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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.
What is 1 mod 1?
If by "mod" you mean "modulo," then your question is meaningless, because "mod 1" is meaningless. For example, 18 mod 5 = 3, because you subtract the maximum number of multiples of 5 and the remainder is 3. But by definition any whole number modulo 1 would always be 0.
How do you get the 'mod' of a number?
"mod" is short for "modulo", and indicates the remainder after division of the first number by the second. For example, 11 mod 2 = 1 (11 / 2 has an integer quotient of 5, with 1 left over).
What is the equivalent number on the 12-hour clock 329?
329 5 (mod 12)
Whose number 5 on Miami Heat?
Chris bosh
Visual basic code for making change in dollars quarters dimesnickelsand pennies?
intNumOfQuarters = intNumOfPennies \ 25 'calculate number of qaurters' intNumOfDimes = (intNumOfPennies Mod 25) \ 10 'calculate number of dimes' intNumOfNickels = ((intNumOfPennies Mod 25) Mod 10) \ 5 'calculate number of nickels' intNumOfPenniesleft = (((intNumOf
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 2 mod 5?
2 mod 5 is the remainder left when 2 is divided by 5.2 = 5*0 + 2 so [the quotient is 0 and] the remainder is 2that is 2 mod 5 = 2.On spreadsheets, the formula is usually "=MOD(2,5)"2 mod 5 is the remainder left when 2 is divided by 5.2 = 5*0 + 2 so [the quotient is 0 and] the remainder is 2that is 2 mod 5 = 2.On spreadsheets, the formula is usually "=MOD(2,5)"2 mod 5 is the remainder left when 2 is divided by 5.2 = 5*0 + 2 so [the quotient is 0 and] the remainder is 2that is 2 mod 5 = 2.On spreadsheets, the formula is usually "=MOD(2,5)"2 mod 5 is the remainder left when 2 is divided by 5.2 = 5*0 + 2 so [the quotient is 0 and] the remainder is 2that is 2 mod 5 = 2.On spreadsheets, the formula is usually "=MOD(2,5)"
What is the additive inverse of 3 in mod 8?
Add 5 in mod 8. (You can always subtract 3 in mod 8 as well, eg 5 + 3 = 0; 0 - 3 = 5.)
What is the probability of getting exactly 5 boys and 5 girls if you have 10 kids?
Here the sample space is(s)=10, =>mod(s)=10 a=be the event of getting exactly 5 boys =>mod(a)=5 b=be the event of getting exactly 5 girls =>mod(b)=5 thus, p(a)=mod(a)/mod(s)=5/10=1/2 p(b)=mod(b)/mod(s)=5/10=1/2 p(5 boys and 5 girls)=p(a)*p(b)=1/2*1/2=1/4
The number you are thinking of is between 12 and 32 when you divide it by 5 it leaves a remainder of 1 when i divide it by 4 it leaves a remainder of2?
To find the number, we need to consider the remainders when the number is divided by 5 and 4. Let's denote the number as x. From the information given, we have two equations: x ≡ 1 (mod 5) and x ≡ 2 (mod 4). By solving these congruences simultaneously using the Chinese Remainder Theorem, we find that x ≡ 21 (mod 20). Therefore, the number you are thinking of is 21.
How do you evaluate 6-6 -5 in the mod 12 system?
(6-6)-5 in mod 12
