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Just use the objective menu and click on each slide to by pass the check on learning for ssd1 come back here to look for answers for the test and if you cant find any use the reference in the top right hand corner and use CTRL F and look for a main word in the question to find the right answer this works for all of SSD1
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What number containing the numbers 1 through 9 with all numbers only used once is divisible by all the numbers from 1 through 9?
36288 WRONG!(1*2*3*4*5*6*7*8*9)/10 WRONG!I obviously misreads the question. after writing a simple visual basic program, see below. I am changing my answer to "does not exist." as the program failed to return a value.Private Sub Command0_Click()Dim i As Longi = 1Do While i Mod 9 > 0 Or i Mod 10 = 0i = i + 1Do While i Mod 8 > 0 Or i Mod 10 = 0i = i + 1Do While i Mod 7 > 0 Or i Mod 10 = 0i = i + 1Do While i Mod 6 > 0 Or i Mod 10 = 0i = i + 1Do While i Mod 5 > 0 Or i Mod 10 = 0i = i + 1Do While i Mod 4 > 0 Or i Mod 10 = 0i = i + 1Do While i Mod 3 > 0 Or i Mod 10 = 0i = i + 1Do While i Mod 2 > 0 Or i Mod 10 = 0i = i + 1LoopLoopLoopLoopLoopLoopLoopLoopMsgBox iEnd Sub
What ones digit when you multiply 100 nines together?
9100 mod 10 = (910 mod 10)10 mod 10 = 110 mod 10 = 1. Thus 1 is the ones digit in 9100.
How do you find pairs of numbers which are equal?
Numbers. i^n = i^(n mod 4). With n = 27, 27 mod 4...
What is the radical of 30?
mod Base
PROVE THAT nnn-n is divisible by 6?
A number is divisible by 6 if it is divisible by 2 and 3. Look at 333-3 which is 330 The sum of the digits is 6 and it is even so it is divisible by 6 Now consider 222-2 which I picked because unlike 333, 222 has even digits. 222-2=220, one again even number so divisible by 2 but NOT divisible by 3 so NOT divisible by 6 So it look like this is not true for all n For any odd n, we have the following 1. nnn-n ends in 0 so it is even if we can show it is divisible by 3 we are done. but 777-7 is 770 which is NOT divisible by 3 so it is NOT true. For some n it is true, but not for all n... Now when will nnn-n be divisible by 3. only when n+n is a multiple of 3, ie n=33,66, 99 an that is it! So we could easily prove that nnn-n is divisible by 6 if and only if n=3,6,or 9 ----------------------------- If by nnn, you mean n3, a proof is as follows: n=0,1,2,3,4, or 5 (mod 6) If n=0 (mod 6), we have (0 (mod 6))((0(mod 6))2-1)=0 (mod 6). [Since the first term is zero] If n=1 (mod 6), we have (1 (mod 6))((1(mod 6))2-1)=0 (mod 6) [Since 1-1=0]. If n=2 (mod 6), we have (2 (mod 6))((2(mod 6))2-1)=(2*3) (mod 6) = 6 (mod 6)=0 (mod 6). If n=3 (mod 6), we have (3 (mod 6))((3(mod 6))2-1)=(3*8) (mod 6) = 24 (mod 6) = 0 (mod 6). If n=4 (mod 6), we have (4 (mod 6))((4(mod 6))2-1)=(4*15) (mod 6) = 60 (mod 6) = 0 (mod 6). If n=5 (mod 6), we have (5 (mod 6))((5(mod 6))2-1)=(5*24) (mod 6) = 120 (mod 6) = 0 (mod 6). If you're not comfortable with the modular arethmetic, you can substitue 6m+_, where the blank is each of the numbers 0 through 5 (since every number can be expressed either as a multiple of six, or as a multiple of six plus some number between 1 and 5 --the remainder when the number is divided by six). Taking our example with 5, you would get: (n)(n2-1) can be written as (6m+5)((6m+5)2-1), where m is an integer. Simplifying this, you get: (6m+5)((6m+5)2-1) (6m+5)((6m2+60m+25-1) 6m*6m2+6m*60m+6m*25-6m+5*6m2+5*60m+5*25+5(-1) 6m*6m2+6m*60m+6m*25-6m+5*6m2+5*60m+5*24 Since m is an integer and each term is divisible by 6, (n)(n2-1) is divisible by 6 for integers that can be expressed as 6m+5. You would then repeat the process for each of 0 through 4 to complete the proof. Clearly, if you are comfortable with it, modular arithmetic is the less cumbersome way to proceed.
