N = 77777777, where the digit 7 repeats itself 429 times. What is the remainder left when N is divided by 1144?
pls provide detailed solution
I think the answer should be 777.
1144=8*11*13 Lets take the number to be *1000 + 777 Now we know any number having digits repeated 6n times will be divisible by 11 and 13. So 7777...426times will be divisible by 11 and 13. Again 1000 is divisible by 8.
So 7777...426times will be divisilble by 1144. Remaining will be 777 which is less than 1144 and hence should be the remainder.
Privileged Number is a natural number which has two prime numbers as its neighbors on the number line. For example, 4 and 12 are privileged numbers. What is the mean of all privileged numbers less than 100? A. 30.5 B. 34 C. 28.5 D.26
A) 30.5 4,6,12,18,30,42,60,72 - priveleged numbers
1. The single digits a and b are neither both nine nor both zero. The repeating decimal 0.abababab... is expressed as a fraction in lowest terms. How many different denominators are possible? A.4 B.6 C.5 D.3
2. What is the value of n such that n! = 3! 5! 7! A.10 B.11 C.8 D.9
1. The single digits a and b are neither both nine nor both zero. The repeating decimal 0.abababab... is expressed as a fraction in lowest terms. How many different denominators are possible? A.4 B.6 C.5 D.3
2. What is the value of n such that n! = 3! 5! 7! A.10 B.11 C.8 D.9
1)4 2)10
pls verify
1. C(5)........................... 3,9,11,33,99...All factors of 99 except one. 2. This one I am a bit confused. I also think the answer should be n=10, because RHS has two 5 and hence n! should be ending in 00. But here is the explanation provided from the place where I found it.
Since R.H.S. has 7! Multiplied by a positive quantity, L.H.S will be greater than 7!. As R.H.S. does not have the prime factor of 11, L.H.S. will be less than 11!. As R.H.S. has only one power of 5 (in 5!), L.H.S. will be less than 10!. Therefore, only possibilities are 8! And 9!. Considering powers of 3 on both sides, n = 9.
I didnt get it very well. Can someone please explain if I am doing something wrong.:banghead::banghead::banghead:
1. The single digits a and b are neither both nine nor both zero. The repeating decimal 0.abababab... Is expressed as a fraction in lowest terms. How many different denominators are possible? A.4 b.6 c.5 d.3
2. What is the value of n such that n! = 3! 5! 7! A.10 b.11 c.8 d.9
1. C(5)........................... 3,9,11,33,99...All factors of 99 except one. 2. This one I am a bit confused. I also think the answer should be n=10, because RHS has two 5 and hence n! should be ending in 00. But here is the explanation provided from the place where I found it.
Since R.H.S. has 7! Multiplied by a positive quantity, L.H.S will be greater than 7!. As R.H.S. does not have the prime factor of 11, L.H.S. will be less than 11!. As R.H.S. has only one power of 5 (in 5!), L.H.S. will be less than 10!. Therefore, only possibilities are 8! And 9!. Considering powers of 3 on both sides, n = 9.
I didnt get it very well. Can someone please explain if I am doing something wrong.:banghead::banghead::banghead:
i got my first 1 wrong...blunder...i wrote 2*3=5..lol.. anyways...2nd que is an easy 1...it goes like dis... 3!*5!*7!= 7!*(2*3*4*5)*(2*3)=7!*(4*2*3*3*5*2)=7!*8*9*10=10!... cheers...!!
1. C(5)........................... 3,9,11,33,99...all factors of 99 except one. 2. This one i am a bit confused. I also think the answer should be n=10, because rhs has two 5 and hence n! Should be ending in 00. But here is the explanation provided from the place where i found it.
since r.h.s. Has 7! Multiplied by a positive quantity, l.h.s will be greater than 7!. As r.h.s. Does not have the prime factor of 11, l.h.s. Will be less than 11!. As r.h.s. Has only one power of 5 (in 5!), l.h.s. Will be less than 10!. Therefore, only possibilities are 8! And 9!. Considering powers of 3 on both sides, n = 9.
i didnt get it very well. Can someone please explain if i am doing something wrong.:banghead::banghead::banghead:
N = 77777777, where the digit 7 repeats itself 429 times. What is the remainder left when N is divided by 1144?
pls provide detailed solution
I think the answer should be 777.
1144=8*11*13 Lets take the number to be *1000 + 777 Now we know any number having digits repeated 6n times will be divisible by 11 and 13. So 7777...426times will be divisible by 11 and 13. Again 1000 is divisible by 8.
So 7777...426times will be divisilble by 1144. Remaining will be 777 which is less than 1144 and hence should be the remainder.
Let me know the OA.
Girish....what is the OA for this problem? Can u guys post some more solutions to this? I am not sure whether I am on the right track with this or not....
m nt getting this 1...can u xplain why did u take hcf of powers..?? i mean any explaination by binomial theorem etc...
we have to find HCF of the nos lets try for (a^2-1) & (a^3-1) we get HCF as (a^1-1) also for 2 & 3 we have HCF 1 try for (a^2-1) & (a^4-1) we get HCF as (a^2-1) also for 2 & 4 we have HCF 2
for (a^8-1)= (a^4-1)*(a^4+1) = (a^2-1)*(a^2+1)*(a^4+1) & (a^6-1)==(a^2-1)(a^4+a^2+1)....apply (a^3-b^3) formula we get HCF as (a^2-1) also for 8 & 6 we have HCF 2
so we conclude this method of taking HCF of powers is applicable for (a^n-1) as 1 can take any power and can be used for expansion using (a^2-b^2) or (a^3-b^3) or (any other) formulae to find HCF of the terms
Girish....what is the OA for this problem? Can u guys post some more solutions to this? I am not sure whether I am on the right track with this or not....
the options weren't given...777 is the right answer... options weren't given... i needed the approach...thnx!!!!
P is the product of first 30 multiples of 30. N is the total number of factors of P. In how many ways N can be written as the product of two natural numbers such that the HCF of these two natural numbers is 19? (a)3 (b)4 (c)5 (d)6
2. What is the value of n such that n! = 3! 5! 7! A.10 B.11 C.8 D.9 ans is 10:thumbsup: in 10! power of 2 is 8 , power of 3 is 4 & power of 5 is 2 & power of 7 is 1.... and all the condition is satisfied with the que i.e3!*5!*7!.....=2^8*3^4*5^2*7^1
How did u get these numbers? There are 24 prime numbers within hundred. Did you filter each out manually?
What if the number had exceeded say 200 or 500?
All such numbers have to be multiples of 6 because in a series of three consequtive numbers two of the numbers wil be multiples of 2 & 3, exceptions being wen the number in the middle is a multiple of 6.....after that u have to check if neighbours are prime or not
Hey people! Can someone tell me how to obtain the remainder when 128^1000 is divided by 153? Answer is 52... Losing sleep over this question someone please help
1. C(5)........................... 3,9,11,33,99...All factors of 99 except one. 2. This one I am a bit confused. I also think the answer should be n=10, because RHS has two 5 and hence n! should be ending in 00. But here is the explanation provided from the place where I found it.
Since R.H.S. has 7! Multiplied by a positive quantity, L.H.S will be greater than 7!. As R.H.S. does not have the prime factor of 11, L.H.S. will be less than 11!. As R.H.S. has only one power of 5 (in 5!), L.H.S. will be less than 10!. Therefore, only possibilities are 8! And 9!. Considering powers of 3 on both sides, n = 9.
I didnt get it very well. Can someone please explain if I am doing something wrong.:banghead::banghead::banghead:
bolded part is the place where i think you are going wrong... there are two powers of 5 in 7!*5!*3!..
