Prove: If 3 divides a^2, then 3 divides a.?
Given that 3 divides a²: Assume that 3 does not divide a. Then a must be in the form 3k+1 or 3k+2, where k is a non-negative integer. Case 1: a = 3k + 1 thus a² = (3k + 1)² =9k² + 6k + 1 = 3(3k² + 2k) + 1 = 3x + 1 *CONTRADICTION* 3 does not divide (3x + 1) = a² Case 2: a = 3k + 2 thus a² = (3k + 2)² =9k² + 12k + 4 =9k² + 12k + 3 + 1 = 3(3k² + 4k + 1) + 1 = 3y + 1 *CONTRADICTION* 3 does not divide (3y + 1) = a² Therefore, 3 MUST divide a if 3 divides a² As an observation, the square of a number not divisible by 3 will result in a number that will give a remainder of 1 when divided by 3.
What is 2 plus 2 divided by 2?
Most of the people will solve it like this2+2÷22+2=44÷2=2But it is wrong the correct way to do it is sokve with( BODMAS)BO=brackets openD=divisionM=multiplicationA=additionS=subtractionSequence viseSollution =2+2÷22÷2=1So, 2+1=3
What will be the number if seven is added to a certain number, the sum is multiplied by five and the product is divided by nine, then three is subtracted from the quotient, and the remainder left is twelve?
1) Go in the reverse order of the steps as performed12--> 3 was subtracted => Add 3 gives 1515--> Was divided by 9=> Multiply by 9 gives 135135--> Was multiplied by 5=> Divide by 5 gives 2727--> 7 was added=> Subtract 7 gives 20OR2) Use algebraLet the number be xAdd 7--> x+7Multiply by 5--> 5x+ 35Divide by 9--> 5x/9 + 35/9Subtract 3--> 5x/9 + 8/9 = 12=>5x=100=> x=20
How do you divide in different bases, for example, 245 base 9 divided by 7 base 9?
Using base-9, throughout, we need to know our 7 times table.[math]1 \times 7 = 7[/math][math]2 \times 7 = 15[/math][math]3 \times 7 = 23[/math][math]4 \times 7 = 31[/math][math]5 \times 7 = 38[/math][math]6 \times 7 = 46[/math][math]7 \times 7 = 54[/math][math]8 \times 7 = 62[/math][math]9 \times 7 = 70[/math]Now we will perform long division.How many times does 7 go into 24? Well, from our table, 3 × 7 = 23; subtracting this from 24 leaves a remainder of 1. Carry down the next digit.How many times does 7 go into 15? Well, from our table, 2 × 7 = 15; subtracting this from 15 leaves a remainder of 0.Let’s check our answer by converting to decimal. From now on, base-9 numbers will be designated with a ‘9’ as a subscript.[math]245_{9} = 2 \times 9^2 + 4 \times 9^1 + 5 \times 9^0 = 2 \times 81 + 4 \times 9 + 5 = 162 + 36 + 5 = 203[/math][math]\frac {203}{7} = 29[/math][math]32_{9} = 3 \times 9^1 + 2 \times 9^0 = 3 \times 9 + 2 = 29[/math]Just as easy as using decimal numbers.
Limits of Riemann Sums problem?
the question asks : find a formula for the Riemann sum obtained by dividing the interval [a,b] into n equal sub intervals and using the right-hand endpoint for each Ck. Then take a limit of these sums as n-->infinity to calculate area under the curve. f(x)=2(x^3) over the interval [0,1] I dont get what it waaantsss :'( please answer it with enough explanations cuz I'm totally lost over this... Thank you
What is the greatest 2-digit number divisible by 2?
First, clarity. I’m assuming here that ‘divisible by 2’ and ‘2-digit number’ imply that this is an integer and that one of its factors is 2 (it divides by 2 without a decimal).The answer’s pretty obvious.[math]98[/math].The only 2-digit number greater than it, 99, is odd, and thus not divisible by 2. The next number after 99 is 100, which, if you notice, is not 2 digits.But why stick to base 10? A larger base puts more value into a single digit. For instance, what about… Base 64? This is how YouTube encodes their video addresses, because a few characters hold huge numbers. It consists of the letters A-Z, then a-z, then 0–9, then 2 other characters (I’ll use + and /).The LARGEST, two digit number in base 64 is… (drumroll please)[math]/+[/math], equivalent to [math]4094[/math]! The next even number, 4094, is BAA, a 3-digit number.But what if we went bigger?Base 64 is the largest established base, but there could theoretically be larger ones. In fact, if there are an infinite number of possible symbols, then you could create a base TREE(3) or larger that could store a truly massive number.But the answer you were looking for is probably [math]98[/math].
What is the number of ways in which n identical objects can be divided into r groups where each group can have any number of objects, including 0? The formula is (n+r-1) C (r-1).
Consider that the number of identical objects is n. Now consider that all these n objects are placed in a row. Let us assume some separators that separates the n objects. Now as we want to separate them into r group [as blank groups are also allowed], take the separators as objects. Now we need r-1 separators to make r groups. Therefore total number of objects is n+r-1. There will be r-1 places for the separators to occupy. Therefore we can arrange the separators in (n+r-1) C (r-1) ways, which is the answer to the question asked.
How can we calculate the logarithms by hand without using any calculator?
Here is how to calculate logarithms by hand using only multiplication and subtraction.And this procedure produces digit by digit, so you can stop whenever you have enough digits.Before we do that, let’s give an example so it will be easier to understand:Log 2 = 0,30103…If we take the 10th power:log (2^10) = 10 X 0,30103… = 3.0103…then the digits in the logarithm have all shifted 1 place to the left.But also: 2^10 = 1024We can see that the log of 1024 must be 3 point something because 1024 is slightly over 1000.If we can find the 10th power on a normal calculator we know he first digit of the logarithm by looking at the number before the decimal point:0-9 => 010-99 => 1100-999 => 21000-9999 => 3etc.So log 1024 is 3.????. We have retrieved the first digit of log 2.If we subtract 3 from this number we can go through the previous procedure again to retrieve the next number.The 10th power is easily calculated on paper by multiplying a number by itself 3 times:2 X 2 =44 X 4 = 1616 X 16 = 256And the multiplying 256 with the result of the first calculation: 4256 X 4 = 1024So 4 multiplications in total.I divide 1024 by 1000 to get 1.024Doing the same procedure I get:1.2676506So the next digit is 0.And then taking 1.2676506 through this procedure I get:10.715806So the next digit is 1.I divide 10.715806 by 10 to get 1.0715806And then taking 1.0715806 through this procedure I get:1.995063So the next digit is 0.And then taking 1.995063 through this procedure I get:999.00506So the next digit is 2.However since it is so close to 1000 we can round the digit to 3.Now we have 3.0103
If n is a natural number, then what is the unit digit in 6^n-5^n?
The unit digit of [math]6^1-5^1[/math] is clearly [math]1[/math]. We now assume [math]n>1[/math], and consider [math]6^n-5^n[/math], which is equal to[math](6-5)(6^{n-1}+5\times 6^{n-2}+5^2\times 6^{n-3}+\cdots+5^{n-2}\times 6+5^{n-1}).[/math]The unit digit of [math]5^a\times 6^b[/math] is [math]0[/math] whenever [math]a[/math] and [math]b[/math] are both positive, since [math]5\times 6=30[/math]. Thus, the unit digit of [math]6^n-5^n[/math] is the same as that of [math]6^{n-1}+5^{n-1}[/math]. But, in turn, the unit digit of [math]6^{n-1}+5^{n-1}[/math] is the same as that of [math]6^{n-1}-5^{n-1}[/math], since [math]6^{n-1}+5^{n-1}=6^{n-1}-5^{n-1}+2\times 5^{n-1}[/math], and the latter term [math]2\times 5^{n-1}[/math] contains [math]2\times 5[/math], which has a zero unit digit. Hence the unit digit of [math]6^n-5^n[/math] is the same as the unit digit of [math]6^{n-1}-5^{n-1}[/math]. Applying this argument repeatedly to [math]6^{n-1}-5^{n-1}[/math] until we reach [math]6^1-5^1[/math], we conclude that the unit digit of [math]6^n-5^n[/math] is [math]1[/math] for all natural numbers [math]n[/math].The above is an alternative way of understanding how and why induction works, for those people who still have slight doubts about the validity of the induction process. (If you didn’t realize, the above is a proof by induction, written in an unorthodox way.)