Can you replace the question mark with the correct number, given the pair of numbers exhibits similar relationship?
? : 3839 :: 11 : 1209
16
Formular used :
pow(x,3) - pow(x,2) - 1
11*11*11 - 11*11 - 1
1331 - 121 - 1
1209
Similarly, we can solve
pow($x,3) - pow($x,2) - 1 = 3839
This condition holds true for number-16.
Which number should replace the question mark in the picture below ?
26
14 = 1^2 + 2^2 + 3^2
21 = 4^2 + 1^2 + 2^2
29 = 4^2 + 3^2 + 2^2
26 = 3^2 + 4^2 + 1^2
The square root of number 121 is "11". What is the square root of number "12345678987654321." ?
111111111
It is a very popular square root series i.e.
Square root fo number 121 is 11
Square root fo number 12321 is 111
Square root fo number 1234321 is 1111 and so on....
In "Clash Royal" App Game, Warrior "The Bomber" is a very good defensive options. It can be purchased from the card shop.The cost of the bomber is quite interesting.The Cost of 1st bomber is: 2The Cost of 2nd bomber is: 4The Cost of 3rd bomber is: 6The Cost of 4th bomber is: 8The Cost of 5th bomber is: 10i.e to buy 5 bombers you need 2+4+6+8+10 = 30 coins.How many bombers can you buy using 1000 coins ?
62
The Cost of 1st bomber is: 2 i.e Total coins spend: 2
The Cost of 2nd bomber is: 4 i.e Total coins spend: 6
The Cost of 3rd bomber is: 6 i.e Total coins spend: 12
The Cost of 4th bomber is: 8 i.e Total coins spend: 20
The Cost of 5th bomber is: 10 i.e Total coins spend: 30
The Cost of 6th bomber is: 12 i.e Total coins spend: 42
The Cost of 7th bomber is: 14 i.e Total coins spend: 56
The Cost of 8th bomber is: 16 i.e Total coins spend: 72
The Cost of 9th bomber is: 18 i.e Total coins spend: 90
The Cost of 10th bomber is: 20 i.e Total coins spend: 110
The Cost of 11th bomber is: 22 i.e Total coins spend: 132
The Cost of 12th bomber is: 24 i.e Total coins spend: 156
The Cost of 13th bomber is: 26 i.e Total coins spend: 182
The Cost of 14th bomber is: 28 i.e Total coins spend: 210
The Cost of 15th bomber is: 30 i.e Total coins spend: 240
The Cost of 16th bomber is: 32 i.e Total coins spend: 272
The Cost of 17th bomber is: 34 i.e Total coins spend: 306
The Cost of 18th bomber is: 36 i.e Total coins spend: 342
The Cost of 19th bomber is: 38 i.e Total coins spend: 380
The Cost of 20th bomber is: 40 i.e Total coins spend: 420
The Cost of 21st bomber is: 42 i.e Total coins spend: 462
The Cost of 22nd bomber is: 44 i.e Total coins spend: 506
The Cost of 23rd bomber is: 46 i.e Total coins spend: 552
The Cost of 24th bomber is: 48 i.e Total coins spend: 600
The Cost of 25th bomber is: 50 i.e Total coins spend: 650
The Cost of 26th bomber is: 52 i.e Total coins spend: 702
The Cost of 27th bomber is: 54 i.e Total coins spend: 756
The Cost of 28th bomber is: 56 i.e Total coins spend: 812
The Cost of 29th bomber is: 58 i.e Total coins spend: 870
The Cost of 30th bomber is: 60 i.e Total coins spend: 930
The Cost of 31st bomber is: 62 i.e Total coins spend: 992
There is a unique number which when multiplied by any number from 1 to 6, we will get the new number that contains same digits only.
Can you find that number ?
142857
142857 * 1 = 142857
142857 * 2 = 285714
142857 * 3 = 428571
142857 * 4 = 571428
142857 * 5 = 714285
142857 * 6 = 857142
These number are considered as cyclic numbers.
Before the start of the car race, Tony Stewart and Jamie Mcmurray have the same amount of fuel in the car. With this fuel Tony Stewart can drive for 4 hours while Jamie can drive five hours.
After a time they realize that amount of fuel left in Tony car is 1/4th of the fuel in Jamie car.
For long they are racing ?
15/4 hours
Explanation:
Let us say they drive for "t" hours and initial fuel is "f"
Amount of fuel used by Tony car is ft/4
Amount of fuel used by Jamie car is ft/5
As per given condition,
f - ft/5 = 4(f-ft/4)
Solving, t = 15/4 hours
What will be the best approach to finding all the prime numbers less than 75 that leave an odd reminder when we divide them with 5?
Except 2 and 3, all the prime numbers are in a form of (6k +1) or (6k -1)
Therefore, you can use the formula to write down the prime numbers as:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 61, 67, 71, 73
Now, you can separate numbers as per the requirement that are:
3, 11, 13, 23, 31, 41, 43, 53, 61, 71 and 73
Can you solve the equation by finding the value ofA) HorseB) Cowboy bootC) Horseshoe
Horse = 10, Cowboy-Boot = 1 ,HorseShoe= 2
= Step1 =
3Horse = 30
=> Horse = 10 .....(A)
= Step2 =
1Horse + 2HorseShoe + 2HorseShoe = 18
10 + 4HorseShoe =18
HorseShoe = 2 .....(C)
= Step3 =
2HorseShoe - 2Cowboy-Boot = 2
4 - 2Cowboy-Boot = 2
=> Cowboy-Boot = 1 ... (B)
= Step4 =
Cowboy-Boot+Horse*HorseShoe = 1+10*2 =>21
Chances of Mr. Button winning the local lottery in 8%. All participants lined up and Mr. Button is 3rd in the row. The first two participants lose the lottery.
What is the chance of Samuel Mr. Button now?
8%
The winning chance probability is still 8% as the outcome of Mr.Bitton winning the lottery is a separate event from rest losing the lottery.
You need to fill number in the bricks in the image below such that the top brick is sum of two brick below it.
It can be solved as shown below.