it will bw 14. Did it logically, no specific approach.
LHS eq. has to be less than 5, so added 4,3,2,1 ( from eq.) to 5... Then subtracted 1, as middle sign is <.>
160000 = 2^4*10^4
Both m and n should not be as multiples of 10. So one number must be multiple of 5.
= 2^8*5^4
So m + n = 256+625 = 881
Option C
@akansh_1 said: Solve: ||||x – 1| – 2| – 3| – 4| a. 4 b. 10 c. 14 d. 15 (Please tell me how to approach such multiple Modulus sums)
Are you sure about I guess 4,10 and 14 will satisfy in that case. If it is = then the solutions for x are -13, 15. From option 15 will be the answer in that case.
@kambarish- That's correct.. Nice shortcut way
@HydRocker - Ans is 14.
@HydRocker - Could you please tell me what was ur approach (for both the cases). Thanks!
@akansh_1 said: Solve: ||||x – 1| – 2| – 3| – 4| a. 4 b. 10 c. 14 d. 15 (Please tell me how to approach such multiple Modulus sums)
In this question, you can just keep opening the mod signs. Removing the first mod sign, we have two cases:
Case 1. |||x-1|-2|-3| -4
OR
Case 2. -|||x-1|-2|-3| +4 -1
Case 2 is always true, as |y|>=0 for any y. So we can discard it as it does not provide any extra information, and proceed with just Case 1. Keep opening the mod signs, and each time you will get just one meaningful inequality, while the other one states the obvious like Case 2 above.
Final result: x
Yeah, it was from either Proc-5 or Proc-6. Case-2, which was so obvious, added to my confusion when I was solving then. Thanks lot :)
@RoadKill said:In this question, you can just keep opening the mod signs. Removing the first mod sign, we have two cases:Case 1. |||x-1|-2|-3| -4ORCase 2. -|||x-1|-2|-3| +4 -1Case 2 is always true, as |y|>=0 for any y. So we can discard it as it does not provide any extra information, and proceed with just Case 1. Keep opening the mod signs, and each time you will get just one meaningful inequality, while the other one states the obvious like Case 2 above.Final result: x
@[262236:akansh_1] Hmm. Waise, the question could also be "what is the highest integer that satisfies the inequality?"; answer would be the same..
13 zeroes in the factorial of n -- possibe value of n?
I need the method
@[475264:Vegeta555]
Number of zeroes in n! is given by
[n/5] + [n/25] + [n/125] + [n/625] + ..... where [.] is the greatst integer functn.
So, for above expression to be equal to 13, do some hit and trial, and you will get the value of n between 55 to 59, ie each one of 55,56,57,58 and 59 contains 13 Zeroes.
******************
Behind the Curtain :
You know that N! = 1.2.3.4.....N , a product of numbers from 1 to N.
Note that in any product if you have to determine the number of zeroes you will look for either 10s or any numbers that result in 10s, the simplest such numbers being 5s and 2s, since 5*2 = 10.
So if we count pairs of 5s and 2s in a product, we will be able to determine the number of 10s and hence the number of Zeroes in that product.
Secondly note that there is an abundance of 2s in numbers, every even number has a 2.
But 5 comes rare, one after every 5th number.
So in essence every 5 you will find can find a 2 of its own from somewhere and result into 10.
This reduces our job to count only the number of 5s in a product.
Now we have a simple formula that helps us to count the number of Ks in N!.
This formula is
[N/K] + [N/K^2] + [N/K^3] + ....
where [.] denotes the greatest integer functn, so that you have to take only the integer parts of the division.
Apply this formula to determine the number of 5s in any product to determine the number of Zeroes in that product.
Find the remainder when (6^83 + 8^83)/ 49.
Answer is 35. Approach pls
@frostie said: Find the remainder when (6^83 + 8^83)/ 49.Answer is 35. Approach pls
re writing the above question as (7-1)^83 + (7 +1)^83 /49
Expanding according to binmoial theorem everything will be divisible by 49 except for the last two terms.. but again the last term will get cancelled so the second last term of the series matters
which will be 83C1*7... * 2 for the second series
So reminder of 83* 14 /49 which will be 5*7 = 35
@[460820:frostie]:u can use normal toiteint method, bt as said by @[228054:manoj_msr], dat apporach is better for dis q...
@[460820:frostie]
Although the binomial theorem is the conventional way as well as I guess, the expected way by the Qn maker to solve this Qn.
But if it happens, that the remarkable thing with all the numbers in the Qn that they are expressible in terms of 7, misses your mind and you are stuck with Euler to solve this Qn. It can be solved like this.
********************************
Things to Notice
First note that there are two more special things about this Qn
1) The numbers 6 , 8 and 49.
49 = 6*8 + 1
2) The power 83 = E(49)*2 - 1
*********************************
Solve Now
Now, Euler(49) = 7^2-7 = 42
So, Rem[ 6^42 / 49 ] = 1
Rem[ 6^84 / 49 ] = 1
Knowing that 49 = 6*8 + 1
Rem[ 6^83 / 49] = 49-8 = 41 .............{ EUREKA ! }
Similarly,
Rem[ 8^42 / 49 ] = 1
Rem[ 8^84 / 49 ] = 1
and by Eureka Method, Rem[ 8^83 / 49 ] = 49 - 6 = 43
So the Qn reduces to
REM[ (6^83 + 8^83) / 49 ]
= REM[ 6^83/49 ] + REM[ 8^83/49 ]
= REM[ (41 + 43) / 49 ]
= REM[ 84/49 ]
= 35 .........Answer
**********************************
The NewFound Trick
The new shortrick found while solving this Qn ,dont know if its an extension of the official Euler Theorem.
I will generalize it here.
Let B is Base, P is power, D is divisor.
For an expression like,
Rem [ B^P / D ]
If B and D are coprime ,
Euler(D) = E
and
D = B*(some number k ) + 1
then , by EUler's
Rem[ B^P / D ] = 1
Rem[ B^2P/ D ] = 1
and by extensn of Euler's
Rem[ (B^(P-1) ) / D ]
= Rem[ (B^(2P-1) ) / D ]
= Rem[ (B^(3P-1) ) / D ]
= Rem[ (B^(nP-1) ) / D ]
= D - k
***********
Examples :
Rem[ 3^18/19 ] = 1
So, Rem[ 3^17/19 ] = 19 - 6 = 13
Rem[ 12^36/37 ] = 1
= Rem[ 12^72/37 ]
= Rem[ 12^108/37 ]
= 1
So, Rem[ 12^35/37 ]
= Rem[ 12^71/37 ]
= Rem[ 12^107/37 ]
= 37 - 3
= 34
***********
A natural number N has €˜k €™ distinct prime factors. If the total number of factors of N is 72, then what is the product of all the possible values of k?
@rahul005 said: A natural number N has €˜k €™ distinct prime factors. If the total number of factors of N is 72, then what is the product of all the possible values of k?
120?
correct...pls show detailed process.
@[410661:rahul005]:72=2^3*3^2..now 72 can be expressed as a product of maximum 5 distinct numbers and minimum 1..so product of all is 5!=120
Awesomeeeeeeeeeeeeee!!!!!!!!!!!!!!!!!!!!!!!!!!! 😃 😃 😃 😃 :)
@[568667:Psychamour]
1-)Rem[ 3^18/19 ] = 1
2)Rem[ 12^36/37 ] = 1
How come the remainder is 1...where as in 1st taking negative remainder would be -16 and in 2nd negative rem would be -25.....
Please correct me....
How come the remainder is 1...where as in 1st taking negative remainder would be -16 and in 2nd negative rem would be -25.....
Please correct me....