 43. hence the ten's digit = 4 this method works for all integers the only difficult part being that of finding the least nonzero power that leaves a reaminder of 1

## I guess, I have more

I guess, I have more mathematical ans: 7^99 => 7 x 7^98 => 7 x 49^49 Now, we will make use of Bionomial Theorem. The funda is that one of the two terms will be 1. 49 => 50-1 So, 7 x (50-1)^49 ------ (i) Now, if you want to find last two digit, it's equivalent to dividing a no by 100 and finding the remainder. Similarly, if you want to find third last digit, it's equivalent to dividing by 1000 and finding the remainder. Expanding (i) 7[ 49 C 50^49 - 49 C 1 . 50 ^ 48..... 49 c 48. 50 - 49 C49]---(ii) Except the last two terms, all the terms will be divisible by 100, as 50 has power of 2 or higher. So, our concern is with last two terms only. last two terms equal to = 49 x 50 -1 = 2449. This multiplied by 7 gives 17143. This on being divided by 100 will be give remainder 43. Hence the last two digits will be 43. ANY CLAPS FOR ME??? PLZ RESPOND.

## I guess, i answered the

I guess, i answered the secondharsa's query?

## Thanks manu, you deserve more than few claps :-)

May be a call from ---- I was cracking my head on desk for last few hours :-)) Any way its a good concept you should add this to math e book else I 'll do that for you.

## solution

7^99= 7 x(7^98)= 7 x (49)^49 =7 x [(50-1)^49]=7 x [50^49- 49C1(50)^48+.........-(49)C47(50)^2 +49C48(50)-1] (Binomial Expansion) Now ,except the last two terms in the expression each term will have atleast two zeroes (at unit and tens place) i.e. the whole of expression barring last two terms is divisible by 100 therefore, the expansion can be written as = 7x[100k +49(50)-1] =7 X [100k+2449]=700k+ 17143 (where k is some integer) In the above expression(700k+17143), last two digits of 700k would be 00 , hence the addition will give 3 at units place and 4 at tens place respectively

## Solution

Question: What is the digit at the ten’s place of (7)^99? Solution: The digit in the ten’s place of a number can be calculated by finding out the remainder when the given number is divided by 100. It is clear that the remainder will be either a single digit number or a two – digit number. In the former case, the digit in the ten’s place will be 0 otherwise the digit in the ten’s place of the given number will be the same as the digit in the ten’s place of the remainder obtained when the number is divided by 100. (7)^99 = (7^96) x (7^3) = {(7^4)^24} x (343)} = {(2401)^24} x (343) = {(2400 + 1)^24} x (300 + 43) Now it is clear that (2400 + 1)^24 when divided by 100, the remainder will be 1 and the remainder when 343 is divided by 100, the remainder is 43. Hence the remainder when (7)^99 is divided by 100 is (1) x (43) = 43. Hence the digit in the ten’s place of the number (7)^99 is 4. Thank You. Ravi Raja