1. A number is having exactly 72 factors. What can be the maximum and minimum number of prime factors of this number?
a.71 &1 b. 6 & 2 c. 5 & 2 d. None of These
2. If we write all 1852 natural no. side by side starting from 1,we will get a very large no. Find the remainder when first 1750 digits of this no. are divided by 16.
a.30 b.20 c.10 d. None of These
3. N = 22225555 + 55552222 . What is the remainder when N is divided by 7?
a 6 b. 5 c. 4 d. 0
4. M = Product of all even numbers from 1 to 25.
N = Product of all odd numbers from 26 to 200.
M
a. 42 b. 25 c. 21 d. none of these
Q1:-......
ReplyDeleteAnswer-(a)
Q2:-......
Answer-(b)
Q3:-......
Answer:-(d)
Q4:-......
Answer:-(b)
This comment has been removed by the author.
Deleteq3....ans(d)
ReplyDeleteans1. c)
ReplyDeleteans2. d) (13)
ans4. d) (23)
ans1. d) (5,1)
ReplyDeleteans2. d)
ans3. d)
ans4. d) (23)
ans1= a
ReplyDeleteans2=none of these, 4
ans3=d
ans4=d(23)
Sol 1 72 = 2x2x2x3x3 a^1 b^1 c^1 d^2 e^2
ReplyDeleteTherefore Max No of prime factors is 5.
Min no of prime number is 1 a^71
Ans none of these
Sol 2.First 1750 digits
ReplyDelete12345678910111213141516171819---9899100101101---
Single digits numbers 9 digits used 9
Souble sigits 90 , digits used 90x2=180
Total digits used 9+180 = 189.
Remaining numbers 1750-189=1561 = 1560+1.
In group of three 100101101---- group of three .
1560/3=520 . thus 520 group is 100+(520-1)1 = 619
Thus last four digits 6196 , when divided by 16 leaves remainder 4
A= 2222^5555 + 5555^2222 when divided by 7
ReplyDelete2222=7K+3 & 5555= 7m+4 thus
3^5555 /7 = 3^5550 X 3^2 X 3^3 = (3^3)^1850X9X27
(28-1)^even X 2 x 6 = 1 x 12 = 5( Remainder )
4^2220 X 4^2 = (4^3)^740 X 16 =(63+1)^740 X 2 =1X2 =2
Thus required remainder = 5 + 2 = 7 = 0 Thus remainder is 0.
M = Product of all even numbers from 1 to 25.
ReplyDeleteN = Product of all odd numbers from 26 to 200.
M N will give how many zeros?
a. 42 b. 25 c. 21 d. none of these
We are looking only for 2 & 5.
M Will give 2 & 5
2X4X6X8x10X12x14X16x18x20X22X24 = 2^12 X 2^10 X 5^2 = 2^22 X 5^2
N 27X29X31x--------199 will give 5 from odd multiples of 5.
35 X 45x 55 X 65 X 75 X ------ X 195 which will contain 5^21
Thus M X N will contain 2^22 X 5^23 Thus zero will be 22. None of these
(((32)^32)^32)/7
ReplyDeleteremainder for the given problem
plz. give the solution
(2222^5555+5555^2222)/7
ReplyDelete2222/7 remainder is 3
5555/7 remainder is 4
=> (3^5*1111)/7 + (4^2*1111)/7
=> (5^1111)/7 + (2^1111)/7
=> (5^1111+2^1111)/7
=> (5+2)(5^1110-(5^1109)*2 + (5^1108)*2^2-.................)/7
5+2=7 hence remainder is 0
Q. 888.....8(48 times 8)/19 .what is the remainder ?
ReplyDelete3.
ReplyDelete(2222^5555 + 5555^2222)/7 ... [1]
now,
2222 ≡ 3 (mod 7)
5555 ≡ 4 (mod 7)
this implies:
[1] ≡ (3^5555 + 4^2222)/7 ... [2]
now,
E(7) = 6 (Euler's totient)
since (2222, 7) and (5555, 7) are co-prime,
2222^6 ≡ 1 (mod 7)
5555^6 ≡ 1 (mod 7)
furthermore,
5555 ≡ 5 (mod 6)
2222 ≡ 2 (mod 6)
this implies,
[2] ≡ (3^5 + 4^2)/7 or (243 + 16)/7 or 259/7
therefore, since 259 is divisible by 7, the remainder is 0.