Calendar Math Tricks

• Ordinary Year

A year having 365 days is called an Ordinary Year, which has

(52 complete weeks + 1 Extra/Odd day)

• Leap Year

A Leap Year has 366 days (including 29th of February) and

(52 complete weeks + 2 Extra/Odd days)

1. A leap year is divisible by 4, like 1984, 1988, 2008, 2012 etc.
2. A century is divisible by 400, like 1600, 2000, 2400 etc.
3. Centuries like 1700, 1800 are not divisible by 400.

• Odd Days

• Number of odd days in Ordinary Year

365 days = (52 weeks*7) + 1 odd day.

• Number of days in a Leap Year

366 days = (52*7) + 2 odd days.

• Number of odd days in a Century (100 year)

100/4 = 24 leap year + (100 - 24 = 76) 76 ordinary year.

(76*1 + 24*2)[since ordinary year has 1 odd day and leap yeas has 2 odd days]

124 days = 7 week*17 + 5 odd days.

• Number of Odd days in 200 year

= 5*2 [since 100 year has 5 odd days and here it is 200 year]

10 odd days = 1 week + 3 odd days.

• Number of Odd days in 300 year

5*3 = 15 odd days 

=2 week + 1 odd day.

• Number of Odd days in 400 year

5*4 + 1 = 21 odd days

=3 weeks + 0 odd days.

[since 400th year is a leap year, so 1 more day is added]

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HCF and LCM

• Find the largest number which divide 1356, 1868 and 2764 leaving the same remainder in each case.

Ans: From the Solution Methods the required number is,
HCF of (1868-1356), (2764-1356) and (2764-1868)
HCF of 512, 1408 and 896

So the HCF = 8*8 = 64. Since there are only two 8s in common.

• Find the greatest number that divides 130, 305 and 245 leaving remainders 6, 9 and 17 respectively.

Ans: From the Solution Methods required number is,
HCF of [(130-6), (305-9), (245-17)]
HCF of 124, 296, 228

So the HCF is 4.

• Find the number which when divided by 12, 16 and 18 leaves 5 as remainder in each case.

Ans: From the Solution Methods we see that,

(LCM of 12, 16, 18) + 5

So LCM = 2*2*3*4*3 = 144

Required number is 144 + 5 = 149.

• Find the largest possible number of 5 digits which is exactly divisible by 32, 36 and 40.

Ans: From the Solution Methods we see that,

LCM of 32, 36 and 40 = 1440.

The greatest 6 digit number is 99999

So the required number = 9999-639 = 99360.

• Find the greatest 4 digit number which when divided by 10, 15, 21 and 28 leaving remainders 4, 9, 15 and 22.

Ans: From the Solution Methods we get,

LCM of 10, 15, 21 and 28 = 420.

The largest 4 digit number is 9999

So four digit number divisible by 10, 15, 21 and 28 = 9999-339 = 9660.

Also k = 10-4 = 15-9= 21-15= 28-22 = 6

Required number is 9660-6 = 9654.

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LCM and HCF

LCM and HCF of Fractions

• LCM of fractions = (LCM of numerators)/(HCF of denominators)

• HCF of fractions = (HCF of numerators)/(LCM of denomenators)

Solution Methods:

• Product of two numbers = HCF of the numbers * LCM of the numbers.

• The greatest number which divides the numbers x, y and z, leaving remainders a, b and c respectively is given by

HCF of (x-a), (y-b), (z-c)

• The least number which when divided by x, y and z leaves the remainders a, b and c respectively is given by

LCM of (x, y, z) - k

where,   K = (x-a) = (y-b) = (z-c)

• The least number which when divided by x, y and z leaves the same remainder k in each case, is given by

LCM of (x, y, z) + k

• The greatest number that will divide x, y and z leaving the same remainder in each case is given by

HCF of (x-y), (y-z), (z-x)

• The greatest n - digit number which when divided by x, y and z:

      (a) Leaves no remainder

Requires number = n-digit greatest number - R

      (b) Leaves remainder k

 Required number = [n-digit greatest number - R] + k

where R is the remainder obtained when n-digit greatest number is divided by the LCM of x, y and z.

• The smallest n-digit number which when divided by x, y ans z leaves

      (a) No remainder

Required number = [n-digit smallest number  + (L-R)]

     (b) Remainder k

Required number = [n-digit smallest number + (L-R)] + k

where, R is the number obtained when n-digit smallest number is divided by LCM of x, y, and z. L is the LCM of x, y and z.

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Number Series

• n² Series

Find the missing term in the series 4, 16, 36, 64, ..., 144.

Ans: This is a series of squares of consecutive even numbers.

2² = 4, 4² = 16, 6² = 36, 8² = 64, 10² = 100

• (n² + 1) Series

Find the missing term in the series 10, 17, ..., 37.

Ans: Series pattern

3² + 1 = 10, 4² + 1 = 17, 5² + 1 = 26

• (n² - 1) Series

Find the missing term in the series 0, 3, 8, ..., 24.

Ans: Series pattern

1² - 1 = 0, 2² - 1 = 3, 3² - 1 = 8, 4² - 1 = 15

• (n² + n) Series

Find the missing term in the series 12, 20, 30, 42.

Ans: Series pattern

3² + 3 , 4² + 4, 5² + 5, 6² + 6

• (n² - n) Series

Find the missing term in the series 42, 30, ..., 12, 6.

Ans: Series pattern

7² - 7, 6² - 6, 5² - 5, 4² - 4, 3² - 3

There are some other series that follow the same rule like n³, (n³ + 1), (n³ - 1), (n³ + n), (n³ - n). I will suggest do it yourself.

• Alternating Series

Find the next term in the series 15, 14, 19, 11, 23, 8, ...
Ans: Series pattern

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Boat and Stream Formulas

• If speed of stream is ‘a’ and a boat (swimmer) takes ‘n’ times as long to row up as to row down the river, then

Speed of Boat (swimmer) in still water = a*(n+1)/(n-1)

• A person can row at a speed of ‘x’ in still water. If stream is flowing at a speed of y, it takes time ‘T’ to row to a place and back, then

Distance between two places = T*(x² – y²) / 2x

• A man rows a certain distance downstream in ‘x’ hour and returns the same distance in ‘y’ hour. When the stream flows at the rate of ‘a’ km/h, then

Speed of the man in still water = a*(x+ y) / (y – x)

• If boat’s speed in still water is ‘a’ km/h and river is flowing with a speed of ‘b’ km/h, then average speed in going to a certain place and coming back to starting point is given by

(a + b)(a – b) / a km/h

• When boat’s speed in still water is ‘a’ km/h and river is flowing with a speed of ‘b’ km/h and time taken to cover a certain distance upstream is ‘T’ more that the time taken to cover the same distance downstream, then

Distance = (a² – b²)*T / 2b

• If a man covers ‘I’ km distance in ‘t₁’ hour along the direction of the river and he covers same distance in ‘t₂' hour against the direction of the river, then

Speed of man = 1/2 * (I/t₁ + I/t₂)

Speed of stream = 1/2 * (I/t₁ – I/t₂)

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Basic Number Theory

• Square of every even Number = An even Number

• Sum of first ‘n’ natural numbers = {n(n+1)}/2

• Sum of first ‘n’ odd numbers = n²

• Sum of first ‘n’ even numbers = n(n+1)

• Sum of square of first ‘n’ natural numbers = {n(n+1)(2n+1)}/6

• Sum of cubes of first ‘n’ natural numbers = {n(n+1)/2}²

• There are 15 prime numbers between 1 & 50 and 10 prime numbers between 50 & 100

• If n divides p & q, then n divides their sum and difference also

      Like: 6 divides 12 & 18, then 12+18 = 30 and 18-12 = 6 are also divided by 6

• For any natural number n, (n³ – n) is divisible by 6

• The product of three consecutive natural numbers is always divisible by 6

• (x^m – a^m) is divisible by (x – a) for all values of m

• (x^m – a^m) is divisible by (x + a) for even values of m

• (x^m + a^m) is divisivlke by (x + a) for odd values of m

• Number of prime factors of a^p b^q c^r d^s is p+q+r+s, where a, b, c, d are prime numbers

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How to find Unit Place Digit Phase 2

We have already found the units place digit for some numbers like 0, 1, 5, 6 and 2. So far we have seen that there are some numbers for which we don’t need any explanation to find out their units place digit.

Here we come to find the units place digit of 3, 4, 8, and 7.

Let’s take a look:

If Units place digit is 4 or 8, then we should first divide the power by 4 and then represent in the form of 4⁴ or 8⁴. In that case the value of 4⁴ and 8⁴ will be 6. [Because 4*4*4*4 = 256 or 8*8*8*8 = 4096]

       (124)³⁷²    = (4) ^(372/4)  

                         = (4⁴)⁹³

                         = 4⁴ = 256                                   

       (128)³²⁵     = (8) ^(324/4)

                          = (8⁴)⁸¹

                          = 8⁴ = 4096

So the unit place digit is 6.

Again, if the unit place digit is 3 or 7, then we should first divide the power by 4 and then represent in the form of 3⁴ and 7⁴. In that case the value of 3⁴ and 7⁴ will be 1.                                                                  

        (2467)¹⁵³   = (7) ^(153/4)                    

                          = (7⁴)³⁸ * 7¹                                 

                           = 7⁴ * 7                                        

                          = 1 * 7 = 7                                  

                      [7⁴ = 7*7*7*7 = 2401]                               

       (293)¹⁴¹     = (3) ^(141/4)

                          = (3⁴)³⁵ * 3¹

                          = 3⁴ * 3

                          = 1 * 3 = 3

                      [3⁴ = 3*3*3*3 = 81]

 So the unit place digits are 7 and 3 respectively.

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How to find Unit place Digit

Finding Unit place digit is very easy when it comes to multiplication of nay two or more numbers. Multiply the last digits of the numbers and you get the result. But what if a number is given with power of two or three or more digits to it. It’s not as hard as you imagine.

Let’s take a look:

If the unit place digit is 0 or 1 or 5 or 6 then the resultant Unit place digit remains same.

        (479)¹¹⁵¹   =  Its Unit digit is 6.

        (155)¹²⁰    =   Its Unit digit is 5.

        (191)¹⁹     =   Its Unit digit is 1.

        (900)⁵¹     =   Its unit digit is 0.

If Unit place digit is 2, then the power of the number is first divided by 4 and there after represented in the form of 2⁴

(572)⁴⁴²    =   (2) ^(442/4) [taking unit place digit or the last digit of 572] 

                  =   (2⁴)¹¹⁰ * 2²

                  =   2⁴ * 2²

                  =   16 * 8

There taking the multiplication the Unit place digit is 8.  (16*8=48)

To find Unit Place Digits of other numbers visit: Unit Place Digit Phase 2

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Daily GK update 14th May 2018

1. Petrol price hiked after 19 days:

Ending the 19 days freeze on the hike in petrol and diesel prices, the oil marketing companies on Monday May 14, 2018 increased the petrol price by 17 paisa a liter while the diesel price was hiked by 21 paisa.

2. Fortis board went against Financial Advisers on Munjal-Burman bid:

The much wooed Fortis Healthcare may have selected a suitor at last. The recent selection of the winning bid did not get the nod of FHL’s independent directors and as a result the fissure in the boardroom has widened further.

3. WPI inflation shoots to 3.18% in April on costlier fuel:

Inflation based on whole sale price index spiked to 3.18% per cent in April on increasing prices of petrol and diesel as well as fruits and vegetables. The whole sale price index stood at 2.47 per cent in March and 3.85 per cent in April last year.

4. Walmart may launch IPO for Flipkart:

Walmart Inc said on Saturday in a filing with a U.S. regulator that it may take India’s Flipkart public in as early as four years, detailing for the first time a potential listing time line for Walmart’s largest ever acquisition.

5. Finance Ministry mulling innovative ways to deal with banks’ NPA provision:

The FM is examining a proposal to find innovative ways for dealing with burden of NPA provisions by issuing provision Shore-up certificates to banks.

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Square of 666 Or Square of 666

The traditional method of finding Squares has now become extinct. There are lot more easier way to find out the Squares. We are going to find out the Square of 666 or 6666 with a lot more easiest way.

Let’s take a look:

(666)² = 443556 

(6666)² = 44435556

Note: To find out this square we have used and what you always have to keep in mind is - ‘FOTFIS’. Where,

‘FO’ - Four or 4 ; ‘T’ – Three or 3 ; ‘FI’ – Five or 5 ; ‘S’ – Six or 6.

In the answer the number of digit 4 and the number of digit 5 will be same and the number of each of these digits 4 & 5 will be one less than the number of digit 6 in the question. Again the number of digits 3 & 6 will be one and they will be placed just as shown in the example.

For more visit : AcademiaBlogs

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