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FAQs
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What number would follow 0, 9, 1, 9, 4, 2, 3, 2, 1, 2, 8, 6, 3, 8, 3, 7, 6, 7, 9, 3, 5, 3?
Any number you want. Pick a number at random, call it s, add it to the end of your list, and form the list of ordered pairs (1, 0), (2, 9), (3, 1), … , (23, s), where the x-values are consecutive integers and the y-values are the numbers in your list. Fit a polynomial to the list of ordered pairs: one is guaranteed to exist. Then you will have a formula that will give you all the terms in your list for x = 1, 2, 3, … , 22, and will give you whatever number you picked for x = 23. -
How many 3 digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 if repetitions of digits are allowed?
This question can be answered using two methods. Let's start with the simplest one.Method 1:The number is three digits, so for them let's take three blanks _ _ _The first blank can be filled using any of the digits from 1–9 because if we use zero to fill the first blank the number becomes of two digits. Hence we have 9 ways to fill the first blank.Now, the second blank can be filled by any of the remaining 10 digits because repetition is allowed and thus the digit selected for the first blank can also be selected. So 10 ways.Similarly 10 ways for the third blank.So total number of combinations become 9 x 10 x 10 = 900Hence the answer is 900 such number can be formed.Method 2:Since the first digit cannot be zero, we have 9C1 ways to select the first digit (one digit selected from a set of nine distinct digits). (9C1 = 9)Now, for the remaining two places we can have zero as well. Hence we have 10C1 ways to select a digit for tens and ones place each. (10C1 = 10)Hence total number of combinations become 9C1 x 10C1 x 10C1 = 9 x 10 x 10 = 900Hence the answer is 900 such numbers can be formed.Hope it helped you! :) -
How many numbers can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9?
Not as many as you might think [math]\ddot\smallfrown[/math] — a countable infinity at most, if we allow repetition of digits.Even assuming we also allow you a decimal point, you will have difficulty forming the number [math]\sqrt2[/math] as you will never quite write down the countable infinity of digits required. The uncountably infinite collection of all Irrational numbers are well beyond your signNow.You also stand no chance of forming my favourite number [math]\varsigma=\sqrt{\omega}[/math].And Complex numbers like [math]3+2i[/math] have that pesky little [math]i[/math] in them that defeats your digits. Not to mention Quaternions and the like if you include those in your class of “numbers”…In fact decimal digits are a rather restricted way of representing numbers, even if they do more than satisfy the vast majority of humanity [math]\ddot\smallsmile[/math] -
How many 3-digit odd numbers can be formed with 0, 1, 2, 3, 4, 5, 6, 7, 8, 9?
There are 5 ways to select the units digit. I.e. 1,3,5,7,92. After the units digit is selected there are 9 numbers left from which to choose the 10s digit3. The hundreds digit can be selected 8 ways4. Total =5×9×8=360 -
Is it possible to make 10000 out of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 using only the sum operator and using all the digits exactly once?
No, at least not without doing something cute and tricksy, like turning a digit sideways or whatever.The remainder of any number with respect to 9 is the modulo 9 sum of the digits of that number. So for instance, the remainder of 12345 with respect to 9 is (1+2+3+4+5) mod 9 = 15 mod 9 = 6. (12345 = 1371*9 + 6). The remainder of the sum of two numbers with respect to 9 will be the modulo 9 sum of their individual remainders. So the remainder of any number formed using the digits 0...9 and the sum operator with respect to 9 will be the modulo 9 sum of the digits 0...9. Which is 0. So, however you arrange the digits, the sum will be divisible by 9.10,000 is not divisible by 9, so it cannot be formed using such an arrangement of digits.(If "cute" solutions are permitted, then all bets are off; for instance, I can turn the digit '1' sideways to form a minus sign, and get 4032+5976-8=10,000.) -
Could you analyze this list of numbers:1-6-1-8-0-3-3-( number is obscured appears to be a 9)-8-8-7-4-9-8-9-14-11-9-4?
Well. It is The Golden Ratio number of subdivisions . And this number is an amazing number, for it was applied in Aesthetics, Architecture, Painting, Book design, Design and so on. If you are interested in how to apply this number to the field mentioned above, you can see on the website Golden ratio . Here, I want to introduce another beautiful character for you. This number also is the Limit of consecutive quotients of Fibonacci number. -
How does the series 97, 33, 17, 8, 4, 3, 2, 1, 1, 1, 0, 0, 1, 0 continue?
First, a comment, if you venture into the complex numbers, you must consider cases involving +/-. However, you have not been doing that so far. After all, there could have been two values for term 4 and further, namely, 8 or -8.If the sequence was given and your task was to provide the next three terms, it is more likely that rather than just subtract, instead find the difference which is always positive. Doing so, the next terms would be 1, 1, 1, 0, 0, 1, 0, 1, 0 and these 8 would repeat. -
How come [math]\zeta(-2) = 0[/math]? If you plug in negative two, shouldn't it turn out to be [math]1 + 4 + 9 + 16 + ...[/math]?
The zeta function of [math]s[/math] equals [math]\sum_{n=1}^\infty \frac{1}{n^s}[/math] only when [math]s>1[/math], or more generally when [math]s[/math] is complex with real part greater than [math]1[/math]. The formula for the zeta function at other places is very different.The series [math]\sum_{n=1}^\infty \frac{1}{n^s}[/math] doesn’t converge when [math]s \le 1[/math], so there’s no point in comparing the value of [math]\zeta[/math] with those non-convergent series.One of the ways to establish the value of [math]\zeta[/math] when [math]\Re(s)<1[/math] is using the functional equation:[math]\displaystyle \zeta(s) = 2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)[/math]which tells you that [math]\zeta(-2)[/math] is a product of various things, including [math]\zeta(3)[/math], [math]\Gamma(3)[/math] and [math]\sin(-\pi)=0[/math]. That’s why it is [math]0[/math]. -
How many 5 digit numbers are divisible by 4 out of 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9?
All numbers ending with 00,04, 08, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96 are divisible by 4 because if last two digits is divisible by 4, then entire number is divisible by 4.Using CombinatoricsTen thousand'th digit can be 9 waysThousand'th digit can be 10 waysHundred'th digit can be 10 waysIf Tenth digit is 0, then unit place 3 ways.=> 9 x 10 x 10 x 1 x 2 numbersIf Tenth digit is 1, then unit place 2 ways.=> 9 x 10 x 10 x 1 x 2 numbersIf Tenth digit is 2, then unit place 3 ways.=> 9 x 10 x 10 x 1 x 3 numbersIf Tenth digit is 3, then unit place 2 ways.=> 9 x 10 x 10 x 1 x 2 numbersIf Tenth digit is 4, then unit place 3 ways.=> 9 x 10 x 10 x 1 x 3 numbersIf Tenth digit is 5, then unit place 2 ways.= 9 x 10 x 10 x 1 x 2 numbersIf Tenth digit is 6, then unit place 3 ways.=> 9 x 10 x 10 x 1 x 3 numbersIf Tenth digit is 7, then unit place 2 ways.=> 9 x 10 x 10 x 1 x 2 numbersIf Tenth digit is 8, then unit place 3 ways.=> 9 x 10 x 10 x 1 x 3 numbersIf Tenth digit is 9, then unit place 2 ways.=> 9 x 10 x 10 x 1 x 2 numbersTotal Numbers = sum of all above= 9 x 10 x 10 x 25= 22500Alternate Easy MethodFirst number = 10000Last number = 99996Number of numbers divisible by 4 is :(99996 - 10000)/4 + 1 = 22500:-)
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