Number Systems, Progressions

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Twinkle tells Raveena that she has got 3 kids and 2 of these kids are twins, and also that their ages are all integers. She tells Raveena the sum of the ages of her kids and also the product of their ages. Raveena says that she has insufic...
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 If (5.55)^x = (0.555)^y = 1000, then the value of 1/x - 1/y is 

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1/3
2/3
3
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 of 64 boxes of apricots, each box contains at least 60 and at most 81 apricots. no more than three boxes have the same # of apricots. what is the least number of apricots that can be there in all the boxes together? 

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in an infinite geometric progression ,each term is equal to 2 times the sum of the terms that follow.if the first term of the series is 8,find the sum of the series.

I need answer with explanation..

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32/3
34/3
data inadequate
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I need answer with explanation...

Post 6076658568134656
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Shaktig39
@Shaktig39TISS Mumbai  ·  1 karma

97.

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Adil1997
@Adil1997  ·  7 karma

97

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Find the remainder when the 100-digit numbers formed by writing the consecutive natural numbers starting from 1 next to each other is divided by 16 a. 6 b. 8 c. 9 d. 10 pls help me I need answer with explanation

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tanujagrawal23
@tanujagrawal23  ·  0 karma

Remainder when a no. Is divided by 2^n is calculated by seeing the last n digits of a no.

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Nappy
@Nappy  ·  0 karma

Please ignore the comment

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Find the number of numbers between 200 and 300, both included, which are not divisible by 2, 3, 4 and 5. (a) 27 (b) 26 (c) 25 (d) 28 (e) None of these

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Ajindal
@Ajindal  ·  0 karma

i think it should be 'not divisible by 2,3,4 or 5'...

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yashchawla
@yashchawla  ·  532 karma

can we do it by eulers?

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which is more likely to happen? rolling at least one 'six' in four throws of a single die or rolling at least one 'double six' in 24 throws of a pair of dice? pls show solutions.:)

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getgoingmba
@getgoingmba  ·  0 karma

first one is 91/6)^4 and 2ns must be (1-(35/36))^24 or (1/36)^24 or (1/6)^48 so 1st is more likely to occur... IMO

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culdip
@culdip  ·  7,464 karma

1 - (5/6)^4 and  1 - (35/36)^24

check by taking log, which one is greater

log(5/6)^4 = 4*(log5 - log6) = 4*(.699 - .778) = -316

log(35/36)^24 = 24*(.699+.845) - 48*(.778) = 24*1.544 - 24*1.556  = 24*-.012 = -.288

Therefore (35/36)^24 > (5/6)^4

So (1 - (5/6)^4) > (1 - (35/36)^24), So 1st one is more likely to happen

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For how many values of natural number n

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etcebiswa
@etcebiswa  ·  145 karma

no common factor

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120594sayan
@120594sayan  ·  0 karma

every value

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At the farewell party of a certain class, having n students, if each student of the class gave one gift each to exactly k other students of the class, the number of students in the class who received at least one gift each cannot be less than

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rishab619
@rishab619  ·  0 karma

k+1

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Shaktig39
@Shaktig39TISS Mumbai  ·  1 karma

How?

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find last two digits of 37^11 . share approach plz.

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22baba
@22baba  ·  0 karma

This is using the Binomial expansion method where first we did need the manipulation to effect an ending with 9 (i.e., 69) to re-express it as (-1+70) and then proceed with the expansion. Note that higher power terms in 70 can be neglected as they have more than two zeros at the end and contribute nothing to the sought last two digits.

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22baba
@22baba  ·  0 karma

Of course 37^2 = 1369 = _69 before setting off with the Binomial expansion here, just in case that was not clear.

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