Number System - Questions & Discussions

MBA CAT 2022
3156
@notimetowaste said: if u multiply 10!, the last digit u get is 8,(ignoring 0's and taking the nonzero numbers only), then upto 96! we have nine 8's to multiply with 6!(which has last digit 2) of 96! so in totality we have,8^9*2= will give last digit = 6 (u can solve 8^9 using cyclicity of powers)i know this method is not good for bigger numbers if anyone knows to do in a better way, kindly highlight them
thanx... 😃 will post more questions if any difficulty... :)
3157
@YouMadFellow said: www.wolframalpha.com/input/?i=...Ans is -2 or 69.. I guess after the last step k = +2, or -2
That is correct YouMadFellow,I am not able to eliminate one of the solutions. Not sure, how to proceed from here, my soln says +/-2, while the correct solution is -2.
3158

did u get this answer? i guess its in four ways.

@sameersapre23 said: please anyone explain this question its recent aimcat questionin how many ways 600 can be expressed as the sum of two or more consecutive integers.... !!
3159
@janvats said: find the remainder when (38^16!)^1777 is divided by 17 a. 1 b. 16 c. 8 d. 13
ans is a (i.e 1)
3160
@taruniim said: what is the remainder when 9+9^2+9^3............9^(2n+1) is divided by 6.....ans is 36=2 *3..all powers of 9 leave remainder 0 and 1 in case of division by 3 and 2 respectively...so remainder should be of form 1+1+1...(2n+1)times...i.e remainder on divin by 2 should be 1...so answer sud b 1...bt answer is 3...??HELP..
Hi there,
First of all if you get any question in generalized form such as this, the quick and dirty way is to put values and tick the answer. Like here Put n=0, you get answer 3 and done. This is CAT not IIT JEE.

You are right that, 3 divides the expression, but 2 leaves a remainder of 1. Like for 9|2 gives 1 as remainder, and when 9+9^2+9^3|2 leaves 1 as remainder.

Now, E = 3k and E=2k' +1. Since E gives a remainder of 1 when divided by 2, so
3k gives a remainder of 1 when divided by 2, knowing that 3 gives a remainder of 1 when divided by 2, K should also give a remainder of 1 when divided by 2. Hence,
k=2k'' +1
so 3k=6k''+3
Therefore, E=6k''+3.

This is a special case of Chinese Remainder Theorem, if say n1 divides E and n2 gives a remainder of k, then n1*n2 gives a remainder of n1*k while dividing E.


3161

Puys please help:
122333444.... upto 1000 digits when divided by 16 yields?

How do I solve such type of questions - Approach please!

3162
@akansh_1 said: Puys please help: 122333444.... upto 1000 digits when divided by 16 yields? How do I solve such type of questions - Approach please!
To find the remainder of given number is equivalent to find the remainder of last 4-digit of the number

So,Using arithmetic mean,

5*9*1 + [(10 + n)/2]*(n - 10)*2 = 1000

=>(10 + n)*(n - 10) =955

At n = 32,(10 + n)*(n - 10) = 924

So,the remainder 31 digits would be 3 only and 3333 mod 16 = 5
3163
@akansh_1 said: Puys please help: 122333444.... upto 1000 digits when divided by 16 yields? How do I solve such type of questions - Approach please!
1
22
333
4444
55555
...
Till the single digit numbers we have 9*10/2 = 45 digits.
Remaining digits = 1000 - 45 = 955.
10 would occur 10 times.
11 11 times, 12 12 times. and so on.
=> 2*10 + 2*11 + 2*12 + 2*13 + .... + 2*99 = 2(10 + 11 + 12 + 13 + ... +n).
At n = 32. this sum becomes 966.
Hence, we have 966+45 = 1011 digits now.
So last 4 digits would be 2323

For remainder by 16, just consider the last 4 digits.
=> 2323 mod 16 = 3.
3164

Thanks a lot ..Enceladus and mani0303 for the approach.

3165

@[262236:akansh_1] what is the answer?

3166

@[75748:karl]..is there any particular method for solving remainder problems when factorials r involved and v cant apply wilson or fermats theorem?

3167

@[592846:YouMadFellow] couldnt get the right ans of this Qs

the sum of three numbrz in gp is 14 & sum of their squares is 84 find the largest.plz help me out

3168

@[419941:Enceladus]two numbers a & b r such dat their GM is 20% lower than AM find ratio b/w numbrz. couldnt find the right ans

3169

@[592846:YouMadFellow] the sum of first four terms of AP is 28 & sum of first 8 terms of same AP is 88 find sum of first 16 terms of AP

3170

@[592846:YouMadFellow] A SUM of money kept in bank amounts to Rs 1240 in 4 years and Rs 1600 in 10 years at simple interest. find the sum

3171
@notimetowaste said: did u get this answer? i guess its in four ways.
no its not 4 ways it should be 5 ways
3,5,15,25 and one case is that when 2^k will come n "k" is the multiple of 16.
so total ways is 5
3172

@[595194:sahib_sheetu]

p = 1000
r = 6%
3173
@sahib_sheetu said: @YouMadFellow couldnt get the right ans of this Qs the sum of three numbrz in gp is 14 & sum of their squares is 84 find the largest.plz help me out
For this, you can take the numbers in GP as (a/r), a, (ar)

-> Then a+ar+a/r = 14 .. (1)
-> a^2 + (ar)^2 + (a/r)^2 = 84.... (2)

Square the first equation, you will get the (2) in the square of the first equation, replace the value of the second equation, and you will get

a^2(1 + r + 1/r) = 56

Using this, and equation (1) get a = 4 and then put a in (1) to get r = 2
3174
@sahib_sheetu said: @YouMadFellow the sum of first four terms of AP is 28 & sum of first 8 terms of same AP is 88 find sum of first 16 terms of AP
Let the first term be a , and common difference is d

So, sum of first four terms = (2a+3d)*4/2 = 28
=> 2a + 3d = 14

sum of first 8 terms = (2a+7d)*8/2 = 88
=> 2a + 7d = 22

a = 4, d = 2
3175
@sahib_sheetu said: @YouMadFellow A SUM of money kept in bank amounts to Rs 1240 in 4 years and Rs 1600 in 10 years at simple interest. find the sum
Amount = principle + interest.

principle be P, rate of interest be r

1240 = P + P.r.4/100
1600 = P + P.r.10/100

Solve it for P and r.