Ques:- Find the Max. value of n such that
42x57x92x91x52x62x63x64x65x66x67
is perfectly divisible by 42^n .
Please provide the answer with explanation. TIA.
Ques:- Find the Max. value of n such that
42x57x92x91x52x62x63x64x65x66x67
is perfectly divisible by 42^n .
Please provide the answer with explanation. TIA.
Hi
I m not sure but the answer could be n=3.
Since 42^n = (2*3*7)^n
In the expression
42x57x92x91x52x62x63x64x65x66x67
no of 2's & 3's are greater than 7
So number of 7 in the expression will decide the maximum value of n. i.e 3
Ques:- Find the Max. value of n such that
42x57x92x91x52x62x63x64x65x66x67
is perfectly divisible by 42^n .
Please provide the answer with explanation. TIA.
I have got the answer as 3.
I think i am able to find out the approach and solved two similar probs too.. So now i just want to knw that is my approach right?
Following is my approach:-
from 42, i got 2x3x7 , then i had a look on the given multiplication of numbers and able to find out that 7 is the number which will have the minimum appearance. So searched the power of 7 in multiplication and got the answer as 3.
P.S. Please correct me if m wrong somewhere.
Ques:- Find the Max. value of n such that
42x57x92x91x52x62x63x64x65x66x67
is perfectly divisible by 42^n .
Please provide the answer with explanation. TIA.
Hi
I m not sure but the answer could be n=3.
Since 42^n = (2*3*7)^n
In the expression
42x57x92x91x52x62x63x64x65x66x67
no of 2's & 3's are greater than 7
So number of 7 in the expression will decide the maximum value of n. i.e 3
Yes the answer is right.
There are two similar questions:-
Ques1) 570x60x30x90x100x500x700x343x720x81 is perfectly divisible by 30^n.
Ques2)-77x42x37x57x30x90x70x2400x2402x243x343 is perfectly divisible by 21^n.
Both the above ques. can be solved in similar manner.
I hope the approach i mentioned in my last post which is similar to urs is the best fit here. 😃
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, P.2, & 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
Waiting for the Explanation puys.
The Questions is of Progression from Arun Sharma in LOD2 & LOD 3
for question 2 ...try to solve just by simply ...1+ 0.33+ .16+0.1+0.06+0.04+0.03+0.02+.01+0.001+......so u can see that after 7 8 term u can find the pattern and the values are also not increasing by much difference.....so simply add u willl get 1.61...so u can easliy guess the values will not reach more than 2 .......till infinty....it is short cut for this kind of question.....
Q1. The sum of series
1/(sqrt2+sqrt1) + 1/(sqrt2+qrt3) +.......................+ 1/(sqrt120+sqrt121)
Ans = 10, 11 , 12, None of these
My take is 10.
Rationalise the first term 1/sqrt2+sqrt1
we have sqrt2-sqrt1/1
Similarly for second term
sqrt3-sqrt2/1
On adding all terms together all terms cancel out except
-1+sqrt121
-1+11=10 Ans
Q)
n! has 23 zeroer.What is max possible value of n?
q)how to solve below mentioned types of question in quick way
Find max value of n such that
42x57x92x91x52x62x63x64x65x66x67 is perfectly divisible by 42^n.
Q)
n! has 23 zeroer.What is max possible value of n?
q)how to solve below mentioned types of question in quick way
Find max value of n such that
42x57x92x91x52x62x63x64x65x66x67 is perfectly divisible by 42^n.
--> Is the Answer to the Quest 1 is 99! ?
--> The second question does not have any kind of formula on the basis of which we can solve it quickly. Its just a practice of observing the numbers as fast as possible so that we can find out mentally that how many instances of a prime factor are present in the term.
Like in 2nd question we have, 42=2x3x7.
So by looking at the term (42x57x92x91x52x62x63x64x65x66x67) it is apparent that out of prime factors 2,3 n 7 , 7 must be the one with least number of appearances in the term (42x57x92x91x52x62x63x64x65x66x67).
So we just need to find the max of 7s in the above term so that the 42 divides the term perfectly.
7 in term is in 42=7x6, 91=7x13, 63=7x9 .
So max instances of 7 are 3 which is the answer. 42^3
P.S. I hope its pretty clear. But correct if m wrong somewhere.
For the number 2450 find .
1. the sum and number of all factors.
2. the sum and number of all even factors.
3. the sum and number of all odd factors.
4. the sum and number of all factors divisible by 5.
5. the sum and number of all factors divisible by 35.
6. the sum and number of all factors divisible by 245.
Please tell the approach .
--> Is the Answer to the Quest 1 is 99! ?
--> The second question does not have any kind of formula on the basis of which we can solve it quickly. Its just a practice of observing the numbers as fast as possible so that we can find out mentally that how many instances of a prime factor are present in the term.
Like in 2nd question we have, 42=2x3x7.
So by looking at the term (42x57x92x91x52x62x63x64x65x66x67) it is apparent that out of prime factors 2,3 n 7 , 7 must be the one with least number of appearances in the term (42x57x92x91x52x62x63x64x65x66x67).
So we just need to find the max of 7s in the above term so that the 42 divides the term perfectly.
7 in term is in 42=7x6, 91=7x13, 63=7x9 .
So max instances of 7 are 3 which is the answer. 42^3
P.S. I hope its pretty clear. But correct if m wrong somewhere.
yaar,
Answer for
1)"this is not possible "as per arun sharma's quant
2)42^3.
pls let me know if u can try q1 again.
Q)
n! has 23 zeroer.What is max possible value of n?
q)how to solve below mentioned types of question in quick way
Find max value of n such that
42x57x92x91x52x62x63x64x65x66x67 is perfectly divisible by 42^n.
1) There is no such n! that would have 23 zeros.
The reason is that the number of zeros in any factorial increase by 1 with every multiple of 5, by 2 with every multiple of 25 and so on...
As 100! has 24 zeros and since 100 is a multiple of 25 hence (100-1) = 99 would have 22 zeros and not 23.
2) 42 = 7*6.
Since, 7>6.
Hence there would be fewer 7's.
So, just find multiples of 7 in the multiplication.
1) There is no such n! that would have 23 zeros.
.
I spent half an hour on it Yesterday, at Office..As a result Headache..
--------------------------------------------------------------
For the number 2450 find .
2450= 2 x 5^2 x 7^2.......break into prime factors
Now we can form n no. of factors multiplying powers of 2 5 7
1. the sum and number of all factors.
No.= (1+1)(2+1)(2+1) Sum=(2^0 + 2^1)(5^0 + 5^1 + 5^2)(7^0 +7^1+7^2)
2. the sum and number of all even factors.
For even factors the power of 2 must always be 1 in this case
So Sum= (2^1)(5^0 + 5^1 + 5^2)(7^0 +7^1+7^2)
3. the sum and number of all odd factors.
For odd factors the power of 2 must always be 0 in this case
So Sum= (2^0)(5^0 + 5^1 + 5^2)(7^0 +7^1+7^2)
4. the sum and number of all factors divisible by 5.
For factors of 5 power of 5 must be 1 or 2..(2^0 + 2^1)( 5^1 + 5^2)(7^0 +7^1+7^2)
5. the sum and number of all factors divisible by 35.
For 35 .. power 0f 5 , 7 must be >=1..Similarly
6. the sum and number of all factors divisible by 245.
Do post if it seems confusing
Regards,
Never Back Downif N = 77777777, where the digit 7 repeats itself 429 times. What is the remainder left
when N is divided by 1144?
Is the answer 512.
If yes then will explain it, so let me know as i dont have the book.
yes the answer is 512. pls explain.
if N = 77777777, where the digit 7 repeats itself 429 times. What is the remainder left
when N is divided by 1144?
1144 = 8*11*13.
777 mod 8 = 1.
429 mod 6 = 3.
777 mod 11 = 7.
777 mod 13 = 10.
So, remainder would be of the form 8k+1 = 11p+7 = 13x+10. Now check with options for a number of that form.
1144 = 8*11*13.
777 mod 8 = 1.
429 mod 6 = 3.
777 mod 11 = 7.
777 mod 13 = 10.
So, remainder would be of the form 8k+1 = 11p+7 = 13x+10. Now check with options for a number of that form.
Bhai rd wala part samajhana...yeh kyun liya hai??
smnitb Saysyes the answer is 512. pls explain.
Need to form a committee of 5 from 4 men and 4 women. Committee consists of a president, a vice president and 3 secretaries.
How many ways can the committee be selected with a maximum of 2 women and with a maximum of 1 woman holding one of the two posts on the committee?
Available 4 men(M) and 4 women(W)
Group of 1 President(P) 1 Vice-president(VP) and 3 Secretaries (Sec)
P VP Sec.
1 M M 1M 2W 4C1 * 3C1 * 2C1 * 4C2 = 144
2 M M 2M 1W 4C1 * 3C1 * 2C2 * 4C1 = 48
3 M W 3M 4C1 * 4C1 * 3C3 = 16
4 M W 2M 1W 4C1 * 4C1 * 3C2 * 3C1 = 144
5 W M 3M 4C1 * 4C1 * 3C3 = 16
6 W M 2M 1W 4C1 * 4C1 * 3C2 * 3C1 = 144
1 + 2 + 3 + 4 + 5 + 6 = 512
:)
vyomconfused SaysBhai rd wala part samajhana...yeh kyun liya hai??
Bhai any number repeated 6k times is divisible by 7,11,13.
Bas wohi isliye kara hai. 😃
Bhai any number repeated 6k times is divisible by 7,11,13.
Bas wohi isliye kara hai. :)
==================
bhai
7, 11, 13 se check kyu kiya hai wo to 8, 11, 13 hai, plz btao.
429 ka rmdr kyu nikala ??
==================
bhai
7, 11, 13 se check kyu kiya hai wo to 8, 11, 13 hai, plz btao.
429 ka rmdr kyu nikala ??
Bhai mein toh sirf bata ra tha ki 7,11 aur 13 se divisible hota hai jab koi bi number 6k times repeat ho.
isiliye 429 is 6k+3 form na.
so, 7777...429 times can be written as 7777....(426 times) 777 (3 time). 426 tak sab divisible. bas 777 bacha. uska remainder lelo 11 aur 13 se.
Hope it is clear now. 😃