Hey puys can you clear the concept of mine for this 2 Ques..
1. 32^32^32/9 will leave remainder; choices 4 7 1 2
2. 32^32^32/7 will leave remainder: choices 4 2 1 3 5
N= 2*101*2*10001*2*100....1(15 zeros)*2*100...1(31 zeros)
N= 16*(101)*(10001)*(1000...1 (15zeros ))*(10000..1(31 zeros))
N= 16*(10^2 +1)(10^4 + 1)(10^16 +1)(10^32 +1)
on solving above you will get 16 times 16 in N
So the some of digit will be 16(1 + 6)= 112
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@ Nihilist: dude can you plz explain SUM OF DIGIT MEANS WHT?
-->I mean suppose any no. is "abcd" toh sum of digit means "a+b+c+d " or
wht. AND----> in above solution why u didnot write 16 (which is outside the bracket as "1+6"ot being divided by 9..WHY?) and if i am not wrong u divided all by 9 inside the bracket?>
---> moreover to get the sum of digit we need to divide by 9 but 16(outside the bracket is not divided by 9)
plz explain..
Hey puys can you clear the concept of mine for this 2 Ques..
1. 32^32^32/9 will leave remainder; choices 4 7 1 2
2. 32^32^32/7 will leave remainder: choices 4 2 1 3 5
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1) go by euler: E(9)=6, (32^32)/6 => Rndr= 4, hence : /9 i.e.
(32*32*32*32)/9 => R as (5*5*5*5)/9=> Rndr as 4
similarly your qstion no 2nd can be solve
2) E(7)=6 hence , power Rndr=4 => (32*32*32*32)/7 => (4*4*4*4)/7 hence final Rndr=4
hope it may help you..
Hey puys!
pls help me understand how to check whether a number is a perfect square or not.
Which of the following is not a perfect square?
a - 100856, b - 325137, c - 945729, d - All, e - None
Thanks!
Hey puys!
pls help me understand how to check whether a number is a perfect square or not.
Which of the following is not a perfect square?
a - 100856, b - 325137, c - 945729, d - All, e - None
Thanks!
325137 cant b a prfct square bcz no sqr end with 7..
for 100856 da digit sum is coming as 1+0+0+8+5+6=20=2+0=2
so it cnt b prfct square,since da digit sum of a prfct sqr is alwys 1,4,7 or 9
if option 1 n 2 are not prfct square going by the options i wud mark my asnwr as option d.
quidditch88 SaysCan someone ans this- a number is mistakenly divided by 5 instead of being multiplied by 5. Find the percentage change in the result due to this mistake.
wat is the ans buddy is that 4%.
Hey puys!
pls help me understand how to check whether a number is a perfect square or not.
Which of the following is not a perfect square?
a - 100856, b - 325137, c - 945729, d - All, e - None
Thanks!
is the ans is C.
Kushwaha Saysis the ans is C.
ans is D..none of them are perfect squares
325137 cant b a prfct square bcz no sqr end with 7..
for 100856 da digit sum is coming as 1+0+0+8+5+6=20=2+0=2
so it cnt b prfct square,since da digit sum of a prfct sqr is alwys 1,4,7 or 9
if option 1 n 2 are not prfct square going by the options i wud mark my asnwr as option d.
bjhai option c explain krdo though I knw that the digit sum is ending with 9 n even know that it is not necessarily happen that the digit sum ending with any of 1,4,7 or 9 will alwys b a perfect square. So how will one get this right ?
Can anyone Explain.
If x be the firt term, y be the nth term and p be the product of n terms of a GP, then value of p^2 will be?
(xy)^n-1, (xy)^n, (xy)^1-n, (xy)^n/2, None of these
Kushwaha Sayswat is the ans buddy is that 4%.
ans is 96%. soln - (5x-x/5)/5x
Q.
The expression
333^555 +555^333
is divisible by
a. 2
b. 3
c. 37
d. 11
e. All of these
Q.
The expression
333^555 +555^333
is divisible by
a. 2
b. 3
c. 37
d. 11
e. All of these
Hi All,
Can anyone please explain abt Euler's Theroem?
Also let me know where should we use it?
Thanks In Advance...
All the Best
Hi All,
Can anyone please explain abt Euler's Theroem?
Also let me know where should we use it?
Thanks In Advance...
All the Best
Euler's theorem says
x^E/N=1
where, x and N are co-primes and E is the Euler's Number of N.
If N=a^m*b^n where a,b are prime factors,
E=N(1-a^-1)(1-b^-1)
=Number of co-primes less than N
=Number of non-multiples of prime factors of N
=N-1(if N is a prime number)
Its used as a shortcut method to solve remainder questions.
Lets try to find the remainder of 8^25/13
Step 1 - Find remainder of 8/13 = 8
Step 2 - Find E of 13 = 12
Step 3 - Find remainder of 25/12 = 1
Step 4 - Original expression is simplified to 8^1/13
Step 5 - Remainder = 8
Hope that serves your purpose.
Thanks a lot man:-)
Q.
The expression
333^555 +555^333
is divisible by
a. 2
b. 3
c. 37
d. 11
e. All of these
My take would be option e.All of these.
chking divisibility of 2:
Rule: number should be an even number
last digit of 333^555 will be 7 (by property of cyclicity)
last digit of 555^ will be 5 (obivously 5 power anything, last digit is 5)
7+5=12 hence 2 will be the last digit which is even.
hence the expression will be divisible by 2.
chking divisibility of 3:
Rule: sum of digits should be divisible by 3
3+3+3=9 divisible by 3
so 333 power anything will be divisible by 3
5+5+5=15 divisible by 3
so 555 power anything will be divisible by 3
And,remember remainders are additive in nature hence 0+0/3 =0
hence the expression will be divisible by 3.
This much work would suffice to prove option should be e.All of these
Some Questions puys.
Q1. The sum of series
1/(sqrt2+sqrt1) + 1/(sqrt2+qrt3) +.......................+ 1/(sqrt120+sqrt121)
Ans = 10, 11 , 12, None of these
Q2. Find Infinite sum of the series 1/1+1/3+1/6+1/10+1/15.............
Ans = 2, 2.25, 3, 4
Q3. How many 3 digit nos have the property that their digits taken from left to right form an AP or a GP?
Ans = 15, 36, 20, 42
Q4. An AP P consists of n erms.
From the progressions three different progressions P1, P2, & P3 re created such that P1 is obtained by the 1st,4th,7th....terms of P, P2 has the 2nd,5th,8th .......terms of P and P3 hs the 3rd,6t,9th....terms of P. It is found that of P1,P2 and P3 two progression have the propertythat their average is ielfa term of the original Progression P. Find possible values of n.
Ans = 20, 26, 36, both 1 & 2
Q.
The expression
333^555 +555^333
is divisible by
a. 2
b. 3
c. 37
d. 11
e. All of these
The number obviously will be divisible by 11 and 2 because sum of two odd nos. is always even...I shall be glad if anyone could explain why 37 which happens to be a prime number...
My take would be option e.All of these.
..............................................
Thanks a lot...