some dr raju's prblem / need explanation

[tushar4036]

Need clear explanation/solution

1.What is the remainder of the expression(7^0+7^1+7^2+……+7^20)when divided by14?

2.col:A (1/25+1/26+1/27+1/28+1/29+1/30)
COL:B 0.2
answer a

3.given standard deviation of three numbers x,y,z as ‘d’
Col A:standard deviation of x+1,y+1 and z+1
Col B:d+1

4.given three points (5,9),(x,1),(4,5).if these points lie on a same line ,find the value of x.
5.If the arithmetic mean of a series 2,x,y,7 is 3,then what is the median?

[walt526]

1. Remainder is 1 since 7^0=1. 7^2+7^3+...+7^20 is evenly divisible by 14 since it will be divisible by both 2 and 7.

3. B. Adding 1 to each of the three variables will not affect the stdev. So Column A=d. Obviously, d+1>d.

4. Slope=Rise/Run
Thus, Slope=(9-5)/(5-4)=4
Thus, 4=(9-1)/(5-x)
Thus, x=(8/4)+5=2+5=7

5. I think that this problem requires the assumption that x and y are positive integers. Assuming that assumption is correct

Mean=3=(2+x+y+7)/4
Thus, 12=9+x+y
Thus, 3=x+y
WLOG: (1,2)

If (1,2), median={1,2,2,7}=2

[totymody]

Q2-
Answer is : A
for column A : we have six terms >= (1/30), if we assume that all terms are (1/30), then the summation = (6/30) = 0.2
but actually (1/25, 1/26, 1/27, 1/28, 1/29) > 1/30
So the summation is > 0.2



Q4-
X should = 3 ( you missed a negative sign Walt)
Slope=(9-5)/(5-4)=4
then, 4= (9-1)/(5-x)
5-x=(9-1)/4
5-x=2
x=3

[walt526]

Yep. That's what happens when I try to combine steps.

[tushar4036]

thanks totymody & walt526

still confused about 1.would u clarify elaborately..................

[munimM]

thanks tusar

[walt526]

Re: #1

To be divisible by 14 means to be divisible by both 2 and 7.

7^1+7^2+...+7^20 is clearly divisible by 7 (just factor out a 7).

Each 7^X term will be odd (odd*odd=odd). The sum of 20 odd numbers will be even (i.e., divisible by 2) since 20 is even. Thus, 7^1+7^2+...+7^20 is divisible by both 2 and 7, which means it is divisible by 14.

7^0=1. 1+something divisible by 14 cannot be divisible by 14. And in fact, the remainder is 1.

[Md. Minuddin]

1.What is the remainder of the expression(7^0+7^1+7^2+……+7^20)when divided by14?

This question can be explain in the following way too:

7^2=7*7=7*(6+1)=7*6+7; If we divided it by 14 then reminder will be 7

7^3=7^2*(6+1)=7^2*6+7^2=7^2*6+7*6+7; if we divided it by 14 then also reminder will be 7

So sum of 7^2+7^3 is divisible by 14

Similarly every pair of the series is divisible by 14 except 7^0=1
Hence reminder will be 1

[bhavana87]

If n is less than 0 and ab equal to 1, then
Col A: a^n
Col B: 1/b^n can anyone solve this??

[bhavana87]

Given that, if 252^5 is divisible by 6^n; n is a positive integer.
Col A: The largest possible integer value of ‘n’
Col B: 10