What are the odd numbers from 1-100?
Using Iverson notation (APL)Find all the integers that are not evenly divisible by 2 by dividing by two, truncating the decimals, multiplying back by 2 and showing out the ones that are the same(N≠2×⌊N÷2)/N←⍳⎕⎕:1001 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99Or you could do it this way, but you have to cheat with the 1. (throw away values that are zero mod 2 (but you have to add in the initial one).1,(0≠N|2)/ N←⍳⎕⎕:1001 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 7273 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100Or you could do it this way …(generate all the even numbers between 2 and 100 and add negative 1)¯1+2×⍳⎕⎕:501 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99Note:This⎕:100Is asking for what the top number is.
The sum of 3 consecutive odd numbers whose sum is 42?
The question is impossible, i think you have the right answer. Why is it impossible? any 3 consecutive ODD numbers will equal an odd number. 7+9+11= 27 13+15+17= 45 21+21= 44 + 25 = 69 See where i'm getting at? If you take an odd number then add another odd number to it it becomes an even number ... but if you add ANOTHER odd number to that even number, the number will become odd again. PROBLEMS YOU MAY HAVE HAD: a) it should have been 3 consecutive EVEN numbers, 12+14+16 b) it should have been 3 consecutive numbers, 13+14+15
Can you add 3 odd numbers to get 30?
There is No WAY 3 odd number can sum up to 30 but there are a few solutions thinking out of the box.It says using 1, 3, 5, 7, 9, 11, 13, 15 (repetition allowed), it doesn’t restrict from using any arithmetic operators (factorial, iota, n-base, units) with these numbers.[math]3! + 24 = 30[/math] , since [math]24[/math] = [math]11 + 13[/math] or [math]9 + 15[/math][math]5! + 0 = 30[/math], [math]X\iota ^2 + X = 0[/math], where [math]X \in \left \{ 1,3,...,15 \right \}[/math][math]1 day + 3 hr + 3 hr = 30 hrs[/math]Note: [math]3! = 3 \times 2 \times 1 = 6 \Rightarrow[/math] (inverted 9)[math]13 + 11 + 6 = 30[/math]Think other than simply adding 3 odds, try base conversions and different units like time units etc. Good Luck.
The sum of three consecutive even numbers is 42. Find the middle number of the three?
Since they are even consecutive they will have common difference 2let the consecutive numbers be x,x+2,x+4 respectivelyhence by given conditionx + (x + 2) + (x + 4) = 423x + 6 = 423(x + 2) = 42x + 2=42/3x + 2=14x = 12therefore the numbers arex=12x+2=12+2=14x+4=12+4=16and therefore the middle number is 14
A, B, C and D are four consecutive odd numbers and their average is 42. What is the product of B and D?
As per given statement,Average of Consecutive numbers a,b,c,d is 42So, It can be written in equation form as(a+b+c+d)/4=42thus,a+b+c+d= 168Now considering consecutive odd numbers , Just take two odd consecutive odd numbers from Preceding and Succeeding odd numbers of 42That is 39,41,43 and 45.So when the numbers are arranged in Ascending order, ThenA=39;B=41, C=43 and D=45Now product of B and D is41*45= 1845The answer is 1845
What is the sum of numbers from 1 to 42?
Ah! That's something known as a Telescopic sum or you can also call it to be an Arithmetic Progression.There's an interesting method to do it(first one)! Here are two methods :First one :Let n = 1+2+3+4+5+………+41+42. - (I)Well it would be no harm if you write the series in reverse order, just like thatn = 42+41+40+39…….+2+1. - (II)So, add up (I) and (II) and you get :2n = 43+43+43+43…..(upto 42 times)=> n = (43×42) / 2=> n = 903It's simple and easy!So basically, the formula for solving these series sums where there's common difference is one is :Sum = [n(n+1)]/2 ( where n is the last term)Second one :You can do it by Arithmetic Progressions too! Here's the formula for that!In our case, the sum would be:S = 1/2×42(2×1 + (42 - 1)×1) = 903Or by the second formula,S = 1/2×42(1+42) = 903So that's more or less the same thing which I showed in different ways.I hope I helped! :)
What is the smallest positive integer n so that 14n, 16n, 18n, and 20n all have the same number of divisors?
Not possible to have integer in place on n because as the numbers are in increasing order the product should be in increasing order.
I am a multiple of 7 and a factor of 210. The product of my two digit is odd. What number am I?
It is 35.7*5=35, so a multiple of 7.35*6=210, so a factor of 2103*5=15