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@bpolagani** Remainder when 2^1100 is divided by 101***Fermat's theorem* says that if p is prime and **p does not divide a, then a^(p-1) when divided by p the remainder is 1.**

**Remainder when 971(30^99+61^100)*(114 8)^56 is divided by 31**

Since the divisor 101 is prime make use of the Fermat's rule.

Hence, 2^100 divided by 101, the remainder is 1 => 2^1100 when divide by 101 the remainder is 1.

Consider 30^99 + 61^100

30 divided by 31, the remainder is -1 and hence 30^99 divided by 21 the remainder is -1

61 divided by 31, the remainder is -1 and hence 61^100 divided by 31, the remainder is 1.

Thus the remainder of 30^99 + 61^100 divided by 31 is -1+1 = 0

Hence the remainder of the given product must also be 0

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@swahamohapatra said: need help!!A clock set right at 12 noon on monday loses 1/2% on the correct time in the 1st wk but gains 1/4% on the true time during 2nd wk. the tttime shown on monday after two wks will be12:25:1212:50:2411:34:48none

gains 0.5%(X) and loses 0.25%(X) is as good as loosing 0.25%(X)

Hence in two weeks the clock looses 0.25%(7x24x60) = 25.2 minutes = 25 mins 12 sec

Thus, after two weeks the clock shows 12:00 noon - 25 mins 12 sec = 11:34:48

- 2 Likes

@neerajagayatri said:There are 4 identical oranges, 3 identical mangoes and 2 identical apples in the basket. the number of ways in whihc we can select one or more fruits from the basket isAns:60, 59, 57, 55, 56

Please help in solving this

With oranges i have 5 chances namely -> picking one orange or two oranges or three oranges or four oranges and not picking up an orange at all i.e. 5 chances

Similarly with mangoes 4 chances and with apples 3 chances.

Total chances = 5X4X3 = 60. But I cannot leave out all of them since i need to pick at least one fruit.

Hence, the number of ways = 60-1 = 59

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It looks correct for me as well...

Suresh sir,please take a look on this prob...

Same here as well...

Even if you consider only integral percentages it comes down to 98

- 1 Like

culdip bhai ka question

A straight stick is broken randomly in 3 pieces, what is the probability that a triangle can be made ?

with approach...

Two cuts will give us three pieces

If both the cuts are on the same half of the stick, one resulting piece will have a length more than or equal to half of the length of stick thus violating the rule that the sum of two sides must be greater than the third side.

Thus if both the cuts are on the same half of the stick, no triangle can be formed. For this the probability is 1/2

Now, though both the cuts are on different halves of the stick, if the distance between them is more than or equal to half the length of the stick, no triangle can be formed, for which the probability is 1/4

Thus, 1/2+1/4 = 3/4 of the cases, no triangle can be formed

Hence, probability of forming a triangle = 1/4

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Q.) A 5- digit numbernis such that if we put the digit 1 at the beginning of the number, we get a number that is three times smaller than what we get if we put the digit 1 at the end of the numbern. What is the sum of the digits of this numbern?a.) 23........... b.) 25..........c.) 26...........d.)27

O.A. is 26

Good Night puys

Number is 42857

Let the 5 digit number be n

Adding 1 at the beginning, the resultant number can be written as 10^5+n

Adding 1 at the end, the resultant number can be written as 10n+1

Given 3(10^5+n) = 10n+1

=> 7n = 299999 => n = 42857

Hence sum = 26

Bed Time...

- 1 Like

q.) a puzzle type question

forty

+ ten

+ ten

--------

sixty

if f,o,r,t,y,e.n,s,i,x are the digits then find all the digits.

sabko pranaam _____/\_____ ....va se bore hoke back to quant thread

29786

850

850

-------

31486

Another easy one :DQ.) How many pair of positive integers (a, b) are there such that their LCM is 2012?

P.S. Yaar Allan VA ki tayyari ho ri hai Zor Shor se.....Aaj bore ho gaya to socha doston se mil loon;)

2012 = 2^2 * 503

Let a = 2^x1 * 503^y1 and b = 2^x2 * 503^y2

(x1,x2) => 5 chances

(y1,y2) => 3 chances

Total 5*3 = 15 numbers

O.A. for the previous one is31

Q.) The ratio of the sum of the first p term to the sum of the first q terms of an arithmetic progression is p^2/q^2. Find the ratio of the twelfth term to the fifteenth term of the same AP .a.) 4/5.......... b.) 23/29.......... c.) 25/31...........4)Cannot be determined.

23/29

s12/s15 = 12^2/15^2

=>12/2/15/2= 144/225

=>d = 2a

T12/T15 = a+11d/a+14d = 23a/29a = 23/29

- 1 Like

Itna sannata kyu hai bhai :|Q.) If x and y are positive numbers and sqrt(x^2 + 16y) + sqrt(y^2 + 16x) = 45 and x - y = 9 then find the value of x + y .

a.) 15........b.) 21........c.) 31..........d.) 36

31 I just made use of the choices

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