CAT4MBA

Q1.

break 36 into factors i.e 9*4

9 willl divide every factor, calculate remainder when whole term is to be divided by 4.....

now this will be easy....veryy easy infact

Q2..

i cnt solve...um vry weak in these type of questinnzzz....may b AM>GM is to be usedd...ravi or neeel cn throw sum lite on it

Q3...n^7-n cn be written as n(n^6-1) which further can be written as n((n^3)^2-1)== n(n^3+1)(n^3-1)

above equation will be divisible  for all +ve integers...hence remainder =0

Q2) & Q3) both are very easy...c i'll tell u

In Q2)use options to solve..infact, its always benefitial to solve such qns via options
consider option a)a=b=c=d...let a=b=c=d=1

Then L.H.S = 1^2+1^2+1^2+1^2 = 4
& R.H.S = 1*1+1*1+1*1+1*1 = 4

since L.H.S. =R.H.S.
Hence
optin (a) is the correct answer

In Qn 3) put values of n= 2,3,4..

For every value of "n" it will be div by 42, so rem =0

for first Q

ans will be 0

it is very easy Q and can be solved by factorising 36 as 9*4

now we have to check remainder just by 4

remainder comes out to be 0.

for 2nd Q

ans will be 1st option

clearly  yaar ,simply when we put a=b=c=d in above Q it simply satisfies ,in no time.

for third Q

mearly by observing in no time ,remainder comes out to be 0.

Q1. What is the reminder when 27 + 27² + 27³ + 27⁴ + . . . . + 27⁶ is divided by 36 ?
a. 9

b. 81

c. 27

d. 243

Solution:

Note that the given expression can also be written as follows:

27 + 27² + 27³ + 27⁴ + . . . . + 27⁶

= (3³) + (3³)² + (3³)³ + (3³)⁴ + . . . . + (3³)⁶

= 3³ + 3⁶ + 3⁹ + 3¹² + 3¹⁵ + 3¹⁸

Now, note that each of the given terms is divisible by 9 and in that case, the remainder is 0 and the quotient is:

= 3¹ + 3⁴ + 3⁷ + 3¹⁰ + 3¹³ + 3¹⁶

Now, when we divide this number by 4, then note that the remainder from the terms are 3 and 1 respectively depending upon whether the power od 3 is odd or even. That is, when the power of 3 is odd, the remainder is 3 and when it is even, the remainder is 1. Now, altogether, there being 6 terms, there are three terms whose remainder is 1 and three terms whose remainder is 3, and on adding all these remainders, the sum of the remainders is 12, which again when divided by 4, leaves no remainder.

Hence, the given number, that is, 27 + 27² + 27³ + 27⁴ + . . . . + 27⁶ when divided by 4 and 9, that is by 36, leaves 0 as the remainder.

Thank You.

Ravi Raja

Q2. If a² +  b²  + c²  +  d²  = ab + bc + cd+ da, then which of the following is ture?

a. a = b = c = d

b. 4a = 3b = 2c = d

c. a = 2b = 3c = 4d

d. None of these

Solution:

Here the best way is to use options:

According to option (a), we see that if we take a = b = c = d = 1, say, and substitute the values in the given equation, a² +  b²  + c²  +  d²  = ab + bc + cd+ da, then we see that the equality holds.

According to option (b), we see that if we take a = 3, b = 4, c = 6 and d = 12, say, and substitute the values in the given equation, a² +  b²  + c²  +  d²  = ab + bc + cd+ da, then we see that the equality does not hold, since the left hand side becomes: 9 + 16 + 36 + 144 = 205, whereas the right and side becomes: 12 + 24 + 72 + 36 = 144, which is NOT equal to the value obtained from the expression from the left hand side and hence option (b) is ruled out.

In a similar manner,we will be able to rule out option (c) too and no need to think about the last option since option (a) satisfies the given equation for ALL values of a, b, c and d satisfying a = b = c = d. Hence, option (a) is the answer.

Thank You.

Ravi Raja

Q3. What is the remainder when n ⁷ – n is divided by 42?

Solution:

Using Fermat's Theorem, we know that: "If 'p' is a prime number and 'n' uis a number not divisible by 'p', then (n^p - n) when divided by 'p', the remainder is always 1".

Using the above theorem, we can say that n⁷ - n is divisible by 7 and now, we check whether it is divisible by 6 or not.

n⁷ - n = n(n⁶ - 1) = n(n³ - 1)(n³ + 1) = n(n - 1)(n² + n + 1)(n + 1)(n² - n + 1) = (n - 1)n(n + 1)(n² + n + 1)(n² - n + 1)

Now, note that (n - 1), n and (n + 1) are three consecutive integers and is thus divisible by 3! = 6, since we know that the product of n consecutive integers is divisible by n! (n - factorial).

So, the given expression is divisible by both 6 and 7 and thus by 42.

Hence, the remainder when n ⁷ – n is divided by 42, the remainder is 0.

Thank You.

Ravi Raja

Hello Himanshu,

I have posted the solutions of all the three problems. You can go through the solutions and ask if you do not understand anything in the steps or in my method that I have used. Will try to explain it to you using some other method, if possible.

Thank You.

Ravi Raja

@ Ravi

well Ravi ..if u r refeerring to me....thn thnx a lot fa ur concern...i ve solved first n thrd questn but fa solving 2nd question...i need to know approach....a GENERAL APPROACH  FOR SOLVING SUCH PROBLEMS..

option wala trick se th easy h....i'l b thnkful 2 u if ya tell me a general approach.....

thnxx..

i dont think solving genrally would help, 'cos a similar question on ratio and proporation was asked in cat2006,

people who solved it using algebra arrived at one correct  anwser but missed out on another which incidently satiesfied the question.