Q9

Hi

can anyone solve 9th ques

thanks

Question 1

Q1. The number FIVE as written using block capitals contains exactly 10 strokes or segments of a straight line. Find a number which when written out in words (using no tricks) contains as many strokes as the number says. Solution: TWENTY – NINE is the number which is written with as many strokes as the value of the number it states. 2 + 4 + 4 + 3 + 2 + 3 + 3 + 1 + 3 + 4 = 29 Thank You. Ravi Raja

Question 2

Q2. The number of 1's in the binary notation of (2^89) – 1 is (a) 89 (b) 88 (c) 90 (d) 1 Solution: (2^1) – 1 when written in the binary notation is written as 1 and thus contains one 1. (2^2) – 1 when written in the binary notation is written as 11 and thus contains two 1s. (2^3) – 1 when written in the binary notation is written as 111 and thus contains three 1s. (2^4) – 1 when written in the binary notation is written as 1111 and thus contains four 1s. …………………… …………………… …………………… …………………… and so on. Now, make a note of the pattern that is being followed and we can conclude that (2^89)- 1 when written in the binary notation, will contain eighty – nine 1s. Thank You. Ravi Raja

Question 3

Q3. The highest power of 2 in 10! + 11! + 12! + 13! + ...+ 1000! is

(a) 8

(b) 9

(c) 10

(d) 11

Solution:

The highest power of 2 contained in the sum of a set of factorials of different numbers is equal to the highest power of 2 contained in the least of those numbers because the highest power of 2 contained in the rest of them will always be greater than or equal to the highest power of 2 contained in the least of them.

Hence in the given problem, the highest power of 2 will be equal to the highest power of 2 contained in 10! and that is equal to 8.

Thank You.

Ravi Raja

Question 4

Q4. How many natural numbers between 1 and 900 are NOT multiples of any of the numbers 2, 3, or 5?

a. 650

b. 660

c. 240

d. 250

Solution:

The natural numbers between 1 and 900 that are divisible by 2 are: 2, 4, 6, 8, 10, 12, 14, 16, 18, … , 900 and there are altogether 450 such numbers.

The natural numbers between 1 and 900 that are divisible by 3 are: 3, 6, 9, 12, 15, 18,  …, 900 and there are altogether 300 such numbers.

The natural numbers between 1 and 900 that are divisible by 5 are: 5, 10, 15, 20, 25, 30, 35, 40, …, 900 and there are altogether 180 such numbers.

The natural numbers between 1 and 900 that are divisible by both 2 and 3 are: 6, 12, 18, 24, 30, 36, …, 900 and there are altogether 150 such numbers.

The natural numbers between 1 and 900 that are divisible by both 3 and 5 are: 15, 30, 45, 60, 75, 90, …, 900 and there are altogether 60 such numbers.

The natural numbers between 1 and 900 that are divisible by both 5 and 2 are: 10, 20, 30, 40, 50, 60, …, 900 and there are altogether 90 such numbers.

The natural numbers between 1 and 900 that are divisible by 2, 3 and 5 are: 30, 60, 90, 120, 150, 180, …, 900 and there are altogether 30 such numbers.

So the number of numbers between 1 and 900 (both inclusive), that are divisible by at least one of 2, 3 or 5 is given by: The number of numbers divisible by 2 + The number of numbers divisible by 3 + The number of numbers divisible by 5 – The number of numbers divisible by both 2 and 3 – The number of numbers divisible by both 3 and 5 – The number of numbers divisible by both 5 and 2 + The number of numbers divisible by 2, 3 and 5 = 450 + 300 + 180 – 150 – 90 – 60 + 30 = 660.

Hence the number of natural numbers between 1and 900 that are NOT divisible by any of 2, 3 and 5 are: 900 – 660 = 240, which is the required answer.

Thank You.

Ravi Raja

Question 5

Q5. Find the remainder when 51^138 is divided by 7. a. 2 b. 1 c. 2138 d. 3 Solution: (51)^138 = (49 + 2)^138 So the remainder when 51^138 is divided by 7 is the same as the remainder when 2^138 is divided by 7. Now, 2^138 = (2^3)^46 = (8)^46 = (7 + 1)^46 So the remainder when 2^138 is divided by 7 is the same as the remainder when (7 + 1)^46 is divided by 7 and that is equal to 1. Hence the remainder when 51^138 is divided by 7, the remainder is 1. Thank You. Ravi Raja

Question 6

Q6. Assume that all bricklayers work at the same rate of speed. If it takes nine bricklayers (all working at the same time) fourteen days to do a job, how long would it take for the job to be done by a) Seven bricklayers b) Three bricklayers. Solution: Here we will be using the result (M1 x D1)/(W1) = (M2 x D2)/(W2) Where M denotes the number of men involved in doing the work, D denotes the number of days required to complete the work and W denotes the quantity of work done or to be done (and is generally taken as 1, unless the exact work is specified) Solution (a): M1 = 9, D1 = 14, W1 = 1, M2 = 7, W2 = 1 (as the same work is to be done) (M1 x D1)/(W1) = (M2 x D2)/(W2) or, (9 x 14)/(1) = (7 X D2)/(1) or. D2 = (9 x 14)/(7) = 9 x 2 = 18 days, which is the required answer. Solution (b): M1 = 9, D1 = 14, W1 = 1, M2 = 3, W2 = 1 (as the same work is to be done) (M1 x D1)/(W1) = (M2 x D2)/(W2) or, (9 x 14)/(1) = (3 X D2)/(1) or. D2 = (9 x 14)/(3) = 3 x 14 = 42 days, which is the required answer. Thank You. Ravi Raja

Question 7

Q7. If both 11^2 and 3^3 are factors of the number (a) * (4^3) * (6^2) * (13^11), then what is the smallest possible value of a? 1. 121 2. 3267 3. 363 4. 33 Solution: The expression (a) * (4^3) * (6^2) * (13^11) can also be written as (a) * {(2^2)^3} * {(3 * 2)^2} * (13^11) = (a) * (2^6) * (3^2) * (2^2) * (13^11) = (a) * (2^8) * (3^2) * (13^11) Now, it is given that both 11^2 and 3^3 are factors of the given number. That is, (11^2) * (3^3) is a factor of the given number. Now in the simplified form of the given number, we can see that it already contains the factor 3. Hence ‘a’ has to contain at least 11^2 and 3 as its factors and so, the minimum value of ‘a’ has to be (11^2) * (3) = 121 * 3 = 363. Thank You. Ravi Raja

Question 8

Q8. What is the remainder when 6^2002 is divided by 11? Solution: Fermat’s Theorem states that: “If p be a prime number and ‘a’ be a number not divisible by p, then (a)^(p – 1), when divided by p, the remainder is always 1”. In the given problem, we see that 11 is a prime number and 6 is not divisible by 11 and so we can use Fermat’s Theorem in the given problem. So, (6)^10 when divided by 11, the remainder is 1. Therefore, 6^2000 = {(6)^10}^200 when divided by 11, the remainder will be (1)^200 = 1. Now, 6^2002 = (6^2000)*(6^2), and when 6^2 is divided by 11, the remainder is 3. So the remainder when 6^2002 is divided by 11, the remainder is (1)*(3) = 3. Thank You. Ravi Raja

Question 9

Q9. Let x = 1! + 2! + 3! + 4! + ... + 100!. Unit's digit of x^x^x^x^x^x …… is a. 3 b. 1 c. 1 or 3 depending upon the number of times x appears in the power. d. Can’t be determined. ANS: d Solution: We first calculate the last two digits of the sum of these numbers. 1! = 01 2! = 02 3! = 06 4! = 24 5! = 20 6! = 20 7! = 40 8! = 20 9! = 80 10! = 00 11! = 00 12! = 00 …………………… …………………… …………………… …………………… and so on. Note that for all the numbers from 10! Onwards, the last two digits will be 00. hence we now find the sum of the last two digits of the numbers from 1! to 9! And obtain the last two digits as 13. So, the last two digits of the number x = 1! + 2! + 3! + 4! + ... + 100! Is 13. Now, we know that the cycle of 3 is 3, 9, 7, 1 so to find the digit in the unit’s place of any number ending in 3, we have to express the power of that number in the form 4k + r, where k is the quotient and r is the remainder when the power of the number is divided by 4. We also make a note of a property that if the last digit of a number be a and if it is raised to any power which is of the form 4k + 1, then the last digit of the result will always be a. In the given problem, note that the last digit of the number x is 3 and the last two digits of the number x is 13. So this number x can be expressed in the form 4k + 1 and thus using the above mentioned property, we can conclude that the last digit of the number x^x^x^x^x^x …… is 3. Thank You. Ravi Raja

Question 10

Q10. Let (x, y) be co-prime numbers. Then (a) x + y and x – y have no common factor other than 1 and 3 for all values of x and y. (b) x + y and x – y have no common factor other than 1 and 2 for all values of x and y. (c) x + y and x – y have no common factor other than 1 for all values of x and y. (d) none of the above Solution: Now, note that 1 is a common factor of all pairs of integers (a, b). Hence we just have to check whether 2 or 3 is a common factor of x + y and x – y. Suppose that 3 is a common factor of x + y and x – y. Then both x + y and x – y can be written in the forms 3p and 3q, where p and q are co – prime integers. That is, x + y = 3p, for some integer p and x – y = 3q, for some integer q, where p and q are co – prime integers. Now, by adding and subtracting both the equations, we get: 2x = 3(p + q) ------------- (1) 2y = 3(p – q) ------------- (2) Now, x, y, p and q all being integers, it is clear from both (1) and (2) that both x and y are divisible by 3. But that is not possible since it is given that x and y are co – prime numbers. Hence option (a) is ruled out. Now, similarly, suppose that 2 is a common factor of x + y and x – y. Then both x + y and x – y can be written in the forms 3p and 3q, where p and q are co – prime integers. That is, x + y = 2p, for some integer p and x – y = 2q, for some integer q, where p and q are co – prime integers. Now, by adding and subtracting both the equations, we get: 2x = 2(p + q) 2y = 2(p – q) x = (p + q) ------------- (1) y = (p – q) ------------- (2) Now, p and q being co – prime numbers, we can just check by taking examples and see that both p + q and p – q can have a common factor 2 and that implies that both x and y can have a common factor 2, which again contradicts the given condition of the problem that x and y are co – prime numbers. Hence option (b) is also ruled out. Now, consider the general form that, suppose that k is a common factor of x + y and x – y. Then both x + y and x – y can be written in the forms kp and kq, where p and q are co – prime integers. That is, x + y = kp, for some integer p and x – y = kq, for some integer q, where p and q are co – prime integers. Now, by adding and subtracting both the equations, we get: 2x = k(p + q) ------------- (1) 2y = k(p – q) ------------- (2) Now, x, y, p and q all being integers, it is clear from both (1) and (2) that both x and y are divisible by k, which again is not possible since it is given that x and y are co – prime numbers. Hence k cannot take any value other than 1 and this implies that x + y and x – y have no common factor other than 1 for all values of x and y. Thank You. Ravi Raja

Question 11

Q11. What is the least number that should be multiplied to 100! to make it perfectly divisible by 7^18?

a. 1

b. 7

c. 21

d. 49

Solution:

Check that the highest power of 7 contained in 100! Is:

[ 100/7 ] + [ 100/49 ] + [ 100/343 ] + [ 100/2401 ] + ……

= 14 + 2 + 0 + 0 + ……

= 16

(Note:  Here [ x ] denotes the Greatest Integer Function)

Hence for 100! to be perfectly divisible by 7^18, it should contain two more sevens in it and that can be obtained by multiplying it by 7^2 = 49 and that is the least such number to be multiplied to 100! to make it perfectly divisible by 7^18.

Thank You.

Ravi Raja

Question 12

Q12. A car has traveled 24,000 km and, in that distance, has worn out 6 tyres. Each tyre travelled the same distance. How far did each separate tyre travel? Solution: The number of ways in which 4 tyres can be selected out of 6 are 6C4 = 15 ways and thus each tyre has covered 6/15 of the total distance covered and that is equal to (6 x 24,000)/(15) = 6 * 1600 = 9,600 km. Thank You. Ravi Raja

Question 13

Q13. There were 90 questions in an exam. If 1 mark was awarded for every correct answer and 1/3rd mark was deducted for every wrong answer, how many different net scores were possible in the exam? a. 120 b. 359 c. 358 d. 360 Solution: The possible marks that a student can obtain in this exam are: – 30, – 29.66, – 29.33, – 29, – 28.66, – 28.33, ……, 0, 0.33, 0.66, 1, 1.33, 1.66, 2, 2.33, 2.66, ……, 89.33, 89.66 and 90. So altogether there are 360 possible scores in the given range. But note that it is not possible for the student to get 89.33 and 89.66, because to get a score above 89 and below 90, first of all the student has to attempt all 90 questions correctly and at the same time, the student has to get at least one answer wrong among those 90, which is not possible. Hence the number of different possible scores are 360 – 2 = 358. Thank You. Ravi Raja

Question 14

Q14. If x = 15 x 30 x 45 ... x 1500, then how many zeros are there at the end of x? a. 24 b. 124 c. 97 d. 50 Solution: x = 15 x 30 x 45 ... 1500 Taking 15 common from each of the 100 numbers, we get: x = (15^100) x (1 x 2 x 3 …… x 100) x = {(3 x 5)^100} x (100!) Now, to calculate the number of zeros, we generally calculate the number of fives in N! as the number of two’s is always more than the number of fives in N! for N > 1, but note that in the given number, the number of fives might be more than the number of two’s and so we calculate both the number of two’s and the number of fives and whichever is lesser in number, we take that value as our answer. Check that in 100!, the number of two’s is 97 and the number of fives is 24. So in the given number, we have a total of 97 two’s and (100 + 24) = 124 fives. Hence the number of tens that we can get is 97 and thus the number of zeros at the end of the number will be 97, which is our required answer. Thank You. Ravi Raja

can u tell me wht do u mean

can u tell me wht do u mean by 6^8

ANS TO Q9)...I THNK THE ANS

ANS TO Q9)...I THNK THE ANS IS 3 ..

WHEN U FIND UNIT DIGIT...U ALWYS HV TO CONSIDER THE UNIT DIGIT ONLY..

SO THE SUM OF 1!+2!+...+100!= 1!+2!+3!+4!+0+0.... = 1+2+6+4+0+0+0+0... =3(UNIT DIGIT) & CYCLICITY OF 3 IS 4...

SO (3^N)/4 WILL ALWAYS GVE  UNIT DIGIT AS 3..(AS N IS ODD ALWAYS)..SO ANS=3

it means 6 to the power 8i.

it means 6 to the power 8

i. e 6⁸

@ Kamna

Hello Kamna,

I am Sorry that I forgot to specify that

a^b implies "The number 'a' raised to the power of the number 'b'".

Example:

2^3 = 8

3^4 = 81

5^3 = 125

Thank You.

Ravi Raja

@ Neel

Hello Neel,

I agree with the first part of your answer where you have stated that: "WHEN U FIND UNIT DIGIT...U ALWYS HV TO CONSIDER THE UNIT DIGIT ONLY..

SO THE SUM OF 1!+2!+...+100!= 1!+2!+3!+4!+0+0.... = 1+2+6+4+0+0+0+0... =3(UNIT DIGIT) & CYCLICITY OF 3 IS 4...

But I found something wrong with the second part which says:"SO (3^N)/4 WILL ALWAYS GVE  UNIT DIGIT AS 3..(AS N IS ODD ALWAYS)"

What exactly do you mean when you write (3^N)/4

Is it (3 to the power of N), the whole thing divided by 4 or

Is it 3 to the power of (N divided by 4)

Since there is a difference between the two, the answer in each case will also be different.

May be you meant the right thing but were not able to type it in the right format.

Sorry for this but I just did not understand the last part and thats why I corrected you.

Thank You.

Ravi Raja

yeah..actually wat i meant

yeah..actually wat i meant was:- since cyclicity of 3 is 4(3,9,7,1)...& 3^3^3^3....upto any power of 3 is of the form 4n+1..so rem is alwys=3..i hope my ans is rt, even if m nt able to explain it clearly to u..

@ ravi raja

The highest power of 2 contained in the sum of a set of factorials of different numbers is equal to the highest power of 2 contained in the least of those numbers ...i cudnt undrstand dis logic...!! kan u xplain it more clearly ???????

@ Neel

Hello Neel,

What I meant from my statement: "The highest power of 2 contained in the sum of a set of factorials of different numbers is equal to the highest power of 2 contained in the least of those numbers" is the following.

Lets consider a smaller example.

What is the highest power of 2 that will divide 2! + 3! + 4! + 5! + ..... + 10!

Now, we know that:

the number of times 2 divides 2! is 1 time.

the number of times 2 divides 3! is 1 time.

the number of times 2 divides 4! is 3 times.

the number of times 2 divides 5! is 3 times.

........................................................................

........................................................................

........................................................................

the number of times 2 divides 9! is 7 times.

the number of times 2 divides 2! is 8 times.

So, now when we divide the sum 2! + 3! + 4! + 5! + ..... + 10! by 2 once, the result will be an integer, but we cannot divide the quotient thus obtained by 2 for the second time, becase in the first two terms, 2! and 3!, there will be no more 2s, although we have more then one 2s from 4! onwards.

So, we see that the highest power of 2 contained in the sum of a set of factorials of different numbers is equal to the highest power of 2 contained in 2!, which is nothing but the least of those numbers"

I hope this time I could explain it more clearly. If not, then Please do let me know. Will give it another try and with another example.

Thank You.

Ravi Raja

i think the answer could be

i think the answer could be 9 'cos i think it would also depend on wheteher the sum is even or odd when the common 2's are taken out.

ANSWER LESS THAN 27

The answer could be 4 too.......

c how

FOUR= 3+0+0+1=4

CAN U CLARIFY THE SOLUTION

CAN U CLARIFY THE SOLUTION OF Q13 OF QBM47.

@ Meghar

Hello Meghar,

I am assuming that all the letters in the word will consist of straight lines only and none of them will require curved lines and thus we cannot use words that will require letters like: B, C, D, G, J, O, P, Q, R, S and U and thus arrive at the result 27.

Thank You.

Ravi Raja

Q.9

Clearly by adding last two digit upto 9! we get 13 as last two digit in whatever the final sum will be.now 13 will have cycle of 4 as is clear and digits are 3,9,7,1 andthey will repeat .so our task is to express **13^(**13)^(**13).......as 4n+r.now if we divide 13^13 by 4 we get 1.so13^13 can be written as 4n+1.now 13^(4n+1) divided by 4 again gives remainder as 1 by remainder theoram.so again i13^(4n+1) can be written as 4n"+1.so clearly **13^(**13)^(**13)............  is reduced to 13^(4n+1),which again has remainder as 1.so 3 is the answer according to cyclic behaviour studied.

confused

(51)^138 = (49 + 2)^138
So the remainder when 51^138 is divided by 7 is the same as the remainder when 2^138 .

In the above case how u can say that remainder when 51^138 is divided by 7 is the same as the remainder when 2^138 .

Sol Q12

Hi  Ravi Raja,

               Please find a different approach to Q12.

Total distance covered by the car is 24,000 km.
So,the four wheels together cover
4 x 24,000km.
This is equally shared by all 6 wheel
 Hence Distance traveled by each wheel
 = 4 x 24000/6 = 16000 km.

hi!!!! could any one please

hi!!!!

could any one please tell me more simpliar ways of solving Ques.10

hi!!! neel i have solution

hi!!! neel i have solution for ur doubt

as 51 could be said as (49+2)

and 49 is always divisible by 7

and hence the remainder at this stage remains as 2^138

so we can say that the remainder  of 51^138 and 2^138 is same

10!(1+11+11

10!(1+11+11*12+13*12*11 ....)   =2^8*3^3*5^2*7(12+132+evenno.)      2^8(12(1+11....) =2^8*2^2(odd no.)       

=2^10  answer

thanks

question NO.14

could u plz  tell me how did u calculate the no. of  2 and no. of 5 in 100! in Question no.14

doubt

sir, i didnt follow q no.5......how remainder of (49+2)^138 divided by 7 is same as 2^138 divided by 7?...

 and remainder of (7+1)^138 same as 1^138??

bcz 49 is divisible by 7 and

bcz 49 is divisible by 7 and all the terms of (49 + 2)¹³⁸   when  expanded 'll contain 49 except the term 2¹³⁸

(49 + 2)¹³⁸ = 49 (xyz........) + 2¹³⁸

51^138 when divided by 7, rem??

Instead of complicating the problem. take it in the simplest form. Consider (51/7)r i.e. [(49+2)/7] r which can be simplified further as (49/7)r + (2/7)r which is nothin but (0+2)=2 now see it as 51*51*51*51*51....*51 (138 times) so the rem now is 2*2*2*2*2...*2(138 times) i.e. 2^138... now its easy to find the remainder for this. This was just the basic behind the problem. Nyn

A doubt in your explanation regarding the question no.10.

Could you please explain how you said that the assumed integers P and Q must be co-prime integers.Do we get two co-prime integers when we add and subtract two co-prime integers?

Q10. here it goes

X and Y are co-prime. . That means i. Both are odd ii. Or one even and another odd i. when both are odd X + Y and X – Y ‘ll result an even number and they ‘ll have a common factor of 2 ii. when one even and another odd (4, 9 | 4, 13 | 6, 17) we ‘ll always get an odd numbers and both the numbers ( X + Y, X – Y) cant contain factor of 3 (why ?? think) For X + Y and X – Y to contain factor of 3, both must contain factor of 3 and if that is the case then they cant be co-prime. . Hope it clears. . or too confusing????

bhai yeh tujhe kaha se mil

bhai yeh tujhe kaha se mil gaya ??? .... dude i think the number of different marks = no of wrong answers attempted !!! then how in the world could you come up with such an answer !!

@Nayan

Remainder of 51^138(mod 7) Soln. Durin workin tim wat i got always Euler theorem is very useful nd it really reduces time.... As 51 nd 7 r coprime....Euler of 7 is 7x6/7=6..... as 138%6=0...Accordin to Euler theorem 51^138(mod 7)=1.... If u dnt knw abt ds theorem dn go thru the blogs...Blv me it wud reduce ur time nd REMAINDER PROBLMS will b cakewalk to you.............. Regards, Dipanjan.......

For reference, Euler’s

For reference, Euler’s theorem is available @
http://www.cat4mba.com/node/6104

I think the ans is 12.....

we know that max power of 2 in 12!=10 also 10!+11!=10!*12 which also have max power of 2=10 So we cannnot say that "The highest power of 2 contained in the sum of a set of factorials of different numbers is equal to the highest power of 2 contained in the least of those numbers"

Hello plz explain 97two's in

Hello plz explain 97two's in 100!

alternate solution to Q.4

Hey Raja I've also thought another short method to solve this question. Here it goes:- out of 900 numbers 450 are divisible by 2 and 450 are not divisible by2. So we remain with 450(not divisible by 2) N0w out of these 450, for only 450/3=150, i.e 150 are divisible by 3; 450-150=300 are not divisible by 3 as well. We remain with 300,now 300/5=60, i.e 60 are divisible by 5; 300-60=240 are not divisible by 5 as well. So we've broken down to 240 from 900 which ain't divisible by 2 or 3 or 5.

Hey Raja I've also thought

Hey Raja I've also thought another short method to solve this question. Here it goes:- out of 900 numbers 450 are divisible by 2 and 450 are not divisible by2. So we remain with 450(not divisible by 2) N0w out of these 450, for only 450/3=150, i.e 150 are divisible by 3; 450-150=300 are not divisible by 3 as well. We remain with 300,now 300/5=60, i.e 60 are divisible by 5; 300-60=240 are not divisible by 5 as well. So we've broken down to 240 from 900 which ain't divisible by 2 or 3 or 5.

Q10

hey if we take 3 &5 as co prime nos. then x+y =8
X-y =2 so common factor b/w them is 2 &1 so howz answer is C plz explain this

hi raja i new to this site

hi raja i new to this site but have fun in solving anywayz raja bhai 6 is also possible one because s is on stroke i has 3 and x 2

query

hey jus think over it how can ne manage 2 get a 88.33 this not possible shudnt d right answer be 357 which is not even part of d answer choices ........

inncorect choices

this question if ne body has tried solving it Q13. There were 90 questions in an exam. If 1 mark was awarded for every correct answer and 1/3rd mark was deducted for every wrong answer, how many different net scores were possible in the exam? a. 120 b. 359 c. 358 d. 360 ans: 358 after looking at d soln frm ravi ...........we can agree 89 .66 n 89 .33 r not possible ............even fr that mater 88.33 is not possible ...so ther 1 less d no of possible scores ............so d correct answer is 357 whichis no wer in d choices .........there fore ...the question needs 2 hav d corect options ....

I could not understand this

solution to question 9

the answer would be a)3 because everytime you add x=1!+2!+3!+....100! yould get the last two digits as (1+2+6+24=33) and x^(x*x*....x,ntimes)the result will be 3^1=3

10! + 11! + 12! + .... can be

10! + 11! + 12! + .... can be written as 10!(1 + 11 + 11*12+...) and this can be reduced to 10!(12 + 11*12+ 11*12*13+..)...which is 10!*12(1 + 11 +.....). Therefore the highest power of 2 is 10.

Q14. If x = 15 x 30 x 45 .... 1500,

Q14. If x = 15 x 30 x 45 .... 1500, then how many zeros are there at the end of x?
a. 24
b. 124
c. 97
d. 50

solution:- [1500/30]+[1500/60]+[1500/120]+[1500/240]+[1500/480]+[1500/960]
=50+25+12+6+3+1
=97

Query

The expalnation u gave is only for power of 2 or this will work while finding highest power on any natural number in a factorial series.Please reple asap.

doubt as regards to the solution

though d answer u obtained is right,
i doubt d solution,
there r 361 scores in d series { -30 ......90} for this ques..  n nt 360 as mentiond by u...

and thr cud b  3 scores which can nt b obtained with d given conditions..
viz..89.66,89.33,89(since its nt mentiond if he can skip a ques)
  OR
2 as mentioned by u..
which is the correct solution nw?
do let me knw if my points were correct...
thnk u..