What is the no. of distinct terms in expansion of ( a+b+c)^20 ?
1. 231 2.253 3.242 4.210 5.228
Every term of the polynomial will have the sum of powers of a,b,c as 20. Again, powers of each of a,b and c will range from 0 to 20.
Now, each term will be of the form : (constant)*a^x*b^y*c^z Now, according to the facts stated, x+y+z = 20 (the nos. of solution of this equation will give us the number of distinct terms in the expansion) Note : x,y,z will range from 0 to 20 !!
By PnC (Ball and wall method, although naming this specially is of no use other than just making this clear), no. of solutions are C(22,2) = 231
i have question, a boy multiplied 423 by a certain no. and obtained 65589 as his answer . if only the 5's wrong .what is the correct product?
The product is 60489. Solution We know that the units digit of the number multiplied with 423 should be 3 ( Last digit of the product is 9) Also, the product is of the form 6xy89 So, the only desirable solution comes when it is multiplied by 143.
the power of 21 will be odd .after dividing 21 by 27 we will get 6 as a remainder,now power of 6 is odd,which is divisible by 27,so remainder will be 0.
21^199^7/27 rem = (3^199^7)*(7^199^7)/3^3 rem just cut down 3^3 from 3^199^7, OR, numerator is a multiple of denominator. Hence, remainder will be 0, and the quotient will be {3^(199^7-3)}*(7^199^7).
I am getting answer as 21. If we divide numerator and denominator by 3 and in the final result we multiply the answer by 3.So (7/27)^199^7.Applying fermet theorem we get remainder as 7 so in final result we multiply by 3 getting the answer as 21.Though not sure.
I am getting answer as 21. If we divide numerator and denominator by 3 and in the final result we multiply the answer by 3.So (7/27)^199^7.Applying fermet theorem we get remainder as 7 so in final result we multiply by 3 getting the answer as 21.Though not sure.
u dont need to do fermat and all...secondly, even if u use fermat (broadly let us say, Euler), remainder of the numerator term with a power on "3" will be 0. Hence, whatever remainder of the rest of the terms is, final remainder will be 0 only. May be you took remainder of 3^(199^7-1) by 3^2 as 1. That is why you are getting remainder of 1 with that and remainder of 7 with rest of the term. The remainder of the former will be 0 !!
8^643 = 2^(643*3) =2^(1929) now , 2^(1927)/33 = 2^(1927) / (2^5+1) by remainder theorem, (-1)^(1925)*4 is the remainder , but we had cancelled 4 from numerator and denominator , so we multiply -4 by 4 . So remainder is -16 . i.e 132-16=116.
Using Chinese Remainder Theorem :- 8^643 / (11*12)
8^643 / 11 remainder is 6 8^643 / 12 remainder is 8
12k + 8 = 11l + 6
k = 9
so final remainder should be 12*9+8 = 116
Another Approach :
8^643 = 2^1929 .. also 132=4*33 and 2^1929 = 4* 2^1927 hence the problem comes out to be 2^1927/33...then Euler Thereom of 33 is 20 so 2^20/33...the rem is 1 . then after (2^20)*46 /33 =1.....(since 1920=20*46) so 2^7/33...this rem is 29 . Then we have to multiply the 4 which we cancelled out for numerator and denominator earlier ..... So final ans is 29*4 =116
What is the no. of distinct terms in expansion of ( a+b+c)^20 ?
1. 231 2.253 3.242 4.210 5.228
Ans = 231.
If we take the power of a, b, c as x, y, z, then, x+y+z = 20. So, combinations will be 22C2 = 231. This answer will not work if a = b = c or a=b or a=c or b=c. (:shocked:)
