Powerful Remainder
Divide 7³³ by 10 ...
(if it's too small to see properly, it is 7^33)
What is the remainder?
(if it's too small to see properly, it is 7^33)
What is the remainder?
posted by Zaux at
8:43 PM
Saturday, February 20, 2010
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9 Comments:
7730993719707444524137094407
mod 10 = 7
yep ... too easy for Ragknot
773099371970744452413709440.7
Use these equations for the Area of Cut, the Area of the diversion and the CF ratio.
AC=((bw^2*cs+2*bw*(dh-h1)*(1+cs*ss2)+(dh-h1)^2*(1+cs*ss2)*(ss2+ss3))/(2-2*cs*ss3))
AD=((cr^2*cs+h1^2*(ss1+ss2)*(1+cs*ss2)+2*cr*(h1+cs*h1*ss2))/(2-2*cs*ss1))
CF=AC/AD
That post was for Ross, but I put on the wrong ToM.
It was supposed to be "Ag Engineering Problem"
In mod 10, 7^0 = 1, 7^1 = 7, 7^2 = 9, 7^3 = 3, 7^4 = 1 and then
the pattern just repeats.
So 7^n = 7^(n mod 4) (mod 10) = 7^1 = 7 (mod 10)
No calculator required :)
Small slip. I meant to have said:
7^33 = 7^(33 mod 4) = 7^1 = 7 (mod 10)
(7^33)when divided by 10 gives remainder 3 because for example 49/10 gives remainder 9,343/10 gives remainder 3,.... that means the last digit is the remainder when you divide it with 10. so we need to check out the last digit of 7^33,for this 7^1=7,here the last digit is7 so 7 is remainder when divided by 10,7^2=49,remainder9;7^3=343,remainder3;7^4=2401,remainder1;7^5=16807,remainder7 and so on. but the cycle of 7,9,3,1 keeps on repeating with increase in powers of 7.like for 5th power of 7 we again have remainder7.simillarly for 33rd power of 7 remainder is 3.
Hi dhanashree. Good approach, but you got out of step somewhere. The cycle has length 4, so knock multiples of 4 of 33. As 4*8 = 32, that leaves 1. Hence 7^1 = 7
If use e.g. the Windows calculator you'll find that:
7^33 = 7,730,993,719,707,444,524,137,094,407
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