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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...
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.
What is the radical of 30?
mod Base
How do you download cheats for Minecraft?
you have to downlaod a mod seach for a mod loader and the mod its self
How do you get the Dwarves mod in minecraft?
find the answer ur self
How do you install mod in Counter-Strike 1.6?
depending on the mod you want For like players or a map you can download it at fpsbanana.com and you can get if from them. If you want like zombie mod for custom match you cant play that with bots you have to have your own online server and have to modify it your self.
Is MOD on your caller ID the march of dimes?
Maybe, or it could be Ministry of Defence or Ministry of Overseas Development.
Why does your 76 GM truck want to down shift when it is warm and slip when warm?
i found out my self it was the shift mod.
What do you type in to get mermaid tails and wings in the sims 2?
Whoa that's not possible it may be a mod you can download at: www.modthesims2.com (you have to sign up) -or- make it your self but it's harder lol Whoa that's not possible it may be a mod you can download at: www.modthesims2.com (you have to sign up) -or- make it your self but it's harder lol
What are some good terraria mods?
Thorium mod, Calamity mod, Exodus mod, Fargo's mod, GRealm mod, Sacred tools mod, Spirit mod, and Temor mod.
How do you install Zombie mod in Counter Strike 1.6?
Well it is like a patch you cant really install it you have to have your own server and you can set it up to be a zombie mod best is to search for Zombie mod servers all of their maps start with zm_... name of map. So thats how you know its a zombie mod server but if you dont have a server that you run your self you cant set it up. And no you cant do it with bots.
What are some minecraft mods?
the too many items mod the batman mod the iron man mod the spiderman mod the butter mod the creepy pasta mod the slender mod the contagious zombie mod the battle gear mod the Aether mod the You are herobrine mod the herobrine experience mod the herobrine nightmare mod the trampoline mod the fire charge mod i may have not named all of them. please improve if you know one not listed
What are 37 halo mods for halo trial?
well theres cliff mod rocket mod fishy mod night blood gultch mod jerrassic park mod sniper mod no gun mod spiner mod bumper cars elite mod pelican mod echo 116 and more i dont know because you can make your own mods
What is the difference between Garry's Mod 10 and Gary's Mod?
Garry's Mod 10 is a version of Garry's Mod.
What is remainder when 5 to the power of 30 is divided by 7?
Look at the powers of 5 mod 7: 5¹ mod 7 = 5 5² mod 7 = 5 × (5¹ mod 7) mod 7 = (5 × 5) mode 7 = 25 mod 7 = 4 5³ mod 7 = 5 × (5² mod 7) mod 7 = (5 × 4) mod 7 = 20 mod 7 = 6 5⁴ mod 7 = 5 × (5³ mod 7) mod 7 = (5 × 6) mod 7 = 30 mod 7 = 2 5⁵ mod 7 = 5 × (5⁴ mod 7) mod 7 = (5 × 2) mod 7 = 10 mod 7 = 3 5⁶ mod 7 = 5 × (5⁵ mod 7) mod 7 = (5 × 3) mod 7 = 15 mod 7 = 1 5⁷ mod 7 = 5 × (5⁶ mod 7) mod 7 = (5 × 1) mod 7 = 5 mod 7 = 5 At this point, it is obvious that the remainders will repeat the cycle {5, 4, 6, 2, 3, 1} There are 6 remainders in the cycle, so the remainder of 30 divided by 6 will tell you which remainder to use; if the remainder is 0, use the 6th element. 30 ÷ 6 = 5 r 0 →use the 6th element which is 1, so 5³⁰ ÷ 7 will have a remainder of 1. 1 ≡ 5³⁰ mod 7.
