answerQ1

Q1. 2^100 < 3^100 And 2^3 < 3^2 So (2^3)^100 < (3^2)^100

Answers to some of the questions

M1. Which is greater of the two 2^300 or 3^200
Ans. 2^300 or 3^200 taking 300log2 or 200log3
       Log2=0.3010 and Log3=0.4771 Thus 2^300 = 90.3 and 3^200 = 95.16 Thus 3^200 > 2^300 Ans.

M2. Find the last two digits of 15 x 37 x 63 x 51 x 97 x 17 ? Ans. 35
Ans. 15x37 = 555 = 500+55, similarly 63x51 = 3200+13and 97x17=1600+49
        Multiply 55x13 =700+15 thus multiply 15x49=700+35 Thus the number in last two digit is 35 Ans.

M3. If both 11^2 and 3^3 are factors of the number x * 4^3 * 6^2 * 13^11, then what is the smallest possible value of 'x'?
Ans.Consider 6^2=3^2x2^2 thus the smallest possible value of x = 3x11^2 Ans.

M4.How many numbers are in between 140 and 259(both included) which are divisible by 7?
Ans. Consider the number 140,147,.....259.Here a=140 and d=7 Thus 259=140+(n-1)7,which gives n=18 Ans.

M5.11 is the smallest prime made up of the same digit.

M6.Find the number of numbers between 300 and 400(both included) that are not divisible by 2, 3, 4 and 5.
Ans. Numbers not divisible by 2 are = 301,303,305,307....399 i.e Tn= 301+(n-1)2 for n=1 to n=50 i.e 50 numbers
        In this the number not divisible by 3 are those for which ((n-1)2 +4 ) not divisible by 3 for n=1 to n=50
        which is for n=2,5,8,...,50 it is divisible and balance are not divisible. Here divisible 50=2+(k-1)3 => k=17 thus out of 50 values 17 are
       divisible by 3 thus balance non divisible by 2 & 3 are 33 values and number not divisible by 2 is defintely not divisible by 4.
       The Number ending in are 305,315,325,335,345,355,365,375,385,395 out of these 315,345,375 are divisible by 3 are already taken care off above
       The balance numbers are 10-3 =7 so the total numbers not divisible by 2,3,4,5 are 33 -7 = 26 Nos.

M7. Find the sum of all three-digit numbers that give a remainder of 3 when they are divided by 7. 676
Ans. Numners are 100 to 999 all three digit numbers Now Numbers are Q = 7.676xk+3 forms
        100 <= 7.676xk +3 <= 999 for various values of k Now for 100 = 7.676xk+3 => k=12.63 taking next number i.e. k=13
        for 7.676xk+3 = 999 => k =129.75 thus k varies from k=13 to k=129 and number of terms = 129-13+1=117 terms
        Sum = 7.676( 13+15+...+129)+ 117x3 = 7.676(13+129)x117/2 + 117x3 = 224.523+351=575.523

M8. Explain why a number made up of the same digit can only be prime if the repeating digit is one AND the number of digits is itself prime.

M9.A four digit number A375 is divisible by 11. Find A?
Ans. A+7-3-5=>A-1=0 or A=1 Ans.

M10.Find the sum of all odd three digit numbers that are divisible by 5?
Ans. Odd three digits divisible by  are 105+115+125+135+145+..995 Here 995 = 105+(n-1)10 thus no. of terms = 90
        Sum is 45x(105+995) = 49500 Ans.

M11.Given that n>=2 and an 1 is prime, prove that a = 2, and n must be prime

M12. Which is greater of the two 11^13 or 13^11
Ans. 13 log 11 and other is 11 log 13 Now. 13x1.041 = 13.533
        and 11log13 = 11x1.1139 =12.2529 thus 11^13 > 13^11 Ans.

M13. A five-digit number A290B is divisible by 11 and 13. Find A and B?
Ans    Here A+9+B-2=11 => A+B =4 and A+4x2+9x3+0-4B = A+8+27-4B = A+35-4B=39 thus A-4B=4
         As A+B =4 and A-4B = 4 thus B=0 and A=4 Ans.

M14. A number when divided by a divisor leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?
Ans. Let the number by y and divisor be x then y=px+24 and 2y=qx+11 y=(q-x) - 13
        Thus -13+x = 24 or x = 24+13 = 37 Ans.
M15. How many two digit primes can you find for which their reverse is also prime?
Ans. 

solve only 10 sec

there is a another method to find last digit of question no.2

(15*37*63*51*97*17) divide this no. by 100

(15*37*63*51*97*17)/100

(3*37*63*51*97*17)/20

take remainder

(3*17*3*11*17*17)/20

(51*33*289)/20

take remainder

(11*13*9)/20

(11*117)/20

take remainder

(11*17)/20

187/20 take remainder 7

multiply 7 by  we intially divide equ.by 5. last disit is 35

this process can be done orally and it take only 10 sec.

thanks for the brilliant

thanks for the brilliant method, if u have any such "10 sec" methods pls post it we wil be grateful to u

Q15

8 nos..

11,13,17,37,71,73,79,97

there is no three digit number divisible by 7.676 !!!

i think there is a problem with question number 7 ....!!! i think there is no a three digit number(which implies no decimal place) that is divisible by 7.676 and leaving remainder of 3.

simple explanation 7.676 = 7676/1000.

if you simply the above rational number then you get 1919/250 = 19*101/250.

now the number we want is something that is of the below form and satisfying the condition we want it to

100<= 7.676k + 3 <1000 or

100<= (19*101)*k/250 + 3 <1000

now in above expression k should be an integer .... and the number we are trying to find is a three digit number like 100.00000 or 345.00000 what i want to imply is that the decimal places of the number must be zero !!! (and is not the taken as same in above expression) ...

evaluating above condition we get 13<=k<=129 ... for the number to have no decimal place we need k to be a multiple of 250 or 250/k should give us results like 19n or 101n ... then we go and get a integer value for our expression !!!! which is not possible for any value of k less than 250 !!! so there are no numbers that are divisible by 7.676 and a three digit number !!! hope explanation is correct .. !!!! any problem please shoot it

Q14 .. explanation to that doesnt make sense there is a error !!

we get two equations

y  = nx + 24   (1)    this implies that x is greater than 24.

2y = mx + 11  (2)

multiply equation1 with 2 and subtract from equation2 .. we get

x = 37/(m-2n) ....

... we know both m and n are integer so the expression (m-2n) will be an integer. And for x to be positive integer .. values of (m-2n) needs to be 1 or 37.

for (m-2n) = 1 , we get x = 37 ... which satisfies the condition mentioned along with equation1 ...

for(m-2n)=37, we get x = 1 which is not greater than 24 hence not the solution required.

so the unique solution we get is 37. i think this is the rt explanation

I ve another idea

M14.

A number (a) when divided m gives reminder 24
=> When 2a is divided by m the reminder should be 48 but it’s given as 11 i.e 48 is 11 more than the number. So the number a = 48 – 11 = 37

is it OK?

nice way dude !!!!

hmm well you have got the rt answer .. and i would appreciate your way of arriving at it. but i think there is a problem in this too .....

for example check this out !!

                        actual remainder          remainder according your method       so divisor

6/4 remainder =           2                                        2

2*6/4 remainder =        0                                        2*2 = 4                        2*2-0 = 4

3*6/4               =        2                                        3*2 = 6                        3*2-2 = 4

4*6/4               =        0                                        4*2 = 8                        4*2-0 = 8

when you multiply with 4 the divisor you land up finding is greater than the actual number. But infact i guess we get multiples of actual divisor we are trying to find out. and can devise a method to find it .... nice way yaar thanks buddy !!

I am sorry but

I am sorry but I really couldn’t understand what is the problem with that method . .
 

2*6/4 remainder =   0    2*2 = 4          2*2-0 = 4

Here the reminder is 4 but we are dividing the reminder by 4 . .so the actual reminder is 0. its more like repetitive reminder and similarly for

*6/4     2       3*2 = 6                        3*2-2 = 4

The final reminder is reminder of 6 divided by 4 and that’s 2.

Please let me know if I am missing something
 

simple solution for question 1..

2^300 and 3^200 simplify them to (2^3)^100 and (3^2)^100 which becomes 8^100 and 9^100 which states 2^100 < 3^200 no need to remember log of 2 or 3

M4

i think we can do like dis also 140/7=20 259/7=37 hence total number(including both) = 37-20+1 = 18.....

M15) the ans is 9 nos -

M15) the ans is 9 nos - 11,13,17,31,37,71,73,79,97